Continous Random Variable

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December 17th 2009, 09:17 PM
zorro

Continous Random Variable

Question : A continuous random variable has a probability density function

f(x) = $\begin{cases} \frac{1}{b-a} & a<x<b \\ 0 & otherwise \end{cases}$

Find the cumulative distribution
find the E[X] and Var X

Solution

The random variable is a uniform distribution random variable

Cumulative Distribution function $F_x (x)$ = $\begin{cases}0 & x<a \\ \frac{x-a}{b-a} & a \le x<b \\1 & x \ge b \end{cases}$

But i dont know how to prove the above ......

My work :

$F_x(x)$ = $\int_{ - \infty}^{a} f(t) \ dt$

= $\frac{1}{b-a} . 0$ = 0 ...............Is this correct ?

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

= $\frac{1}{b-a} \int_{- \infty}^{ \infty} x \$ dx .............What should i do now integrate with $- \infty$ and $\infty$ or what

And what about the Var X as well??
December 17th 2009, 09:59 PM
dedust

Quote:

Originally Posted by zorro

Question : A continuous random variable has a probability density function

f(x) = $\begin{cases} \frac{1}{b-a} & a<x<b \\ 0 & otherwise \end{cases}$

Find the cumulative distribution
find the E[X] and Var X

Solution

The random variable is a uniform distribution random variable

Cumulative Distribution function $F_x (x)$ = $\begin{cases}0 & x<a \\ \frac{x-a}{b-a} & a \le x<b \\1 & x \ge b \end{cases}$

But i dont know how to prove the above ......

My work :

$F_x(x)$ = $\int_{ - \infty}^{a} f(t) \ dt$

= $\frac{1}{b-a} . 0$ = 0 ...............Is this correct ?

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

= $\frac{1}{b-a} \int_{- \infty}^{ \infty} x \$ dx .............What should i do now integrate with $- \infty$ and $\infty$ or what

And what about the Var X as well??

for $t < a$
$F_x(x)$ = $\int_{ - \infty}^{x} f(t) \ dt = 0$ ( $f(t) = 0$ for $t < a$ )

for $a \leq t < b$
$F_x(x)$ = $\int_{ - \infty}^{x} f(t) \ dt = \int_{ - \infty}^{a} f(t) \ dt + \int_{a}^{x} f(t) \ dt = \int_{a}^{x} \frac{1}{b-a} \ dt = \frac{x - a}{b - a}$

for $t \geq b$
$F_x(x)$ = $\int_{ - \infty}^{x} f(t) \ dt = \int_{ - \infty}^{a} f(t) \ dt + \int_{a}^{b} f(t) \ dt + \int_{ b}^{x} f(t) \ dt = 1$

$E[X] = \int_{- \infty}^{\infty} x \cdot \ f(x) ~dx = \int_{a}^{b} x \cdot \ f(x) ~dx<br />$ ( $f(x) = 0$ outside the interval $(a,b)$ )

$Var(X) = E[X^2] - {E[X]}^2$
calculate $E[X^2] = \int_{- \infty}^{\infty} x^2 \cdot \ f(x) ~dx$
December 17th 2009, 10:04 PM
CaptainBlack

Quote:

Originally Posted by zorro

Question : A continuous random variable has a probability density function

f(x) = $\begin{cases} \frac{1}{b-a} & a<x<b \\ 0 & otherwise \end{cases}$

Find the cumulative distribution
find the E[X] and Var X

Solution

The random variable is a uniform distribution random variable

Cumulative Distribution function $F_x (x)$ = $\begin{cases}0 & x<a \\ \frac{x-a}{b-a} & a \le x<b \\1 & x \ge b \end{cases}$

But i dont know how to prove the above ......

My work :

$F_x(x)$ = $\int_{ - \infty}^{a} f(t) \ dt$

= $\frac{1}{b-a} . 0$ = 0 ...............Is this correct ?

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

= $\frac{1}{b-a} \int_{- \infty}^{ \infty} x \$ dx .............What should i do now integrate with $- \infty$ and $\infty$ or what

And what about the Var X as well??

$F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi$

Now if $x<a,\ f(x)=0$ , so:

$F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi=\int_{-\infty}^x 0\; d\xi=0$

Now if $a \le x<b,\ f(x)=1/(b-a)$ , so:

$F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi=\int_{-\infty}^a 0\; d\xi+\int_{a}^x \frac{1}{b-a} \; d\xi=\int_{a}^x \frac{1}{b-a} \; d\xi$

and because $f(x)=0$ for $x\ge b$ and is a pdf, for $x\ge b$

$F(x)=F(b)=1$

CB
December 18th 2009, 02:25 PM
zorro

Thank u dedust and Captain black

Quote:

Originally Posted by CaptainBlack

$F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi$

Now if $x<a,\ f(x)=0$ , so:

$F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi=\int_{-\infty}^x 0\; d\xi=0$

Now if $a \le x<b,\ f(x)=1/(b-a)$ , so:

$F_X(x)=\int_{-\infty}^x f(\xi)\; d\xi=\int_{-\infty}^a 0\; d\xi+\int_{a}^x \frac{1}{b-a} \; d\xi=\int_{a}^x \frac{1}{b-a} \; d\xi$

and because $f(x)=0$ for $x\ge b$ and is a pdf, for $x\ge b$

$F(x)=F(b)=1$

CB

Thank u dedust and captain black for helping be solve this problem
Zorro is deeply great full......... amigo's
cheers mite
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