Mathematics

Normal distribution is skewed with zero skewness, so the answer to the most common confusion can normal distribution be skewed is normal distribution is not skewed distribution as the curve of the normal distribution is symmetric without tail whose skewness is zero. The normal distribution curve is bell shaped with symmetry on the curve.

Since the skewness is lack of symmetry in the curve so if the symmetry is present in the curve there is lack of skewness.

How do you tell if the data is normally distributed?

For the data to check whether normally distributed or not just try to sketch the histogram and from the curve of the curve if the symmetry is present in the curve then the data is normally distributed, from the curve of data itself the question can normal distribution be skewed or not cleared if the concept of skewness is clear. Sketching the histogram or curve in each case is tedious or time consuming so instead of that their are number of statistical tests like Anderson-Darling statistic (AD) which are more useful to tell whether data is normally distributed or not.

The data which follows normal distribution have zero skewness in the curve and the characteristics of the curve of the skewed distribution is different without symmetry, this we will understand with the following example:

Example: Find the percent of score lies between 70 to 80 if the score of mathematics of university students are normally distributed with the mean 67 and standard deviation 9?

Solution:

To find the percent of score we follow the probability for the normal distribution discussed earlier in normal distribution, so to do so first we will convert into normal variate and follow the table discussed in normal distribution to find the probability using the conversion

Z=(X-μ)/σ

we want to find the score percent between 70 and 80 so we use random variable values 70 and 80 with the given mean 67 and standard deviation 9 this gives

Z=70-67/9 = 0.333

and

Z=80-67/9 = 1.444

This we can sketch as

the above shaded area shows the region between z=0.333 and z=1.444 from the table of standard normal variate the probabilities are

P(z > 0.333)=0.3707
and
P(z > 1.444)=0.0749
so
p(0.333 < z0.333)-P(z > 1.444)=0.3707-0.0749=0.2958

so 29.58% students will score between 70 to 80 .

In the above example the skewness of the curve is zero and the curve is symmetric, to check the data is normally distributed or not we have to perform the hypothesis tests.

How do you tell if a distribution is skewed left or right?

The distribution is known to be skewed if it is right tailed or left tailed in the curve so the depending on the nature of the curve we can judge whether the distribution is positive skewed or negative skewed. The concept of skewness is discussed in detail in the articles positively and negatively skewed distribution. If the symmetry in the left side lacks the distribution is skewed left and if the symmetry lacks in the right side the distribution is skewed right. The best way to check the distribution is skewed is to check the variation in the central tendencies that is if mean<median<mode then the distribution is left skewed and if mean>median>mode then the distribution is right skewed. The geometrical representation is as follows

The measures to calculate the skewness left or right for the information given in detail in the article of skewness.

What is an acceptable skewness?

Since the skewness as earlier discussed is lack of symmetry so what range is acceptable that must be clear. The question can normal distribution is skewed arise to check whether in the normal distribution is acceptable or not and the answer of the acceptable skewness is in normal distribution because in normal distribution the skewness is zero and the distribution in which skewness is near to zero is more acceptable. So after the testing for skewness if the skewness is nearer to zero then the skewness is acceptable depending on the requirement and range for the client.

In brief the acceptable skewness is the skewness which is nearer to zero as per the requirement.

How skewed is too skewed?

The skewness is the statistical measurement to check the symmetry present in the curve of the distribution and the information and all the measures to check skewness is present or not, depending on that we can find if the distribution is far from zero then too skewed or symmetry is zero then we can say the distribution is too skewed.

How do you determine normal distribution?

To determine the distribution is normal or not we have to look the distribution have the symmetry or not if the symmetry is present and the skewness is zero then the distribution is normal distribution, the detail methods and techniques were already discussed in detail in normal distribution

Do outliers skew data?

In the distribution data if any data follow unusual way and very far or away from the usual data that is known as outlier and in most of the cases the outliers are responsible for the skewness of the distribution and because of the unusual nature of outliers the distribution have skewness, so we can say that in the distribution the outliers skew data. The outliers in all cases will not skew data they skewed data only if they also follow the systematic sequence in continuous distribution to give left or right tailed curve.

In the previous articles the detail discussion of normal distribution and skewed distribution discussed.

  The Hermite polynomial is widely occurred in applications as an orthogonal function. Hermite polynomial is the series solution of Hermite differential equation.

Hermite’s Equation

    The differential equation of second order with specific coefficients as

d²y/dx² – 2x dy/dx + 2xy = 0

is known as Hermite’s equation, by solving this differential equation we will get the polynomial which is Hermite Polynomial.

Let us find the solution of the equation

d²y/dx² – 2x dy/dx + 2ny = 0

with the help of series solution of differential equation

now substituting all these values in the Hermite’s equation we have

This equation satisfies for the value of k=0 and as we assumed the value of k will not be negative, now for the lowest degree term x^m-2 take k=0 in the first equation as the second gives negative value, so the coefficient x^m-2 is

a₀m (m-1)=0 ⇒ m=0,m=1

as a₀ ≠ 0

now in the same way equating the coefficient of x^m-1 from the second summation

and equating the coefficients of x^m+k to zero,

a_k+2(m+k+2)(m+k+1)-2a_k(m+k-n) = 0

we can write it as

a_k+2 = 2(m+k-n)/(m+k+2)(m+k+1) a_k

if m=0

a_k+2 = 2(k-n)/(k+2)(k+1) a_k

if m=1

a_k+2 = 2(k+1-n)/(k+3)(k+2) a_k

for these two cases now we discuss the cases for k

When $m=0, a_k+2= 2(k-n)/ (k+2)(k+1)} a_k$

If, $k=0 a₂ =-2 n/2 a₀=-n a₀$

$k=1, a₃=2(1-n)/6 a₁ =-2(n-1)/3 ! a₁$

If $k=2, a₄ =2(2-n)/12 a₂ =2 (2-n)/12 (-n a₀) =2² n(n-2)/4 ! a₀$

so far m=0 we have two conditions when a₁=0, then a₃=a₅=a₇=….=a_2r+1=0 and when a₁ is not zero then

by following this put the values of a₀,a₁,a₂,a₃,a₄ and a₅ we have

and for m=1 a₁=0 by putting k=0,1,2,3,….. we get

a_k+2 = 2(k+1-n)/(k+3)(k+2)a_k

so the solution will be

so the complete solution is

where A and B are the arbitrary constants

Hermite Polynomial

   The Hermite’s equation solution is of the form y(x)=Ay₁(x)+By₂(x) where y₁(x) and y₂(x) are the series terms as discussed above,

one of these series end if n is non negative integer if n is even y₁ terminates otherwise y₂ if n is odd, and we can easily verify that for n=0,1,2,3,4…….. these polynomials are

1,x,1-2x², x-2/3 x³, 1-4x²+4/3x⁴, x-4/3x³+ 4/15x⁵

so we can say here that the solution of Hermite’s equation are constant multiple of these polynomials and the terms containing highest power of x is of the form 2ⁿxⁿ denoted by H_n(x) is known as Hermite polynomial

Generating function of Hermite polynomial

Hermite polynomial usually defined with the help of relation using generating function

[n/2] is the greatest integer less than or equal to n/2 so it follows the value of H_n(x) as

this shows that H_n(x) is a polynomial of degree n in x and

H_n(x) = 2ⁿxⁿ + π_n-2 (x)

where π_n-2 (x) is the polynomial of degree n-2 in x, and it will be even function of x for even value of n and odd function of x for odd value of n, so

