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

1. ## 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.

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.

2. Hi there hazeleyes,

Here's a kick off...

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

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

3. 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.

4. Thanks alot guys!!

5. ## Re: Show that Y has an exponential distribution from f(x)??

Originally Posted by mr fantastic

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.
Apologies in advance as I am not yet familiar with the Weibull distribution, and also I haven't yet learned to use LaTex...

I follow mr. fantastic's logic up to the point of G(y) = P(-y1/2 <= X <= y1/2)

But shouldn't this mean that G(y) = the integral from 0 to y1/2 of 2e-2x dx = 1 - e-2y1/2?

and therefore, g(y) = dG/dy = y-1/2 e-2y1/2?

6. ## Re: Show that Y has an exponential distribution from f(x)??

Sorry, regarding my last post I realized I did not read the original post carefully enough - I thought the original pdf was f(x) = 2e-2x. In the process however, I think I learned that where X~exp(lambda), Y= X2 is also exp(lambda). Is that correct?

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