Let $X_1$ and $X_2$ are independent $N(0, \sigma^2)$ which means (mean = 0, variance = $\sigma^2$) random variables. What is the distribution of $X_1^2 + X_2^2$?
My approach is that $X_1\sim N(0, \sigma^2)$ and $X_2\sim N(0, \sigma^2)$
Then $X_1^2$ and $X_2^2$ have chi-squared distribution with 1 degree of freedom. (I am not sure the degree of freedom and not sure how to show it as well(please help on this))
Then I found the moment-generating function for $X_1^2$ and $X_2^2$;$$m_{X_1^2} = (1-2t)^{-1/2}$$ and $$m_{X_2^2} = (1-2t)^{-1/2}$$
So the moment generating function for $X_1^2 + X_2^2$ is $$m_{X_1^2}(t) m_{X_2^2}(t) = (1-2t)^{-2/2}$$
So $X_1^2 + X_2^2$ has a chi-squared distribution with 2 degrees of freedom. Is this correct?
If $X$ and $Y$ are independent $N(0,\sigma^2)$ random variables, then for any $z \geq 0$, $$\begin{align} P\{X^2+Y^2 > z\} &= \int_{x^2+y^2>z}f_{X,Y}(x,y)\\ &= \int_{x^2+y^2>z}\frac{1}{2\pi\sigma^2}\exp\left(-\frac{x^2+y^2}{2\sigma^2}\right)\\ &= \int_{\sqrt{z}}^\infty\int_0^{2\pi} \frac{1}{2\pi\sigma^2}\exp\left(-\frac{r^2}{2\sigma^2}\right)\cdot r\,\mathrm d\theta\cdot \mathrm dr\\ &= \int_{\sqrt{z}}^\infty \frac{r}{\sigma^2}\exp\left(-\frac{r^2}{2\sigma^2}\right) \, \mathrm dr\\ &= \left. -\exp\left(-\frac{r^2}{2\sigma^2}\right)\right|_{\sqrt{z}}^\infty\\ &= \exp(-z/2\sigma^2). \end{align}$$ Now, if $Z$ is an exponential random variable with parameter $\lambda$, then $P\{Z > z\} = \exp(-\lambda z)$, and so $X^2+Y^2$ is a $\ldots$