Suppose that a sequence of random variables $Y_n$ convergence in $L^2$ to $Y$, i.e. $$ E|Y_n-Y|^2\to0\quad \text{as}\quad n\to\infty. $$ Moreover, there exist constants $c_0$ and $c_1$ such that $$ 0 < c_0 \leq Y_n, Y \leq c_1. $$ Assume further that $X_n Y_n$ convergence in distribution to $Z$. Is it true that $X_n$ convergence in distribution to $Z/Y$?
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The answer is no. E.g., suppose that $X_n=1$ and $Y_n=Y$ for all $n$, and $X:=Z/Y$, where $Z$ and $Y$ are each uniformly distributed on the interval $[1,2]$ and $Z$ is independent of $Y$. Then, trivially, $E|Y_n-Y|^2\to0$ and $X_nY_n=Y$ converges to $XY=Z$ in distribution. However, the distribution of $X_n=1$ does not converge to the absolutely continuous distribution of $X=Z/Y$.
answered Sep 3, 2019 at 15:58
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$\begingroup$ Thanks. I changed the problem sightly, so the counterexample you provided does not hold any more. $\endgroup$ Commented Sep 3, 2019 at 16:16
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1$\begingroup$ @WenguangZhao : People could be reluctant to answer your questions in the future if you change them to invalidate their answers and thus nullify their work. If you want to ask a different, even if related, question, you can post it separately. Anyhow, in this case my counterexample is still valid even after you edited the question -- just consider it a bit more carefully. $\endgroup$ Commented Sep 3, 2019 at 21:23
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$\begingroup$ Thanks again for your answer. You answer is correct. I re-asked a question in a different formulation. But your answer here really helps a lot. $\endgroup$ Commented Sep 4, 2019 at 8:58