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.