Let $Z\sim N(0,1)$ and $Y=Z^2$. Find $f_Y(y)$ by using moment generating function.
So I have moment generating function $M_Y(t)=E(e^{Z^2t})=\int_{-\infty}^\infty e^{z^2t}\cdot f_{Z^2}(z)dz$
Not sure how to continue from here. I believe for $Z\sim N(0,1)$ I have $f_Z(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}$
Best Answer
$$\mathbb{P}(Y \leq y)=\mathbb{P}(Z^2 \leq y)$$ with $y>0$
We have that $$F_y(y)=\mathbb{P}(Z^2 \leq y)=\mathbb{P}(-\sqrt{y}\leq Z \leq \sqrt{y})=\Phi(\sqrt{y})-\Phi(-\sqrt{y})$$ where $\Phi$ is the cdf of $Z$
Differentiating $F_y$ wrt to $y$, we have $$f_Y(y)=\frac{1}{2\sqrt{y}}f_{Z}(\sqrt{y})+\frac{1}{2\sqrt{y}}f_{Z}(-\sqrt{y})$$
We also know that $$f_{Z}(-\sqrt{y})=f_{Z}(\sqrt{y})$$
Therefore, $$f_Y(y)=\frac{1}{\sqrt{y}}f_{Z}(\sqrt{y})$$