- Let $Z\sim \mathcal{N}(0,1)$. What is the probability density function of $X=Z^2$?
- Let $Z_1,Z_2,...,Z_k\sim\mathcal{N}(0,1)$ be $k$ independent standard normal random variables. What is the probability density function of $Y=Z_1^2+...+Z_k^2$? (Hint: compute $\Gamma(1/2)$ by an adequate change of variables, then use the distribution of $Z^2$ from part one).
My Attempt
We know the p.d.f. for $Z\sim \mathcal{N}(0,1)$ is $$\phi(z)=\frac{1}{\sqrt{2\pi}}e^{-z^2/2}$$ So the c.d.f. for $X=Z^2$ is \begin{align} F_X(x)&=\mathbb{P}(X\leq x)=\mathbb{P}(Z^2\leq x)=\mathbb{P}(z\leq\sqrt{x})=F_Z(\sqrt{x})\\ &=\int_0^\sqrt{x}\frac{1}{\sqrt{2\pi}}e^{-z^2/2}dz\\ &=\frac{2}{\sqrt{\pi}}(1-e^{-x/2}) \end{align} Our p.d.f. is the derivative of this, so: $$\frac{1}{\sqrt{\pi}}e^{-x/2}$$
The part that I am stuck on is the second part. How should I handle $Y$ without knowing the value of $k$?
Quick answer:
$$X=Z^2\sim \chi_{(1)}^2$$
$$Y=\sum_k Z_k^2\sim \chi_{(k)}^2$$
Let's see why:
First your error:
$$F_X(x)=\mathbb{P}(Z^2\leq x)=\mathbb{P}(-\sqrt{x}\leq Z\leq \sqrt{x})=\Phi_Z(\sqrt{x})-\Phi_Z(-\sqrt{x})$$
Derivating you get your density
$$f_X(x)=\frac{1}{2\sqrt{x}}\phi(\sqrt{x})+\frac{1}{2\sqrt{x}}\phi(-\sqrt{x})=$$
$$=\frac{1}{\sqrt{x}}\phi(\sqrt{x})=\frac{1}{\sqrt{x}}\cdot \frac{1}{\sqrt{2\pi}}e^{-x/2}$$
Now consider that
$$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$
Thus substitute in the result and verify that $f_X(x)$ is the density of a $\text{Gamma}\left(\frac{1}{2};\frac{1}{2}\right)$ which is a chi-square with 1 d.o.f.
As the second part is concerned, using the result found in 1. you can immediately get the solution using the additivity property of Gamma distribution which can be easy proved using MGF or Characteristic function