Are squares of two i.i.d normal distributions i.i.d? Let $X,Y\sim N(0,\sigma^2)$ i.i.d. Correct me if I'm wrong, but $X^2,Y^2$ are also I.i.d.
Could I get advice on how to show this, or could someone let me know if this is difficult to show? 
My thought was to find the PDFs of $X^2,Y^2$ and show that the definition of independence. I'm not quite sure how to get the pdfs , though. I was thinking looking at the mgf of $X^2$. However, if I get complicated pdfs then showing independence by definition seems like not the best approach.
Thanks

$X$ and $Y$ are independent is equivalent to saying that for any Borel $f,g$, we have (if the expectations are defined)
$$
\mathbb{E}[f(X)g(Y)] = \mathbb{E}[f(X)]\mathbb{E}[g(Y)] \tag{$\dagger$}.
$$
To show independence of $X^2,Y^2$, we use the same criterion. Let $f,g$ be two arbitrary Borel functions (and let $s$ denote the square function, $s\colon x\mapsto x^2$). Then $f^\prime=f\circ s$ and $g^\prime=g\circ s$ are Borel, so that with $(\dagger)$,
$$
\mathbb{E}[f'(X)g'(Y)] = \mathbb{E}[f'(X)]\mathbb{E}[g'(Y)]
$$
(provided the expectations are well-defined). But that is exactly saying
$$
\mathbb{E}[f(X^2)g(Y^2)] = \mathbb{E}[f(X^2)]\mathbb{E}[g(Y^2)]
$$ 
so (as $f,g$ were arbitrary) $X^2$ and $Y^2$ are independent.
To show that they have the same distribution, you can just say that for any $t \in\mathbb{R}$,
$$
\mathbb{P}\{X^2 \leq t \} = \mathbb{P}\{Y^2 \leq t \}
$$
(same cumulative distribution function); this is trivial if $t<0$, and for $t\geq 0$ we have 
$$
\mathbb{P}\{X^2 \leq t \} = \mathbb{P}\{X \leq \sqrt{t} \}
= \mathbb{P}\{Y \leq \sqrt{t} \}
= \mathbb{P}\{Y^2 \leq t \}
$$
where in the middle we used that $X$ and $Y$ are identically distributed (so that their cumulative distribution functions are equal).
