The following is an exercise in Prof Tao's lecture notes on probability theory.
We assume that a sequence of random variables $\xi_n \rightarrow \xi$ in distribution and also $\nu_n \rightarrow \nu$ in distribution as well.
(i) If $\nu$ equals a constant almost surely, then prove that the product $\xi_n \nu_n$ converges to $\xi \nu$ in distribution.
(ii) Find a counterexample to (i) in the case $\nu$ does not equal a constant a.s.
Any help, hints will be greatly appreciated.
Best Answer
(i) We have to show that $\xi_n(\nu_n-\nu)$ goes to $0$ in probability. Take $\varepsilon\gt 0$; then for each positive $R$, $$\mu\{|\xi_n(\nu_n-\nu)|\gt\varepsilon\}\leqslant \mu\{|\xi_n|\geqslant R\}+\mu\{|\nu_n-\nu|\geqslant \varepsilon/R\}.$$ We then have by portmanteau theorem that for each $R$, $$\limsup_{n\to\infty}\mu\{|\xi_n(\nu_n-\nu)|\gt\varepsilon\}\leqslant\mu\{|\xi|\geqslant R\},$$ and we conclude, as $R$ was arbitrary.
(ii) Take $\xi_n$ and $\nu_n$ discrete random variable with disjoint support and which converge to a continuous distribution (for example, we can use Riemann sums).