Continuous random variable | Its Important distribution

Continuous random variable, types and its distribution

     The random variable which takes the finite or countably infinite values is known as discrete random variable and its pair with probability forms the distribution for the discrete random variable. Now for the random variable who takes the values as uncountable, what would be the probability and remaining characteristics that we are going to discuss.  Thus in brief the continuous random variable is the random variable whose set of values are uncountable. The real life example for the continuous random variable is the life span of electrical or electronic components and arrival of specific public vehicle on the stops etc.

Continuous random variable and probability density function

                Random variable  will be continuous random variable if for a non-negative real valued function f on x and B ⊆  and  

    \[P\left \{ X \in B \right \}=\int_{B}f(x)dx\]

this function f is known as Probability density function  of the given random variable X.

The probability density function obviously satisfies the following probability axioms

    \[1. \ \ f(x)\geq 0\]

    \[2. \ \ \int_{-\infty}^{\infty} f(x)dx=1\]

Since from the axioms of the probability we know that the total probability is one so

\1 = P [X \in (-\infty,\infty)] = \ \ \int_{-\infty}^{\infty} f(x)dx\

For the continuous random variable the probability will be calculated in terms of such function f, suppose we want to find the probability for the continuous interval say [a, b] then it would be

P\left { a\leq X\leq b \right } = \int_{a}^{b} f(x)dx

As we know the integration represents the area under the curve so this probability shows such area for the probability like

Continuous random variable | Its Important distribution
Continuous random variable

by equating a=b the value will be

P{\left { X = a \right }} = \int_{a}^{a} f(x)dx =0

and in similar way the probability for the value less than or equal to specific value by following this will be

P \left { X < a \right } = P \left { X \leq a \right } = F(a) = \int_{-\infty}^{a} f(x)dx

Example: The continuous working time of the electronic component is expressed in the form of continuous random variable and the probability density function is given as

x = \begin{cases} \lambda e^{-x/100} &\ x\geq 0 \0 &\ x\geq 0 \end{cases}

find the probability that the component will work effectively between 50 to 150 hours and the probability of less than 100 hours.

since the random variable represents the continuous random variable so the probability density function given in the question gives the total  probability as

1=\int_{-\infty}^{\infty} f(x)dx= \lambda \int_{0}^{\infty} e^{-x/100}dx

So we will get the value of λ

1=-\lambda (100)e^{-x/100}\lvert_{\infty }^{0}=100\lambda

λ =1/100

for the probability of 50 hrs to 150hrs we have

P\left { 50< X < 150 \right } =\int_{50}^{150} \frac{1}{100}e^{-x/100} dx

=- e^{-x/100}\lvert_{150 }^{50}

=e^{-1/2} -e^{-3/2}\approx .384

in the similar way the probability less than 100 will be

P\left { X< 100 \right }=\int_{0}^{100} \frac{1}{100}e^{-x/100} dx

=- e^{-x/100}\lvert_{0 }^{100}

=1- e^{-1} \approx .633

Example: The computer based device has number of chipsets with lifespan given by the probability density function

f(x) = \begin{cases} 0 &\ x \leq 100 \ \frac{100}{x^{2}} &\ x> 100 \end{cases}

then after 150 hours find the probability that we have to replace 2 chipset from total 5 chips.

let us consider Ei be the event to replace the i-th chipset. so the probability of such event will be

P(E_{i}) =\int_{0}^{150} f(x)dx

=100\int_{100}^{150} x^{-2}dx =\frac{1}{3}

as working of all the chips independent so the probability for 2 to be replace will be

p(X) = \binom{5}{2} (\frac{1}{3})^{2}(\frac{2}{3})^{3}=\frac{80}{243}

Cumulative distribution function

  Cumulative distribution function for the continuous random variable is defined with the help of probability distribution function as

F(x) = P (X\leq x) = \int_{-\infty}^{x} f(u)du

in another form

F(a) = P ( X\in (-\infty, a)) = \int_{-\infty}^{a} f(x)dx

we can obtain the probability density function with the help of distribution function as

\frac{\mathrm{d} }{\mathrm{d} a} F(a)=f(a)

Mathematical Expectation and Variance of continuous random variable

Expectation

The mathematical expectation or mean of the continuous random variable  with probability density function  can be define as

E[X] = \int_{-\infty}^{\infty} xf(x)dx

  • For any real valued function of continuous random variable X expectation will be

E[g(X)] = \int_{-\infty}^{\infty} g(x)f(x)dx

where g is the real valued function.

  1. For any non-negative continuous random variable Y the expectation will be

E[Y] = \int_{0}^{\infty} P \left { Y > y \right } dy

  • For any constants a and b

E[aX + b] = aE[X] + b

Variance

                The variance of the continuous random variable X with the parameter mean or expectation  can be define in the similar way as discrete random variable is

Var(X) = E[(X - \mu )^{2}]

Var(X) = E[X^{2}] - (E[X])^{2}

For any constants a and b

Var(aX + b) = a^{2} Var(X)

   The proof of all the above properties of expectation and variance we can easily obtain by just following the steps we have in discrete random variable and the definitions of expectation, variance and probability in terms of continuous random variable

Example: If the probability density function of continuous random variable X is given by

f(x) = \begin{cases} 2x &\ if \ \ 0\leq x\leq 1 \0 &\ otherwise \end{cases}

then find the expectation and variance of the continuous random variable X.