H_n(-x) = (-1)ⁿ H_n(x)

some of the starting Hermite polynomials are

H₀(x) = 1

H₁(x) = 2x

H₂(x) = 4x² – 2

H₃(x) = 8x³-12

H₄(x) = 16x⁴ – 48x²+12

H₅(x) = 32x² – 160x³+120x

Generating function of Hermite polynomial by Rodrigue Formula

Hermite Polynomial can also be defined with the help of Rodrigue formula using generating function

since the relation of generating function

  Using the Maclaurin’s theorem, we have

or

by putting z=x-t and

for t=0,so z=x gives

this we can show in another way as

differentiating

with respect to t gives

taking limit t tends to zero

now differentiating with respect to x

taking limit t tends to zero

from these two expressions we can write

in the same way we can write

 differentiating n times put t=0, we get

from these values we can write

from these we can get the values

Example on Hermite Polynomial           

1. Find the ordinary polynomial of

Solution: using the Hermite polynomial definition and the relations we have

2. Find the Hermite polynomial of the ordinary polynomial

Solution: The given equation we can convert to Hermite as

and from this equation equating the same powers coefficient

hence the Hermite polynomial will be

Orthogonality of Hermite Polynomial | Orthogonal property of Hermite Polynomial

The important characteristic for Hermite polynomial is its orthogonality which states that

To prove this orthogonality let us recall that

which is the generating function for the Hermite polynomial and we know

so multiplying these two equations we will get

multiplying and integrating within infinite limits

and since

so

using this value in above expression we have

which gives

now equate the coefficients on both the sides

which shows the orthogonal property of Hermite polynomial.

  The result of orthogonal property of Hermite polynomial can be shown in another way by considering the recurrence relation

Example on orthogonality of Hermite Polynomial

1.Evaluate the integral

Solution: By using the property of orthogonality of hermite polynomial

since the values here are m=3 and n=2 so

2. Evaluate the integral

Solution: Using the orthogonality property of Hermite polynomial we can write

Recurrence relations of Hermite polynomial

The value of Hermite polynomial can be easily find out by the recurrence relations

These relations can easily obtained with the help of definition and properties.

Proofs:1. We know the Hermite equation

y”-2xy’+2ny = 0

and the relation

by taking differentiation with respect to x partially we can write it as

from these two equations

now replace n by n-1

by equating the coefficient of tⁿ

so the required result is

2. In the similar way differentiating partially with respect to t the equation

we get

n=0 will be vanished so by putting this value of e

now equating the coefficients of tⁿ

thus

3. To prove this result we will eliminate H_n-1 from

and

so we get

thus we can write the result

4. To prove this result we differentiate

we get the relation

substituting the value

and replacing n by n+1

which gives

Examples on Recurrence relations of Hermite polynomial

1.Show that

H_2n(0) = (-1)ⁿ. 2²ⁿ (1/2)ⁿ

Solution:

To show the result we have

H2n(x) =

taking x=0 here we get

2. Show that

H’_2n+1(0) = (-1)ⁿ 2²ⁿ⁺¹ (3/2)²

Solution:

Since from the recurrence relation

H’_n(x) = 2nH_n-1(X)

here replace n by 2n+1 so

H’_2n-1(x) = 2(2n+1) H_2n(x)

taking x=0

3. Find the value of

H_2n+1(0)

Solution

Since we know

use x=0 here

H_2n-1(0) = 0

4. Find the value of H’_2n(0).

Solution :

we have the recurrence relation

H’_n(x) = 2nH_n-1(x)

here replace n by 2n

H’_2n(x) = =2(2n)H_2n-1(x)

put x=0

H’_2n(0) = (4n)H_2n-1(0) = 4n*0=0

5. Show the following result

Solution :

Using the recurrence relation

H’_n(x) = 2nH_n-1 (x)

so

and

d³/dx³ {H_n(x)} = 2³n(n-1)(n-2)H_n-3(x)

differentiating this m times

which gives

6. Show that

H_n(-x) = (-1)ⁿ H_n(x)

Solution :

we can write

from the coefficient of tⁿ we have

and for -x

7. Evaluate the integral and show

Solution : For solving this integral use integration parts as

Now differentiation under the Integral sign differentiate with

respect to x

using

H’_n(x) = 2_nH_n-1 (x)

and

H’_m(x) = 2mH_m-1 (x)

we have

and since

???? _n,m-1 = ????_{n+1, m}

so the value of integral will be

Conclusion:

The specific polynomial which frequently occurs in application is Hermite polynomial, so the basic definition, generating function , recurrence relations and examples related to Hermite Polynomial were discussed in brief here   , if you require further reading go through

https://en.wikipedia.org/wiki/Hermite_polynomials

For more post on mathematics, please follow our Mathematics page

Locus in 2D Coordinate Geometry

Locus is a Latin word. It is derived from the word ‘Place’ or ‘Location’. The Plural of locus is Loci.

Definition of Locus:

In Geometry, ‘Locus’ is a set of points which satisfy one or more specified conditions of a figure or shape. In modern mathematics, the location or the path on which a point moves on the plane satisfying given geometrical conditions, is called locus of the point.

Locus is defined for line, line segment and the regular or irregular curved shapes except the shapes having vertex or angles inside them in Geometry. https://en.wikipedia.org/wiki/Coordinate_system

Examples on Locus:

lines, circles, ellipse, parabola, hyperbola etc. all these geometrical shapes are defined by the locus of points.

Equation of the Locus:

The algebraic form of the geometrical properties or conditions which are satisfied by the coordinates of all the points on Locus, is known as the equation of the locus of those points.

Method of Obtaining the Equation of the Locus:

To find the equation of the locus of a moving point on a plane, follow the process described below

(i) First, assume the coordinates of a moving point on a plane be (h,k).

(ii) Second, derive a algebraic equation with h and k from the given geometrical conditions or properties.

(iii) Third, replace h and k by x and y respectively in the above said equation. Now this equation is called the the equation of the locus of the moving point on the plane. (x,y) is the current coordinates of the moving point and the equation of the locus must always be derived in the form of x and y i.e. current coordinates.

Here are some examples to make the conception clear about locus.

4+different types of solved problems on Locus:

Problem 1: If P be any point on the XY-plane which is equidistant from two given points A(3,2) and B(2,-1) on the same plane, then find the locus and the equation of locus of the point P with graph.

Solution: 

Assume that the coordinates of any point on the locus of P on XY-plane are (h, k).

Since, P is equidistant from A and B, we can write

The distance of P from A=The distance of P from B

Or, |PA|=|PB|

Or, (h² -6h+9+k² -4k+4) = (h² -4h+4+k² +2k+1)——– taking square to both sides.

Or, h² -6h+13+k² -4k -h²+4h-5-k² -2k = 0

Or, -2h -6k+8 = 0

Or, h+3k -4 = 0

Or, h+3k = 4 ——– (1)

This is a first degree equation of h and k.

Now if h and k are replaced by x and y then the equation (1) becomes the first degree equation of x and y in the form of x + 3y = 4 which represents a straight line.

Therefore, the locus of the point P(h, k) on XY-plane is a straight line and the equation of the locus is x + 3y = 4 . (Ans.)

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Problem 2: If a point R moves on the XY-plane in such way that RA : RB = 3:2 where the coordinates of the points A and B are (-5,3) and (2,4) respectively on the same plane, then find the locus of the point R.

What type of curve does the equation of the locus of R indicate?

Solution: Lets assume that the coordinates of any point on the locus of given point R on XY-plane be (m, n).

Asper given condition RA : RB = 3:2,

we have,

(The distance of R from A) / (The distance of R from B) = 3/2

Or, (m² +10m+34+n² -6n) / (m² -4m+n² -8n+20) =9/4 ———– taking square to both sides.

Or, 4(m² +10m+34+n² -6n) = 9(m² -4m+n² -8n+20)

Or, 4m² +40m+136+4n² -24n = 9m² -36m+9n² -72n+180)

Or, 4m² +40m+136+4n² -24n – 9m² +36m-9n² +72n-180 = 0

Or, -5m² +76m-5n²+48n-44 = 0

Or, 5(m²+n²)-76m+48n+44 = 0 ———-(1)

This is a second degree equation of m and n .

