I am attempting to find the p.d.f of the following random variable $V$:
Given $X_1$ and $X_2$ follow standard normal distributions and are independent. I want to find the p.d.f of $V = X_1^2 + X_2^2$.
My idea is to do let $Z = X_1^2$ and $Y=X_2^2$ and then calculate the p.d.f of $V = Z+Y$.
Proposed Method:
$F_Z(z) = P(Z \le z) = P(X_1^2 \le z) = P(-\sqrt{z} \le X_1 \le \sqrt{z}) = 2\Phi(\sqrt{z}) -1$
We also have a similar relationship with $Y$.
Then we get that $F_V(v) = P(V \le v) = P(Z+Y \le v) = P(Z \le v-y)$ - This is the step I am not sure of. If I can do this then I get that $F_V(v) = 2\Phi(v-y) -1 \implies f_V(v) = 2\phi(v-x_2^2) \implies f_V(v) = 2\phi(x_1^2)$
This seems that it most likely is wrong. Where am I going wrong and is there a way to do this using moment generating functions?
Edit: I am incorrect as when differentiating I have not applied the chain rule, and I have not square rooted the contents of the phi
$V=X_1+X_2$ is the definition of a known rv: a chi-square rv with 2 d.0.f
To prove this is very easy:
First, with a simple transformation verify that $X_1^2\sim \chi_{(1)}^2$
Second, using the properties of Gamma distribution prove which is the distribution of the sum (or via MGF, Characteristic function)
To prove that $X_1^2\sim \chi_{(1)}^2$ you can do as follows
Set $X\sim N(0;1)$ and let's derive the distribution of
$$Y=X^2$$
$$F_Y(y)=\mathbb{P}[Y\leq y]=\mathbb{P}[-\sqrt{y }\leq X\leq \sqrt{y}]=\Phi[\sqrt{y}]-\Phi[-\sqrt{y}]$$
Derivating you get
$$f_Y(y)=\phi(\sqrt{y})\frac{1}{2\sqrt{y}}+\phi(-\sqrt{y})\frac{1}{2\sqrt{y}}=\frac{1}{\sqrt{y}}\phi(\sqrt{y})$$
That is
$$f_Y(y)=\frac{1}{\sqrt{y}\sqrt{2\pi}}e^{-\frac{y}{2}}=\frac{\Big(\frac{1}{2}\Big)^{\frac{1}{2}}}{\Gamma\Big(\frac{1}{2}\Big)}y^{\frac{1}{2}-1}e^{-\frac{y}{2}}$$
Thus
$$Y\sim Gamma\Bigg(\frac{1}{2};\frac{1}{2}\Bigg)=\chi_{(1)}^2$$