Let $X$ be standard normal random variable, i.e., $X ∼ N(0, 1)$. Consider transformed random variable: $Y = X^2$.
(a) Find the probability density function of $Y$.
(b) Find the covariance between $X$ and $Y$.
I'm new to probability and don't really get yet transformation of random variables. If anyone could help me with this one I'd much appreciate.
My work: $F_Y(y)=P(Y\leq y)=P(X^2\leq y)=P(X\leq\sqrt{y})=F_X(\sqrt{y})=\int_{-\infty}^\sqrt{y}\frac{1}{\sqrt{2\pi}}e^{-t^2/2}\,dt$ that's where I'm stuck.
Edit: Is the correlation 0 because of the odd-moments popping in the equation?
You are close on the CDF. The only issue is that you need an absolute value: $$ F_Y(y) = P(|X|\le \sqrt{y}) = P(X\le \sqrt{y})-P(X\le -\sqrt{y}) = F_X(\sqrt{y})-F_X(-\sqrt y).$$ This would be an integral $\int_{-\sqrt y}^{\sqrt y}$ instead of $\int_{-\infty}^{\sqrt{y}}.$
In order to get the PDF, you differentiate the CDF. This can either be done from the integral using the fundamental theorem of calculus and the chain rule, or, equivalently, using the chain rule and the fact that $f_X=F_X'.$ So you have $$ f_Y(y) = \frac{d}{dy}(F_X(\sqrt y)-F_X(-\sqrt{y}) = \frac{1}{2\sqrt{y}}(f_X(\sqrt{y})+ f_X(-\sqrt{y})) = \frac{1}{\sqrt{y}}\frac{e^{-y/2}}{\sqrt{2\pi}}$$ for $y>0$.