Now if m and n are replaced by x and y, the equation (1) becomes the second degree equation of x and y in the form of 5(x²+y²)-76x+48y+44 = 0 where the coefficients of x² and y² are same and the coefficient of xy is zero. This equation represents a circle.

Therefore, the locus of the point R(m, n) on XY-plane is a circle and the equation of the locus is

5(x²+y²)-76x+48y+44 = 0 (Ans.)

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Problem 3: For all values of (θ,aCosθ,bSinθ) are the coordinates a point P which moves on the XY plane. Find the equation of locus of P.

Solution: lets (h, k) be the coordinates of any point lying on the locus of P on XY-plane.

Then asper the question, we can say

h= a Cosθ

Or, h/a = Cosθ —————(1)

And k = b Sinθ

Or, k/b = Sinθ —————(2)

Now taking square of both the equations (1) and (2) and then adding, we have the equation

h²/a² + k²/b² =Cos²θ + Sin²θ

Or, h²/a² + k²/b² = 1 (Since Cos²θ + Sin²θ =1 in trigonometry)

Therefore the equation of locus of the point P is x²/a² + y²/b² = 1 . (Ans.)

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Problem 4 : Find the equation of locus of a point Q, moving on the XY-plane, if the coordinates of Q are

where u is the variable parameter.

Solution : Let the coordinates of any point on the locus of given point Q while moving on XY-plane be (h, k).

Then, h = and k =

i.e. h(3u+2) = 7u-2 and k(u-1) = 4u+5

i.e. (3h-7)u = -2h-2 and (k-4)u = 5+k

i.e. u = —————(1)

and u = —————(2)

Now equating the equations (1) and (2) , we get,

Or, (-2h-2)(k-4) = (3h-7)(5+k)

Or, -2hk+8h-2k+8 = 15h+3hk-35-7k

Or, -2hk+8h-2k-15h-3hk+7k = -35-8

Or, -5hk-7h+5k = -43

Or, 5hk+7h-5k = 43

Therefore, the equation of the locus of Q is 5xy+7x-5y = 43.

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More examples on Locus with answers for practice by your own:

Problems 5: If θ be a variables and u be a constant, then find the equation of locus of the point of intersection of the two straight lines x Cosθ + y Sinθ = u and x Sinθ- y Cosθ = u. ( Ans. x²+y² =2u² )

Problems 6: Find the equation of locus of the middle point of the line segment of the straight line x Sinθ + y Cosθ = t between the axes. ( Ans. 1/x²+1/y² =4/t² )

Problems 7: If a point P is moving in such way on the XY-plane that the area of the triangle made by the point with two points (2,-1) and (3,4). ( Ans. 5x-y=11)

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Basic Examples on the Formulae “Centroid of a Triangle”  in 2D Coordinate Geometry

Centroid: The three medians of a triangle always intersect at a point, located in the interior area of the triangle and divides the median at the ratio 2:1 from any vertex to the midpoint of the opposite side. This point is called the centroid of the triangle.   

Problems 1:  Find the centroid of the triangle with vertices (-1,0), (0,4) and (5,0).

Solution:  We already know,

                                             If  A(x₁,y₁) , B(x₂,y₂) and C(x₃,y₃) be the vertices of a Triangle and G(x, y) be the centroid of the triangle, then Coordinates of G are

and

Using this formula we have , 

(x₁,y₁) ≌(-1,0) i.e. x₁= -1, y₁=0 ;

(x₂,y₂) ≌(0,4) i.e.   x₂=0, y₂=4 and

(x₃,y₃) ≌(5,0)  i.e.   x₃=5, y₃=0

(See formulae chart)

So, the x-coordinate of the centroid G,   

i.e.

i.e. x=4/3

                  and 

the y-coordinate of the centroid G,  

i.e

i.e y=4/3

Therefore, the coordinates of the centroid of the given triangle is . (Ans)

More answered problems are given below for further practice using the procedure described in above problem 1 :-

Problems 2: Find the coordinates of the centroid of the triangle with vertices at the points (-3,-1), (-1,3)) and (1,1).

Ans. (-1,1)

Problems 3: What is the x-coordinate of the centroid of the triangle with vertices (5,2), (10,4) and (6,-1) ?

Ans. 7 

Problems 4: Three vertices of a triangle are (5,9), (2,15) and (11,12).Find the centroid of this triangle.

Ans. (6,12)

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Shifting of Origin / Translation of Axes- 2D Co-ordinate Geometry

Shifting of Origin means to shift the Origin to a new point keeping the orientation of the axes unchanged i.e the new axes remain parallel to the original axes in the same plane. By this translation of axes or shifting of origin process many problems on algebraic equation of a geometric shape are simplified and solved easily.

The formula of ” Shifting of Origin” or “Translation of Axes” are described below with graphical representation.

Formula:

If O be the origin ,P(x,y) be any point in the XY plane and O be  shifted to another point O′(a,b) against which the coordinates of the point P become (x₁,y₁) in the same plane with new axes X₁Y₁  ,Then New Coordinates of P are

x₁ = x- a

y₁ = y- b

Graphical representation for clarification: Follow the graphs

Few solved Problems on the formula of ‘Shifting of Origin’ :

Problem-1 : If there are two points (3,1) and (5,4) in the same plane and the origin is shifted to the point (3,1) keeping the new axes parallel to the original axes, then find the co-ordinates of the point (5,4) in respect with the new origin and axes.

Solution: Comparing with the formula of ‘Shifting of Origin’ described above , we have new Origin, O′(a, b) ≌ (3,1) i.e. a=3 , b=1 and the required point P, (x, y) ≌ (5,4) i.e. x=5 , y=4

Now if (x₁,y₁) be the new coordinates of the point P(5,4) ,then asper formula x₁ = x-a and y₁ =y-b,

we get, x₁ = 5-3 and y₁ =4-1

i.e. x₁ = 2 and y₁ =3

Therefore, the required new coordinates of the point (5,4) is (2,3) . (Ans.)

Problem-2 : After shifting the Origin to a point in the same plane ,remaining the axes parallel to each other ,the coordinates of a point (5,-4) become (4,-5).Find the Coordinates of new Origin.

Solution: Here using the formula of ‘Shifting the Origin’ or ‘Translation of Axes’ , we can say the coordinates of the point P with respect to old and new Origin and axes respectively are (x, y) ≌ (5,-4) i.e. x=5 , y= -4 and (x₁,y₁) ≌ (4,-5) i.e.  x₁= 4, y₁= -5

Now we have to find the coordinates of the new Origin O′(a, b) i.e. a=?, b=?

Asper formula,

x₁ = x- a

y₁ = y- b

i.e. a=x-x₁ and b=y-y₁

Or, a=5-4 and b= -4-(-5)

Or, a=1 and b= -4+5

Or, a=1 and b= 1

Therefore, O'(1,1) be the new Origin i.e. the coordinates of the new Origin are (1,1). (Ans.)

Basic Examples on the Formulae “Collinearity of points (three points)” in 2D Coordinate Geometry

Problems 1:  Check whether the points (1,0), (0,0) and (-1,0) are collinear or not.

Solution:  We already know,

                                            If  A(x₁,y₁) , B(x₂,y₂) and C(x₃,y₃) be any three collinear points, then the area of the triangle made by them must be zero i.e the area of the triangle is ½[x₁ (y₂– y₃) + x₂ (y₃– y₁) + x₃ (y₁-y₂)] =0

(See formulae chart)

Using this formula we have ,

(x₁,y₁) ≌(-1,0) i.e.   x₁= -1, y₁= 0   ;

(x₂,y₂) ≌(0,0)  i.e.   x₂= 0, y₂= 0;

(x₃,y₃) ≌(1,0)  i.e.    x₃= 1, y₃= 0

So, the area of the triangle is = |½[x₁ (y₂–  y₃) + x₂ (y₃–  y₁) + x₃ (y₁-y₂)]| i.e.

