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)

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

Quote:

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.

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

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 $\displaystyle \displaystyle pdf = \frac{dG}{dy} = g(y) = e^{-y}$ for $\displaystyle y \geq 0$ 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(-y^{1/2} <= X <= y^{1/2})

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

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

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= X^{2} is also exp(lambda). Is that correct?