Finding the distribution of $2\theta X_i^2$

Question

I need to find the distribution of $2\theta X_i^2$ in order to show that $\sum_{i=1}^n 2\theta X_i^2$ is a pivot, (and thus $\sum_{i=1}^n 2\theta X_i^2 \sim N(0,1)$). $X_1,X_2,...,X_n$ are i.i.d. with $f(x,\theta)=2x\theta e^{-\theta x^2}1(x>0)$ and $\theta >0$.

To find the distribution of $2\theta X_i^2$, I wanted to use the expected value and variance of $f(x,\theta)=2x\theta e^{-\theta x^2}1(x>0)$.

I found that $E[2\theta X_i^2]=2$ and $Var[2\theta X_i^2]=8-16\theta^2$ by integrating $f(x,\theta)$, however I am not sure this is the right method.

Answer 1

I show you if $X_1,X_2,...,X_n$ are i.i.d with $f(x;\theta)=2x\theta e^{-\theta x^2}\mathbb{1}(x>0)$ then what is the distribution of $X_i^2$ then you can go ahead by yourself.

Let $Y=X^2$

For $y \ge 0$, $F_Y(y)=P(Y<y)=P(X^2<y)=P(-\sqrt{y}<X<\sqrt{y})=P(0<X<\sqrt{y})\\=F_X(\sqrt{y})-F_X(0)=F_X(\sqrt{y})$

We take derivative on both sides. \begin{align*} f_Y(y)&=\frac{1}{2\sqrt{y}}f_X(\sqrt{y})\\&=\frac{1}{2\sqrt{y}}f_X(\sqrt{y})\\&=\frac{1}{2\sqrt{y}}(2\sqrt{y}\theta e^{-\theta y})\\&=\theta e^{-\theta y} \end{align*}

You can see $X_i^2$ has an exponential distribution.

Next, you need to find the sum of i.i.d exponential random variables; I think now you can go ahead by yourself.