(L.H.S) = |½[-1 (0-0) + 0 (0-0) + 1 (0-0)]|

= |½[(- 1)x0 + 0x0 + 1×0]|

= |½[0 + 0 + 0]|

= |½ x 0|

= 0  (R.H.S)

Therefore, the area  of the triangle made by those given points become zero which means they are lying on the same line.

Therefore, the given points are collinear points. (Ans)

More answered problems are given below for further practice using the procedure described in the above problem 1 :-

Problems 2: Check whether the points (-1,-1), (0,0) and (1,1) are  collinear or not.

Ans. Yes

Problems 3: Is it possible to draw one line through three points (-3,2), (5,-3) and (2,2) ?

Ans.No

Problems 4: Check whether the points (1,2), (3,2) and (-5,2),connected by lines, can form a triangle in the coordinate plane.

Ans. No

______________________________

Basic Examples on the Formulae “Incenter of a Triangle” in 2D Coordinate Geometry

Incenter:It is the center of the triangle’s largest incircle which fits inside the triangle.It is also the point of intersection of the three bisectors of the interior angles of the triangle.

Problems 1: The vertices of a triangle with sides  are (-2,0), (0,5) and (6,0) respectively. Find the incenter of the triangle.

Solution: We already know,

If  A(x₁,y₁) , B(x₂,y₂) and C(x₃,y₃) be the vertices, BC=a, CA=b and AB=c , G′(x,y) be the incentre of the triangle,

The co-ordinates of G′ are

and         

(See formulae chart)

Asper the formula we have,

(x₁,y₁) ≌(-4,0) i.e.  x₁= -4, y₁=0 ;

(x₂,y₂) ≌(0,3) i.e.  x₂=0, y₂=3 ;

(x₃,y₃) ≌(0,0)  i.e.   x₃=0, y₃=0

We have now,

a= √ [(x₂-x₁)²+(y₂-y₁)² ]

Or, a= √ [(0+4)²+(3-0)² ]

Or, a= √ [(4)²+(3)² ]

Or, a= √ (16+9)

Or, a= √25

Or, a = 5 ——————(1)

b=√ [(x₁-x₃)²+(y₁-y₃)² ]

Or, b= √ [(-4-0)²+(0-0)² ]

Or, b= √ [(-4)²+(0)² ]

Or, b= √ (16+0)

Or, b= √16

Or, b= 4 ——————–(2)

c= √ [(x₃-x₂)²+(y₃-y₂)² ]

Or, c= √ [(0-0)²+(0-3)² ]

Or, c= √ [(0)²+(-3)² ]

Or, c= √ (0+9)

Or, c= √9

Or, c= 3 ——————–(3)

and ax₁+ bx₂ + cx₃ = (5 X (-4)) + (4 X 0) + (3 X 6 )

= -20+0+18

Or, ax₁+ bx₂ + cx₃ = -2 ——————-(4)

ay₁+ by₂+ cy₃ = (5 X 0) + (4 X 3) + (3 X 0)

= 0+12+0

Or, ay₁+ by₂+ cy₃ = 12 ——————–(5)

a+b+c = 5+4+3

Or, a+b+c = 12 ——————(6)

Using the above equations (1), (2), (3), (4), (5) and (6) we can calculate the value of x and y from

Or, x = -2/12

Or, x = -1/6

and

Or, y = 12/12

Or, y = 1

Therefore the required coordinates of the incenter of the given triangle are (-1/6 , 1). (Ans.)

More answered problems are given below for further practice using the procedure described in above problem 1 :-

Problems 2: Find the coordinates of the incenter of the triangle with vertices at the points (-3,-1), (-1,3)) and (1,1).

Problems 3: What is the x-coordinate of the incenter of the triangle with vertices (0,2), (0,0) and (0,-1) ?

Problems 4: Three vertices of a triangle are (1,1), (2,2) and (3,3). Find the incenter of this triangle.

---

COVARIANCE, VARIANCE OF SUMS, AND CORRELATIONS OF RANDOM VARIABLES

  The statistical parameters of the random variables of different nature using the definition of expectation of random variable is easy to obtain and understand, in the following we will find some parameters with the help of mathematical expectation of random variable.

Moments of the number of events that occur

    So far we know that expectation of different powers of random variable is the moments of random variables and how to find the expectation of random variable from the events if number of event occurred already, now we are interested in the expectation if pair of number of events already occurred, now if X represents the number of event occurred then for the events A₁, A₂, ….,A_n define the indicator variable I_i as

the expectation of X in discrete sense will be

because the random variable X is

now to find expectation if number of pair of event occurred already we have to use combination as

this gives expectation as

from this we get the expectation of x square and the value of variance also by

By using this discussion we focus different kinds of random variable to find such moments.

Moments of binomial random variables

   If p is the probability of success from n independent trials then lets denote A_i for the trial i as success so

and hence the variance of binomial random variable will be

because

if we generalize for k events

this expectation we can obtain successively for the value of k greater than 3  let us find for 3

using this iteration we can get

Moments of hypergeometric random variables

  The moments of this random variable we will understand with the help of an example suppose n pens are randomly selected from a box containing N pens of which m are blue, Let A_i denote the events that i-th pen is blue, Now X is the number of blue pen selected is equal to the number of events A₁,A₂,…..,A_n that occur because the ith pen selected is equally likely to any of the N pens of which m are blue

and so

this gives

so the variance of hypergeometric random variable will be

in similar way for the higher moments

hence

Moments of the negative hypergeometric random variables

  consider the example of a package containing n+m vaccines of which n are special and m are ordinary, these vaccines removed one at a time, with each new removal equally likely to be any of the vaccine that remain in the package. Now let random variable Y denote the number of vaccines that need to be withdrawn until a total of r special vaccines have been removed, which is negative hypergeometric distribution, this is somehow similar with negative binomial to binomial as to hypergeometric distribution. to find the probability mass function if the kth draw gives the special vaccine after k-1 draw gives r-1 special and k-r ordinary vaccine

now the random variable Y

Y=r+X

for the events A_i

as

hence to find the variance of Y we must know the variance of X so

hence

COVARIANCE             

The relationship between two random variable can be represented by the statistical parameter covariance, before the definition of covariance of two random variable X and Y recall that the expectation of two functions g and h of random variables X and Y respectively gives

using this relation of expectation we can define covariance as

   “ The covariance between random variable X and random variable Y denoted by cov(X,Y)  is defined as

using definition of expectation and expanding we get

it is clear that if the random variables X and Y are independent then

but the converse is not true for example if

and defining the random variable Y as

so

here clearly X and Y are not independent but covariance is zero.

Properties of covariance

  Covariance between random variables X and Y has some properties as follows

using the definition off the covariance the first three properties are immediate and the fourth property follows by considering

now by definition

Variance of the sums

The important result from these properties is

as

If X_i ‘s are pairwise independent then

Example: Variance of a binomial random variable

  If X is the random variable

where X_i are the independent Bernoulli random variables such that

 then find the variance of a binomial random variable X with parameters n and p.

Solution:

since

so for single variable we have

so the variance is

Example

  For the independent random variables X_i with the respective means and variance and a new random variable with deviation as

then compute

solution:

By using the above property and definition we have

now for the random variable S

take the expectation

Example:

Find the covariance of indicator functions for the events A and B.

Solution:

for the events A and B the indicator functions are

so the expectation of these are

thus the covariance is

Example:

     Show that

where X_i are independent random variables with variance.

