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.


(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).

  • $\begingroup$ I am sorry, but how do you finish the argument of (i)? $\endgroup$ – a12345 Dec 10 '13 at 0:20
  • $\begingroup$ The sequence $(\{|\xi|\geqslant N\})_{N\geqslant 1}$ is non-increasing, and $\bigcap_N\{|\xi|\geqslant N\}=\emptyset$. $\endgroup$ – Davide Giraudo Dec 10 '13 at 9:09
  • $\begingroup$ Yes, but we want to $ \xi_n \nu_n$ converges to $\xi \nu$ in distribution. $\endgroup$ – a12345 Dec 11 '13 at 1:41
  • $\begingroup$ Instead, you showed that $\xi_n (\nu_n - \nu) $ converges to zero in probability $\endgroup$ – a12345 Dec 11 '13 at 1:42

It is worth noting that there can't be any examples for (ii) with any type of convergence:

  • $v_n$ does not converge to a constant almost surely by task description
  • if $v_n$ was to converge in probability to a constant, then, by Slutsky's theorem, the product $\xi_n v_n$ also converged in distribution
  • if $v_n$ was to converge in distribution to a constant, then it also converged in probability. Continue with the second point from here.

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