Solution:  For the given probability density function

f(x) = \begin{cases} 2x &\ if \ \ 0\leq x\leq 1 \0 &\ otherwise \end{cases}

the expected value by the definition will be

E[X] = \int xf(x)dx = \int_{0}^{1} 2 x^{2} dx =\frac{2}{3}

Now to find the variance we require E[X2]

E[X^{2}] = \int_{-\infty}^{\infty} x^{2} f(x) dx =\int_{0}^{1} 2 x^{3} dx = \frac{1}{2}

Since

Var(X) =E[X^{2}] - (E[X])^{2}

so

Var(X) = \frac{1}{2} -(\frac{2}{3})^{2} =\frac{1}{18}

Uniform random variable

    If the continuous random variable X is having the probability density function given by

f(x) = \begin{cases} 1 &\ 0< x < 1 \\0 &\ otherwise \end{cases}

over the interval (0,1) then this distribution is known as uniform distribution and the random variable is known as uniform random variable.

  • For any constants a and b such that 0<a<b<1

P\left { a\leq X\leq b \right } = \int_{a}^{b} f(x)dx =b-a

  • Instead of the interval (0,1) we can generalize the uniform distribution to any general interval (α , β)  if the probability density function for the random variable is

f(x) = \begin{cases} \frac{1}{\beta -\alpha } &\ if \ \ \alpha <x < \beta \\0 &\ otherwise \end{cases}

Continuous random variable
Continuous random variable: Uniform random variable

Expectation and Variance of Uniform random variable

      For the uniformly continuous random variable X on the general interval (α , β) the expectation by the definition will be

E[X] = \int_{-\infty}^{\infty} xf(x)dx

= \int_{\alpha }^{\beta } \frac{x}{\beta -\alpha } dx

= \frac{\beta ^{2}-\alpha^{2}}{2(\beta -\alpha ) }

= \frac{\beta +\alpha}{2}

and variance we will get if we find first E[X2]

E[X^{2}] = \int_{\alpha }^{\beta } \frac{x^{2}}{\beta -\alpha } dx

= \frac{\beta ^{3}-\alpha^{3}}{3(\beta -\alpha)}

= \frac{\beta ^{2}+\beta\alpha+\alpha^{2}}{3}

so

Var(X)= \frac{\beta ^{2}+\beta\alpha+\alpha^{2}}{3} -\frac{(\alpha +\beta )^{2}}{4}

=\frac{(\beta - \alpha)^{2}}{12}

Example: At a particular station the trains for the given destination arrive with frequency of 15 minutes form 7 A.M. For the passenger who is on the station at a time between 7 to 7.30 distributed uniformly what will be the probability that the passenger get train within 5 minutes and what will be probability for more than 10 minutes.

Solution: As the time from 7 to 7.30 is distributed uniformly for the passenger to be at railway station denote this by uniform random variable X. so the interval will be (0, 30)

Since to get the train within 5 minutes passenger must be at the station between 7.10 to 7.15 or 7.25 to 7.30 so the probability will be

P\left { 10< X < 15 \right } + P\left { 25 < X < 30 \right } =\int_{10}^{15}\frac{1}{30} dx + \int_{25}^{30} \frac{1}{30} dx

=1/3

In similar manner to get the train after waiting more than 10 minutes passenger must be at the station from 7 to 7.05 or 7.15 to 7.20 so the probability will be

P\left { 0< X < 5 \right } + P\left { 15 < X < 20 \right } =\frac{1}{3}

Example: Find the probability for the uniform random variable X distributed over the interval (0,10 )

for X<3, X>6 and 3<X<8.

Solution: since the random variable is given as uniformly distributed so the probabilities will be

P \left \{ X< 3 \right \} = \int_{0}^{3}\frac{1}{10} dx =\frac{3}{10}

P\left { X > 6 \right } = \int_{6}^{10}\frac{1}{10} dx =\frac{4}{10}

P \left {3< X< 8 \right } = \int_{3}^{8}\frac{1}{10} dx =\frac{1}{2}

Example: (Bertrands Paradox) For any random chord of a circle. what would be the probability that the length of that random chord will be greater than the side of the equilateral triangle inscribed in the same circle.

This problems does not have clearance about the random chord so this problem were reformulated in terms of diameter or angle and then answer as 1/3 were obtained.

Conclusion:

   In this article the concept of continuous random variable and its distribution with probability density function were discussed and the statistical parameter mean, variance for the continuous random variable is given. The uniform random variable and its distribution with example is given which is the type of continuous random variable in the successive article we will focus some important types of continuous random variable with suitable examples and properties. ,if you want further reading then go through:

Schaum’s Outlines of Probability and Statistics

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

If you want to read more topics on Mathematics then go through Mathematics Page.

About DR. MOHAMMED MAZHAR UL HAQUE

I am DR. Mohammed Mazhar Ul Haque , Assistant professor in Mathematics. Having 12 years of experience in teaching. Having vast knowledge in Pure Mathematics , precisely on Algebra. Having the immense ability of problem designing and solving. Capable of Motivating candidates to enhance their performance.
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