Solution:

The covariance using the properties and definition will be

Example:

  Calculate the mean and variance of random variable S which is the sum of n sampled values if set of N people each of whom has an opinion about a certain subject that is measured by a real number v that represents the person’s “strength of feeling” about the subject. Let  represent the strength of feeling of person  which is unknown, to collect information a sample of n from N is taken randomly, these n people are questioned and their feeling is obtained to calculate vi

Solution

let us define the indicator function as

thus we can express S as

and its expectation as

this gives the variance as

since

we have

we know the identity

so

so the mean and variance for the said random variable will be

Conclusion:

The correlation between two random variables is defined as covariance and using the covariance the sum of the variance is obtained for different random variables, the covariance and different moments with the help of definition of expectation is obtained  , if you require further reading go through

https://en.wikipedia.org/wiki/Expectation

A first course in probability by Sheldon Ross

Schaum’s Outlines of Probability and Statistics

An introduction to probability and statistics by ROHATGI and SALEH.

For more post on mathematics, please follow our Mathematics page

Mathematical Expectation and random variable    

     The mathematical expectation plays very important role in the probability theory, the basic definition and basic properties of mathematical expectation already we discussed in previous some articles now after discussing the various distributions and types of distributions, in the following article we will get familiar with some more advanced properties of mathematical expectation.

Expectation of sum of random variables | Expectation of function of random variables | Expectation of Joint probability distribution

     We know the mathematical expectation of random variable of discrete nature is

and for the continuous one is

now for the random variable X and Y if discrete then with the joint probability mass function p(x,y)

expectation of function of random variable X and Y will be

and if continuous then with the joint probability density function f(x, y) the expectation of function of random variable X and Y will be

if g is addition of these two random variables in continuous form the

and if for the random variables X and Y we have

X>Y

then the expectation also

Example

A Covid-19 hospital is uniformly distributed on the road of the length L at a point X, a vehicle carrying oxygen for the patients is at a location Y which is also uniformly distributed on the road, Find the expected distance between Covid-19 hospital and oxygen carrying vehicle if they are independent.

Solution:

To find the expected distance between X and Y we have to calculate E { | X-Y | }

Now the joint density function of X and Y will be

since

by following this we have

now the value of integral will be

Thus the expected distance between these two points will be

Expectation of Sample mean

  As the sample mean of the sequence of random variables X₁, X₂, ………, X_n with distribution function F and expected value of each as μ is

so the expectation of this sample mean will be

which shows the expected value of sample mean is also μ.

Boole’s Inequality

                Boole’s inequality can be obtained with the help of properties of expectations, suppose the random variable X defined as

where

here A_i ‘s are the random events, this means random variable X represents the occurrence of the number of events A_i and another random variable Y as

clearly

X>=Y

E[X] >= E[Y]

and so is

now if we take the value of random variable X and Y these expectation will be

and

substituting these expectation in the above inequality we will get Boole’s inequality as

Expectation of Binomial random variable | Mean of Binomial random variable

  We know that the binomial random variable is the random variable which shows number of successes in n independent trials with probability of success as p and failure as q=1-p, so if

X=X₁ + X₂+ …….+ X_n

Where

here these X_i ‘s are the Bernoulli and the expectation will be

so the expectation of X will be

Expectation of Negative binomial random variable | Mean of Negative binomial random variable

  Let a random variable X which represents the number of trials needed to collect r successes, then such a random variable is known as negative binomial random variable and it can be expressed as

here each X_i denote the number of trials required after the (i-1)st success  to obtain the total of i successes.

Since each of these X_i represent the geometric random variable and we know the expectation for the geometric random variable is

so

which is the expectation of negative binomial random variable.

Expectation of hypergeometric random variable | Mean of hypergeometric random variable

The expectation or mean of the hypergeometric random variable we will obtain with the help of a simple real life example, if n number of books are randomly selected from a shelf containing N books of which m are of mathematics, then to find the expected number of mathematics books let X denote the number of mathematics books selected then we can write X as

where

so

=n/N

which gives

which is the mean of such a hypergeometric random variable.

Expected number of matches

   This is very popular problem related to expectation, suppose that in a room there are N number of people who throw their hats in the middle of the room  and all the hats are mixed  after that each person randomly choose one hat then the expected number of people who select their own hat we can obtain by letting X to be the number of matches so

Where

since each person has equal opportunity to select any of the hat from N hats then

so

which means exactly one person on average choose his own hat.

The probability of a union of events

     Let us obtain the probability of the union of the events with the help of expectation so for the events A_i

with this we take

so the expectation of this will be

and expanding using expectation property as

since we have

and

so

this implies the probability of union as

Bounds from Expectation using Probabilistic method

    Suppose S be a finite set and f is the function on the elements of S and

here we can obtain the lower bound for this m by expectation of f(s) where “s” is any random element of S whose expectation we can calculate so

here we get expectation as the lower bound for the maximum value

Maximum-Minimum identity

 Maximum Minimum identity is the maximum of the set of numbers to the minimums of the subsets of these numbers that is for any numbers x_i

To show this let us restrict the x_i within the interval [0,1], suppose a uniform random variable U on the interval (0,1) and the events A_i as the uniform variable U is less than x_i that is

since at least one of the above event occur as U is less than one the value of x_i

and

Clearly we know

and all the events will occur if U is less than all the variables and

the probability gives

we have the result of probability of union as

following this inclusion exclusion formula for the probability

consider

this gives

since

which means

• hence we can write it as

taking expectation we can find expected values of maximum and partial minimums as

Conclusion:

The Expectation in terms of various distribution and correlation of expectation with some of the probability theory concepts were the focus of this article which shows the use of expectation as a tool to get expected values of different kind of random variables, if you require further reading go through below books.

For more articles on Mathematics, please see our Mathematics page.

https://en.wikipedia.org/wiki/Expectation

A first course in probability by Sheldon Ross

Schaum’s Outlines of Probability and Statistics

An introduction to probability and statistics by ROHATGI and SALEH

• Content
• Conditional distribution
• Discrete conditional distribution
• Example on discrete conditional distribution
• Continuous conditional distribution
• Example on Continuous conditional distribution
• Conditional distribution of bivariate normal distribution
• Joint Probability distribution of function of random variables
• Examples on Joint Probability distribution of function of random variables

Conditional distribution

   It is very interesting to discuss the conditional case of distribution when two random variables follows the distribution satisfying one given another, we first briefly see the conditional distribution in both the case of random variables, discrete and continuous then after studying some prerequisites we focus on the conditional expectations.

Discrete conditional distribution

     With the help of joint probability mass function in joint distribution we define conditional distribution for the discrete random variables X and Y using conditional probability for X given Y as the distribution with the probability mass function

provided the denominator probability is greater than zero, in similar we can write this as

in the joint probability if the X and Y are independent random variables then this will turn into

so the discrete conditional distribution or conditional distribution for the discrete random variables X given Y is the random variable with the above probability mass function in similar way for Y given X we can define.

Example on discrete conditional distribution

1. Find the probability mass function of random variable X given Y=1, if the joint probability mass function for the random variables X and Y has some values as

p(0,0)=0.4 , p(0,1)=0.2, p(1,0)= 0.1, p(1,1)=0.3

Now first of all for the value Y=1 we have

so using the definition of probability mass function

we have

and

• obtain the conditional distribution of X given X+Y=n, where X and Y are Poisson distributions with the parameters λ₁ and λ₂ and X and Y are independent random variables

Since the random variables X and Y are independent, so the conditional distribution will have probability mass function as

since the sum of Poisson random variable is again poisson so

thus the conditional distribution with above probability mass function will be conditional distribution for such Poisson distributions. The above case can be generalize for more than two random variables.

