Suppose that we have a random variable $X$ such that $X$ ~ $\textit{N}(0,3)$. We are asked to compute the law of $X^2$ and my answer is different from the solution by $\frac{1}{2}$. Help me please figure out where does it go.
Let $Y$ be a random variable s.t $Y = X^2$. Then $\phi(X^2)=\phi(Y)$.
$\mathbb{P}_{X}$ = $\frac{1}{\sqrt {6\pi}}e^{-\frac{x^2}{6}}$ and $y = x^2$, $dy = 2xdx$ $\Leftrightarrow dx = \frac{1}{2\sqrt{y}}dy$
I'll quote the Transfer Theorem. I study in French, so I haven't quite found the equivalent definition in English textbooks.
Let $(\Omega, \mathfrak{B}, \mathbb{P})$ be a probability space and $X: (\Omega, \mathfrak{B}, \mathbb{P}) \rightarrow \mathbb{R}$ a RV s.t. the law of $X$ is $\mathbb{P}_{X}$.
For any measurable function $\phi:\mathbb{R} \rightarrow \mathbb{R^{+}}$ we have $\int_{\mathbb{\Omega}}\phi(X(\omega))d\mathbb{P(\omega)} = \int_{\mathbb{R}}\phi(x)d\mathbb{P}_{X}$
Using this, we get $E[\phi(Y)]=\int_{\Omega}\phi(y)d\mathbb{P}_{Y}(y)=E[\phi(X^2)]=\int_{\mathbb{R}}\phi(x^2)d\mathbb{P}_{X}(x)=\int_{\mathbb{R}}\phi(x^2)\frac{1}{\sqrt {6\pi}}e^{-\frac{x^2}{6}}dx$.
Using $y=x^2$ we get $\int_{\mathbb{R}}\phi(y)\frac{1}{\sqrt {6\pi}}e^{-\frac{y}{6}}\frac{1}{2\sqrt{y}}\mathbb{I}_{\mathbb{R^{*}_{+}}}(y)dy = \int_{]0; \infty[}\phi(y)\frac{1}{2\sqrt{6 \pi y}}e^{-\frac{y}{6}}dy$
Therefore the law of $Y$ is $d\mathbb{P}_{Y} = \mathbb{I}_{]0;\infty[}(y)\frac{1}{2}\frac{1}{\sqrt{6 \pi y}}e^{-\frac{y}{6}}dy $. I only have the final answer:
Using the Transfer Formula we get the law of $Y=X^2$. $ \mathbb{P}_{Y}(dy) = \mathbb{I}_{]0;\infty[}(y)\frac{1}{\sqrt{6 \pi y}}e^{-\frac{y}{6}}dy$
Where does the $\frac{1}{2}$ go?
The $\frac{1}{2}$ that you got is in fact there, but it is cancelled by an extra factor of $2$ corresponding to the fact that each value of $X^2$ has two equally likely preimages.
Personally I find this way of doing it is unnecessarily confusing; the way I would explain it is like this:
$$F_{X^2}(x)=P(X^2 \leq x)=\begin{cases} 0 & x<0 \\ P(|X| \leq \sqrt{x})=F_X(\sqrt{x})-F_X(-\sqrt{x}) & x \geq 0 \end{cases}$$
where I have used the fact that in your situation $F_X$ is continuous. (Otherwise a small modification is required.) You can then get the density by differentiation.
By the way the English buzzword that corresponds to what you're calling "transfer" is "pushforward". The actual theorem you stated is usually called "the law of the unconscious statistician" in English.