Let $X \sim N(0, 1)$ and let $Y$ be a random variable with density
$$p_{Y}(t) = \begin{cases} te^{-t^2/2} & \text{ if } t \geq 0 \\ 0 & \text{otherwise.}\end{cases} $$
How can I find the distribution of $X \cdot Y$, assuming $X$ and $Y$ are independent? Can I just multiply the two densities together? I tried a lot of things, like working backwards from $P(X \cdot Y \leq t)$ and then trying to plug in the densities of $X$ and $Y$, but I got nowhere. Can someone please help me?
Since $X$ has an even distribution and $Y$'s support is $y\ge0$, we should work out the distribution of $|X|Y$ first, then "reflect" it. Since$$P(|X|Y\le z)=P(\ln |X|+\ln Y\le\ln z),$$so let's start with the MGFs of $U:=\ln|X|,\,V:=\ln Y$: for any $k\in\Bbb R$,$$M_U(k)=\Bbb E|X|^k=\sqrt{\frac{2^k}{\pi}}\Gamma\left(\frac{k+1}{2}\right),\,M_V(k)=\Bbb EY^k=\sqrt{2^k}\Gamma\left(\frac{k+2}{2}\right).$$Hence$$\Bbb E(|X|Y)^k=M_{U+V}(k)=\frac{2^k}{\sqrt{\pi}}\Gamma\left(\frac{k+1}{2}\right)\Gamma\left(\frac{k+2}{2}\right)=\Gamma(k+1)=k!,$$by the Legendre duplication formula. Thus $|X|Y\sim\operatorname{Exp}(1)$, so $XY$ has pdf $\frac12e^{-z}$.
You're also welcome to prove this result using the product distribution result discussed in @ChargeShiver's link. The CDF of $|X|Y$ is then$$\sqrt{\frac{2}{\pi}}\int_0^\infty(e^{-w^2/2}-e^{-w^2/2-z^2/2w^2})dw=\sqrt{\frac{1}{2\pi}}\int_{-\infty}^\infty(e^{-w^2/2}-e^{-w^2/2-z^2/2w^2})dw,$$which by Glasser's master theorem (with $p:=w-\frac{z}{w}$) simplifies to $1-e^{-z}$, i.e. an $\operatorname{Exp}(1)$ distribution as before.
So we have two very different approaches that each need a fairly advanced result, one using the duplication formula, the other the master theorem. I'm not sure whether any solution exists that's a bit more basic.