Continuous conditional distribution

   The Continuous conditional distribution of the random variable X given y already defined is the continuous distribution with the probability density function

denominator density is greater than zero, which  for the continuous density function is

thus the probability for such conditional density function is

In similar way as in discrete if X and Y are independent  in continuous then also

and hence

so we can write it as

Example on Continuous conditional distribution

1. Calculate conditional density function of random variable X given Y if the joint probability density function with the open interval (0,1) is given by

If for the random variable X given Y within (0,1) then by using the above density function we have

• Calculate the conditional probability

if the joint probability density function is given by

To find the conditional probability first we require the conditional density function so by the definition it would be

now using this density function in the probability the conditional probability is

Conditional distribution of bivariate normal distribution

  We know that the Bivariate normal distribution of the normal random variables X and Y with the respective means and variances as the parameters has the joint probability density function

so to find the conditional distribution for such a bivariate normal distribution for X given Y is defined by following the conditional density function of the continuous random variable and the above joint density function we have

By observing this we can say that this is normally distributed with the mean

and variance

in the similar way the conditional density function for Y given X already defined will be just interchanging the positions of the parameters of X with Y,

The marginal density function for X we can obtain from the above conditional density function by using the value of the constant

let us substitute in the integral

the density function will be now

since the total value of

by the definition of the probability so the density function will be now

which is nothing but the density function of random variable X with usual mean and variance as the parameters.

Joint Probability distribution of function of random variables

  So far we know the joint probability distribution of two random variables, now if we have functions of such random variables then what would be the joint probability distribution of those functions, how to calculate the density and distribution function because we have real life situations where we have functions of the random variables,

If Y₁ and Y₂ are the functions of the random variables X₁ and X₂ respectively which are jointly continuous then the joint continuous density function of these two functions will be

where Jacobian

and Y₁ =g₁ (X₁, X₂) and Y₂ =g₂ (X₁, X₂) for some functions g₁ and g₂ . Here g₁ and g₂ satisfies the conditions of the Jacobian as continuous and have continuous partial derivatives.

Now the probability for such functions of random variables will be

Examples on Joint Probability distribution of function of random variables

1. Find the joint density function of the random variables Y₁ =X₁ +X₂ and Y₂=X₁ -X₂ , where X₁ and X₂ are the jointly continuous with joint probability density function. also discuss for the different nature of distribution .

Here we first we will check Jacobian

since g₁(x₁, x₂)= x₁ + x₂  and g₂(x₁, x₂)= x₁ – x₂ so

simplifying Y₁ =X₁ +X₂ and Y₂=X₁ -X₂ , for the value of X₁ =1/2( Y₁ +Y₂ ) and X₂ = Y₁ -Y₂ ,

if these random variables are independent uniform random variables

or if these random variables are independent exponential random variables with usual parameters

or if these random variables are independent normal random variables then

• If X and Y are the independent standard normal variables as given

calculate the joint distribution for the respective polar coordinates.

We will convert by usual conversion X and Y into r and θ as

so the partial derivatives of these function will be

so the Jacobian using this functions is

if both the random variables X and Y are greater than zero then conditional joint density function is

now the conversion of cartesian coordinate to the polar coordinate using

so the probability density function for the positive values will be

for the different combinations of X and Y the density functions in similar ways are

now from the average of the above densities we can state the density function as

and the marginal density function from this joint density of polar coordinates over the interval (0, 2π)

• Find the joint density function for the function of random variables

U=X+Y and V=X/(X+Y)

where X and Y are the gamma distribution with parameters (α + λ) and (β +λ) respectively.

Using the definition of gamma distribution and joint distribution function the density function for the random variable X and Y will be

consider the given functions as

g₁ (x,y) =x+y , g₂ (x,y) =x/(x+y),

so the differentiation of these function is

now the Jacobian is

after simplifying the given equations the variables x=uv and y=u(1-v) the probability density function is

we can use the relation

• Calculate the joint probability density function for

Y₁ =X₁ +X₂+ X₃ , Y₂ =X₁– X₂ , Y₃ =X₁ – X₃

where the random variables X1 , X2, X3 are the standard normal random variables.

Now let us calculate the Jacobian by using partial derivatives of

Y₁ =X₁ +X₂+ X₃ , Y₂ =X₁– X₂ , Y₃ =X₁ – X₃

as

simplifying for variables X₁ , X₂ and X₃

X₁ = (Y₁ + Y₂ + Y₃)/3 , X₂ = (Y₁ – 2Y₂ + Y₃)/3 , X₃ = (Y₁ + Y₂ -2 Y₃)/3

we can generalize the joint density function as

so we have

for the normal variable the  joint probability density function is

hence

where the index is

compute the joint density function of Y₁ ……Y_n and marginal density function for Y_n where

and X_i are independent identically distributed exponential random variables with parameter λ.

for the random variables of the form

Y₁ =X₁ , Y₂ =X₁ + X₂ , ……, Y_n =X₁ + ……+ X_n

the Jacobian will be of the form

and hence its value is one, and the joint density function for the exponential random variable

and the values of the variable X_i ‘s will be

so the joint density function is

Now to find the marginal density function of Y_n we will integrate one by one  as

and

like wise

if we continue this process we will get

which is the marginal density function.

Conclusion:

The conditional distribution for the discrete and continuous random variable with different examples considering some of the types of these random variables discussed, where the independent random variable plays important role. In addition the  joint distribution for the function of joint continuous random variables also explained with suitable examples, if you require further reading go through below links.

For more post on Mathematics, please refer to our Mathematics Page

Wikipediahttps://en.wikipedia.org/wiki/joint_probability_distribution/” target=”_blank” rel=”noreferrer noopener” class=”rank-math-link”>Wikipedia.org

A first course in probability by Sheldon Ross

Schaum’s Outlines of Probability and Statistics

An introduction to probability and statistics by ROHATGI and SALEH

Content

• Jointly distributed random variables
• Joint distribution function | Joint Cumulative probability distribution function | joint probability mass function | joint probability density function
• Examples on Joint distribution
• Independent random variables and joint distribution
• Example of independent joint distribution
• SUMS OF INDEPENDENT RANDOM VARIABLES BY JOINT DISTRIBUTION
• sum of independent exponential random variables
• sum of independent Gamma random variables
• sum of independent exponential random variables
• Sum of independent normal random variable | sum of independent Normal distribution
• Sums of independent Poisson random variables
• Sums of independent binomial random variables

Jointly distributed random variables

     The jointly distributed random variables are the random variable more than one with probability jointly distributed for these random variables, in other words in experiments where the different outcome with their common probability is known as jointly distributed random variable or joint distribution, such type of situation occurs frequently while dealing the problems of the chances.

Joint distribution function | Joint Cumulative probability distribution function | joint probability mass function | joint probability density function

    For the random variables X and Y the distribution function or joint cumulative distribution function is

where the nature of the joint probability depends on the nature of random variables X and Y either discrete or continuous, and the individual distribution functions for X and Y can be obtained using this joint cumulative distribution function as

similarly for Y as

these individual distribution functions of X and Y are known as Marginal distribution functions when joint distribution is under consideration. These distributions are very helpful for getting the probabilities like

and in addition the joint probability mass function for the random variables X and Y is defined as

the individual probability mass or density functions for X and Y can be obtained with the help of such joint probability mass or density function like in terms of discrete random variables as

and in terms of continuous random variable the joint probability density function will be

where C is any two dimensional plane, and the joint distribution function for continuous random variable will be

the probability density function from this distribution function can be obtained by differentiating

and the marginal probability from the joint probability density function

as

and

with respect to the random variables X and Y respectively

Examples on Joint distribution

1. The joint probabilities for the random variables X and Y representing the number of mathematics and statistics books from a set of books which contains 3 mathematics, 4 statistics and 5 physics books if 3 books taken randomly

• Find the joint probability mass function for the sample of families having 15% no child, 20% 1 child, 35% 2 child and 30% 3 child if the family we choose randomly from this sample for child to be Boy or Girl?

