$L^2$ convergence of product of $L^2$ random variables Assume $X_n\to X$ in $L^2$ and $Y_n\to Y$ in $L^2$ and further $X_nY_n\in L^2$. Does this imply $XY\in L^2$?
Hoelder of course gives $XY\in L^1$ and we can also conclude $X_nY_n\to XY$ in $L^1$. This does not use the assumption $X_nY_n\in L^2$. Is this assumption enough to conclude $XY\in L^2$? What additional assumptions might be sufficient?
Additionally being dominated, i.e. $|X_nY_n|\leq Z$ for another $L^2$ random variable $Z$, should be enough using dominated convergence. This assumption is however quite strong.
I may be able to assume that $Y$ has a normal distribution with $0$ mean. Does this help?
 A: A simple probabilistic assumption you could add, although rather strong, is the independence of $XY$. This immediately gives $\|XY\|_2 = \|X\|_2 \|Y\|_2 < \infty$.
Another thing you can assume is that $\sup_n \| X_nY_n\|_2$ is finite. The $L^2$ boundedness of $X_nY_n$ implies that a subsequence converges weakly in $L^2$. Therefore its weak limit is some element of $L^2$. But a weak $L^1$ limit is also the weak $L^2$ limit which is unique, and hence $XY$ is in $L^2$.
Regarding if $Y$ being normal can help (and my probability theory is rusty, so do read this paragraph with a careful eye); I think it is just shy of helping. What you can get from Hölder's inequality is that $\|XY\|_{L^p} \le \|X\|_{L^2}\|Y\|_{L^q}$ for any $p,q\in[1,\infty]$ such that $\frac1p = \frac12 + \frac1q$. In order to push all the way to $p=2$, one needs to use $q=\infty$ but a normal r.v. is not a.s. bounded. However, it has finite moments of all (finite) orders, which gives $XY\in L^p$ for all $p$ strictly larger than $2$. Alternatively, if you also have that $X\in L^r$ for some $r\in(2,\infty]$, then getting $p=2$ is possible since $\frac12 = \frac1r+\frac1p$ has the solution $\frac1p:=\frac12-\frac1r\in(0,1]$.
The result is not true without further assumptions. Let $X_n=Y_n = \min(|f|,n)$ where $f$ is in $L^2$ and not in $L^4$. Then $X_nY_n$ belongs to $L^\infty$ and hence all $L^p$, but the ($L^1$) limit is $|f|^2$ which is in $L^1$ and not $L^2$.
A concrete example of the above would be $f(x)=x^{-1/4}$ on $[0,1]$ with uniform measure.
