# Show that Y has an exponential distribution from f(x)??

• January 3rd 2011, 11:41 AM
hazeleyes
Show that Y has an exponential distribution from f(x)??
hey guys, I'm just wondering if you guys can help me with this question, thank you in advance. (Bow)

A random variable, X, has a probability density function (p.d.f.) given by

f(x)=2xe^(-x^2)
where x>0

Y is a random variable such that Y = X^2.

Show that Y has an exponential distribution and state its mean.

thanks.(Happy)
• January 3rd 2011, 12:29 PM
pickslides
Hi there hazeleyes,

Here's a kick off...

$\displaystyle f(Y) = f(X^2) = 2(X^2)e^{-(X^2)^2}$

for the mean $\displaystyle E(Y) = \int_{-\infty}^{\infty}Y\times f(Y)~dY$
• January 3rd 2011, 01:47 PM
mr fantastic
Quote:

Originally Posted by pickslides
Hi there hazeleyes,

Here's a kick off...

$\displaystyle f(Y) = f(X^2) = 2(X^2)e^{-(X^2)^2}$

for the mean $\displaystyle E(Y) = \int_{-\infty}^{\infty}Y\times f(Y)~dY$

Sorry, but that's not correct.

There are several approaches that can be taken.

The OP has carelessly neglected to mention that the support of X is $x \geq 0$ (so that f(x) = 0 for x < 0). The support of Y will be $y \geq 0$.

The most basic approach to finding the pdf of Y is to calculate the cdf of Y and then recall that the pdf is the derivative.

$\displaystyle cdf = G(y) = \Pr(Y \leq y) = \Pr(X^2 \leq y) = \Pr(-\sqrt{y} \leq X \leq \sqrt{y}) = \int_0^{\sqrt{y}} 2 x e^{-x^2} \, dx$.

Differentiating this is simple - either integrate (use a substitution) directly and then differentiate, or use the chain rule and the Fundamental Theorem of Calculus.

You should get $\displaystyle pdf = \frac{dG}{dy} = g(y) = e^{-y}$ for $y \geq 0$ and zero elsewhere. Calculating the mean of Y is trivial.

Some trivia: The random variable X follows a Weibull distribution (http://en.wikipedia.org/wiki/Weibull_distribution) with parameters $k = 2$ and $\lambda = 1$. Since $Y = X^2 = X^k$ it is well known that Y follows an exponential distribution.
• January 3rd 2011, 02:49 PM
hazeleyes
Thanks alot guys!! (Bow)