The joint probability we will find by using the definition as

and this we can illustrate in the tabular form as follows

• Calculate the probabilities

if for the random variables X and Y the joint probability density function is given by

with the help of definition of joint probability for continuous random variable

and the given joint density function the first probability for the given range will be

in the similar way the probability

and finally

• Find the joint density function for the quotient X/Y of random variables X and Y if their joint probability density function is

To find the probability density function for the function X/Y we first find the joint distribution function then we will differentiate the obtained result,

so by the definition of joint distribution function and given probability density function we have

thus by differentiating this distribution function with respect to a we will get the density function as

where a is within zero to infinity.

Independent random variables and joint distribution

     In the joint distribution the probability for two random variable X and Y is said to be independent if

where A and B are the real sets. As already in terms of events we know that the independent random variables are the random variables whose events are independent.

Thus for any values of a and b

and the joint distribution or cumulative distribution function for the independent random variables X and Y will be

if we consider the discrete random variables X and Y then

since

similarly for the continuous random variable also

Example of independent joint distribution

1. If for a specific day in a hospital the patients entered are poisson distributed with parameter λ and probability of male patient as p and probability of female patient as (1-p) then show that the number of male patients and female patients entered in the hospital are independent poisson random variables with parameters λp and λ(1-p) ?

consider the number of male and female patients by random variable X and Y then

as X+Y are the total number of patients entered in the hospital which is poisson distributed so

as the probability of male patient is p and female patient is (1-p) so exactly from total fix number are male or female shows binomial probability as

using these two values we will get the above joint probability as

thus probability of male and female patients will be

and

which shows both of them are poisson random variables with the parameters λp and λ(1-p).

2. find the probability that a person has to wait for more than ten minutes at the meeting for a client as if each client and that person arrives between  12 to 1 pm following uniform distribution.

consider the random variables X and Y to denote the time for that person and client between 12 to 1 so the probability jointly for X and Y will be

calculate

where X,Y and Z are uniform random variable over the interval (0,1).

here the probability will be

for the uniform distribution the density function

for the given range so

SUMS OF INDEPENDENT RANDOM VARIABLES BY JOINT DISTRIBUTION

  The sum of independent variables X and Y with the probability density functions as continuous random variables, the cumulative distribution function will be

by differentiating this cumulative distribution function for the probability density function of these independent sums are

by following these two results we will see some continuous random variables and their sum as independent variables

sum of independent uniform random variables

   for the random variables X and Y uniformly distributed over the interval (0,1) the probability density function for both of these independent variable is

so for the sum X+Y we have

for any value a lies between zero and one

if we restrict a in between one and two it will be

this gives the triangular shape density function

if we generalize for the n independent uniform random variables 1 to n then their distribution function

by mathematical induction will be

sum of independent Gamma random variables

    If we have two independent gamma random variables with their usual density function

then following the density for the sum of independent gamma random variables

this shows the density function for the sum of gamma random variables which are independent

sum of independent exponential random variables

    In the similar way as gamma random variable the sum of independent exponential random variables we can obtain density function and distribution function by just specifically assigning values of gamma random variables.

Sum of independent normal random variable | sum of independent Normal distribution

                If we have n number of independent normal random variables Xi , i=1,2,3,4….n with respective means μi and variances σ2i then their sum is also normal random variable with the mean as Σμi  and variances Σσ2i

    We first show the normally distributed independent sum for two normal random variable X with the parameters 0 and σ² and Y with the parameters 0 and 1, let us find the probability density function for the sum X+Y with

in the joint distribution density function

with the help of definition of density function of normal distribution

thus the density function will be

which is nothing but the density function of a normal distribution with mean 0 and variance (1+σ2) following the same argument we can say

with usual mean and variances. If we take the expansion and observe the sum is normally distributed with the mean as the sum of the respective means and variance as the sum of the respective variances,

thus in the same way the nth sum will be the normally distributed random variable with the mean as Σμ_i  and variances Σσ²_i

Sums of independent Poisson random variables

If we have two independent Poisson random variables X and Y with parameters λ₁ and λ₂ then their sum X+Y is also Poisson random variable or Poisson distributed

since X and Y are Poisson distributed and we can write their sum as the union of disjoint events so

by using the of probability of independent random variables

so we get the sum X+Y is also Poisson distributed with the mean λ₁ +λ₂

Sums of independent binomial random variables

                If we have two independent binomial random variables X and Y with parameters (n,p) and (m, p) then their sum X+Y is also binomial random variable or Binomial distributed with parameter (n+m, p)

let use the probability of the sum with definition of binomial as

which gives

so the sum X+Y is also binomially distributed with parameter (n+m, p).

Conclusion:

The concept of jointly distributed random variables which gives the distribution comparatively for more than one variable in the situation is discussed in addition the basic concept of independent random variable with the help of joint distribution and sum of independent variables with some example of distribution is given with their parameters, if you require further reading go through mentioned books. For more post on mathematics, please click here.

https://en.wikipedia.org

A first course in probability by Sheldon Ross

Schaum’s Outlines of Probability and Statistics

An introduction to probability and statistics by ROHATGI and SALEH

Content

1. Special form of Gamma distributions and relationships of Gamma distribution
2. Gamma distribution exponential family
3. Relationship between gamma and normal distribution
4. Poisson gamma distribution | poisson gamma distribution negative binomial
5. Weibull gamma distribution
6. Application of gamma distribution in real life | gamma distribution uses | application of gamma distribution in statistics 
7. Beta gamma distribution | relationship between gamma and beta distribution
8. Bivariate gamma distribution
9. Double gamma distribution
10. Relation between gamma and exponential distribution | exponential and gamma distribution | gamma exponential distribution
11. Fit gamma distribution
12. Shifted gamma distribution
13. Truncated gamma distribution
14. Survival function of gamma distribution
15. MLE of gamma distribution | maximum likelihood gamma distribution | likelihood function of gamma distribution
16. Gamma distribution parameter estimation method of moments | method of moments estimator gamma distribution
17. Confidence interval for gamma distribution
18. Gamma distribution conjugate prior for exponential distribution | gamma prior distribution | posterior distribution poisson gamma
19. Gamma distribution quantile function
20. Generalized gamma distribution
21. Beta generalized gamma distribution

Special form of Gamma distributions and relationships of Gamma distribution

  In this article we will discuss the special forms of gamma distributions and the relationships of gamma distribution with different continuous and discrete random variables also some estimation methods  in sampling of population using gamma distribution is briefly discuss.

Gamma distribution exponential family

  The gamma distribution exponential family and it is two parameter exponential family which is largely and applicable family of distribution as most of real life problems can be modelled in the gamma distribution exponential family and the quick and useful calculation  within the exponential family can be done easily, in the two parameter if we take probability density function as

if we restrict the known value of α (alpha) this two parameter family will reduce to one parameter exponential family

and for λ (lambda)

Relationship between gamma and normal distribution

  In the probability density function of gamma distribution if we take alpha nearer to 50 we will get the nature of density function as

even the shape parameter in gamma distribution we are increasing which is resulting in similarity of normal distribution normal curve, if we tend shape parameter alpha tends to infinity the gamma distribution will be more symmetric and normal but as alpha tends to infinity value of x in gamma distribution will tends to minus infinity which result in semi infinite support of gamma distribution infinite hence even gamma distribution becomes symmetric but not same with normal distribution.

poisson gamma distribution | poisson gamma distribution negative binomial

   The poisson gamma distribution and binomial distribution are the discrete random variable whose random variable deals with the discrete values specifically success and failure in the form of Bernoulli trials which gives random success or failure as a result only, now the mixture of Poisson and gamma distribution also known as negative binomial distribution is the outcome of the repeated trial of Bernoulli’s trial, this can be parameterize in different way as if r-th success occurs in number of trials then it can be parameterize as

and if the number of failures before the r-th success then it can be parameterize as

and considering the values of r and p

the general form of the parameterization for the negative binomial or poisson gamma distribution is

and alternative one is

this binomial distribution is known as negative because of the coefficient

and this negative binomial or poisson gamma distribution is well define as the total probability we will get as one for this distribution

The mean and variance for this negative binomial or poisson gamma distribution is

the poisson and gamma relation we can get by the following calculation

Thus negative binomial is the mixture of poisson and gamma distribution and this distribution is used in day to day problems modelling where discrete and continuous mixture we require.

Weibull gamma distribution

   There are generalization of exponential distribution which involve Weibull as well as gamma distribution as the Weibull distribution has the probability density function as

and cumulative distribution function as

where as pdf and cdf of gamma distribution is already we discussed above the main connection between Weibull and gamma distribution is both are generalization of exponential distribution the difference between them is when power of variable is greater than one then Weibull distribution gives quick result while for less than 1 gamma gives quick result.

     We will not discuss here generalized Weibull gamma distribution that require separate discussion.

application of gamma distribution in real life | gamma distribution uses | application of gamma distribution in statistics 

  There are number of  application where gamma distribution is used to model the situation such as insurance claim to aggregate, rainfall amount accumulation, for any product its manufacturing and distribution, the crowd on specific web,  and in telecom exchange etc. actually the gamma distribution give the wait time prediction till next event for nth event. There are number of application of gamma distribution in real life.

beta gamma distribution | relationship between gamma and beta distribution

    The beta distribution is the random variable with the probability density function

where

which has the relationship with gamma function as

and beta distribution related to gamma distribution as if X be gamma distribution with parameter alpha and beta as one and Y be the gamma distribution with parameter alpha as one and beta then the random variable X/(X+Y) is beta distribution.

or If X is Gamma(α,1) and Y is Gamma (1, β) then the random variable X/(X+Y) is Beta (α, β) 

and also

bivariate gamma distribution

     A two dimensional or bivariate random variable is continuous if there exists a function f(x,y) such that the joint distribution function

where

and the joint probability density function obtained by

there are number of bivariate gamma distribution one of them is the bivariate gamma distribution with probability density function as

double gamma distribution

  Double gamma distribution is one of the bivariate distribution with gamma random variables having parameter alpha and one with joint probability density function as

this density forms the double gamma distribution with respective random variables and the moment generating function for double gamma distribution is

relation between gamma and exponential distribution | exponential and gamma distribution | gamma exponential distribution

   since the exponential distribution is the distribution with the probability density function

and the gamma distribution has the probability density function

clearly the value of alpha if we put as one we will get the exponential distribution, that is the gamma distribution is nothing but the generalization of the exponential distribution, which predict the wait time till the occurrence of next nth event while exponential distribution predict the wait time till the occurrence of the next event.

fit gamma distribution

   As far as fitting the given data in the form of gamma distribution imply finding the two parameter probability density function which involve shape, location and scale parameters so finding these parameters with different application and calculating the mean, variance, standard deviation and moment generating function is the fitting of gamma distribution, since different real life problems will be modelled in gamma distribution so the information as per situation must be fit in gamma distribution for this purpose various technique in various environment is already there e.g in R, Matlab, excel etc.

shifted gamma distribution

     There are as per application and need whenever the requirement of shifting the distribution required from two parameter gamma distribution the new generalized three parameter or any another generalized gamma distribution shift the shape location and scale , such gamma distribution is known as shifted gamma distribution

truncated gamma distribution

     If we restrict the range or domain of the gamma distribution for the shape scale and location parameters the restricted gamma distribution is known as truncated gamma distribution based on the conditions.

survival function of gamma distribution

                The survival function for the gamma distribution is defined the function s(x) as follows

mle of gamma distribution | maximum likelihood gamma distribution | likelihood function of gamma distribution

we know that the maximum likelihood take the sample from the population as a representative and this sample consider as an estimator for the probability density function to maximize for the parameters of density function, before going to gamma distribution recall some basics as for the random variable X the probability density function with theta as parameter has likelihood function as

this we can express as

and method of maximizing this likelihood function can be

if such theta satisfy this equation, and as log is monotone function we can write in terms of log

and such a supremum exists if

now we apply the maximum likelihood for the gamma distribution function as

the log likelihood of the function will be

so is

and hence

This can be achieved also as

by

and the parameter can be obtained by differentiating

gamma distribution parameter estimation method of moments | method of moments estimator gamma distribution

   We can calculate the moments of the population and sample with the help of expectation of nth order respectively, the method of moment equates these moments of distribution and sample to estimate the parameters, suppose we have sample of gamma random variable with the probability density function as

we know the first tow moments for this probability density function is

so

we will get from the second moment if we substitute lambda

and from this value of alpha is

and now lambda will be

and moment estimator using sample will be

confidence interval for gamma distribution

   confidence interval for gamma distribution is the way to estimate the information and its uncertainty which tells the interval is expected to have the true value of the parameter at what percent, this confidence interval is obtained from the observations of random variables, since it is obtained from random it itself is random to get the confidence interval for the gamma distribution there are different techniques in different application that we have to follow.

gamma distribution conjugate prior for exponential distribution | gamma prior distribution | posterior distribution poisson gamma

     The posterior and prior distribution  are the terminologies of Bayesian probability theory and they are conjugate to each other, any two distributions are conjugate if the posterior of one distribution is another distribution, in terms of theta let us show that gamma distribution is conjugate prior to the exponential distribution

if the probability density function of gamma distribution in terms of theta is as

assume the distribution function for theta is exponential from given data

so the joint distribution will be

and using the relation

we have

which is

so gamma distribution is conjugate prior to exponential distribution as posterior is gamma distribution.

gamma distribution quantile function

   Qauntile function of gamma distribution will be the function that gives the points in gamma distribution which relate the rank order of the values in gamma distribution, this require cumulative distribution function and for different language different algorithm and functions for the quantile of gamma distribution.

generalized gamma distribution

    As gamma distribution itself is the generalization of exponential family of distribution adding more parameters to this distribution gives us generalized gamma distribution which is the further generalization of this distribution family, the physical requirements gives different generalization one of the frequent one is using the probability density function as

the cumulative distribution function for such generalized gamma distribution can be obtained by

where the numerator represents the incomplete gamma function as

using this incomplete gamma function the survival function for the generalized gamma distribution can be obtained as

another version of this three parameter generalized gamma distribution having probability density function is

where k, β, θ are the parameters greater than zero, these generalization has convergence issues to overcome the Weibull parameters replaces

using this parameterization the convergence of the density function obtained so the more generalization for the gamma distribution with convergence is the distribution with probability density function as

Beta generalized gamma distribution

   The gamma distribution involving the parameter beta in the density function because of which sometimes gamma distribution is known as the beta generalized gamma distribution with the density function

with cumulative distribution function as

which is already discussed in detail in the discussion of gamma distribution, the further beta generalized gamma distribution is defined with the cdf as

where B(a,b) is the beta function , and the probability density function for this can be obtained by differentiation and the density function will be

here the G(x) is the above defined cumulative distribution function of gamma distribution, if we put this value then the cumulative distribution function of beta generalized gamma distribution is

and the probability density function

the remaining properties can be extended for this beta generalized gamma distribution with usual definitions.

Conclusion:

There are different forms and generalization of gamma distribution and Gamma distribution exponential family as per the real life situations so possible such forms and generalizations were covered in addition with the estimation  methods of gamma distribution in population sampling of information, if you require further reading on Gamma distribution exponential family, please go through below link and books. For more topics on Mathematics please visit our page.

https://en.wikipedia.org/wiki/Gamma_distribution

A first course in probability by Sheldon Ross

Schaum’s Outlines of Probability and Statistics

An introduction to probability and statistics by ROHATGI and SALEH