Given $Z_n\rightarrow 0$ in probability and $W$ a random variable, proving $WZ_n\rightarrow 0$

Question

Given $Z_n\rightarrow 0$ in probability and $W$ a random variable, I need to prove $WZ_n\rightarrow 0$.

I was given a hint: show that for every $\delta,\epsilon<0$ we have $$\{|WZ_n|\geq\epsilon\}\subset\{|Z_n|\geq\delta\}\cup\{|W|\geq\frac\epsilon\delta\}$$

I'm not sure how to get to this conclusion, and I'm not even sure how it will help-

If we prove the hint, we get $$\lim_{n\rightarrow\infty} P(|WZ_n|\geq\epsilon)\leq\lim_{n\rightarrow\infty} P(|Z_n|\geq\delta)+ P(|W|\geq\frac\epsilon\delta)$$

I understand that $\lim\limits_{n\to\infty} P(|Z_n|\geq\delta)=0$, but why does $P(|W|\geq\frac\epsilon\delta)=0$?

I do know $ P(|W|\geq\frac\epsilon\delta)\leq\frac{E|W|\delta}\epsilon$, from Markov's inequality.

Answer 1

First off, the equality should be an inequality:

$$\lim_{n\rightarrow\infty} P(|WZ_n|\geq\epsilon)\leq \lim_{n\rightarrow\infty} P(|Z_n|\geq\delta)+ P(|W|\geq\frac\epsilon\delta)=0+P(|W|\geq\frac\epsilon\delta).$$

Notice that $\delta>0$ is arbitrary and the left hand side is independent of $\delta$. Thus, take a second limit of $\delta\rightarrow 0$.

To prove the hint, it's easier to prove the complement:

$$\{|W|\geq \epsilon/\delta\}^c\cap \{|Z_n|\geq \delta\}^c\subseteq \{|WZ_n|\geq \epsilon|\}^c,$$

which when simplified should be apparent.

Answer 2

I believe is easier to use the subsequence definition of convergence in probability.

$\{Z_n W\}$ is clearly a sequence of random variables (since it is product of r.v's). To show that $\{Z_n W\} \xrightarrow{ P} 0$, we will show that for any subsequence $\{Z_{n_k} W\}$ there is a further subsequence $\{n_\ell\}$ of $\{n_k\}$ such that $\{Z_{n_\ell}W\}\xrightarrow{a.e.} 0$.

Let $\{Z_{n(k)} W\}$ a subsequence. Since $Z_n \xrightarrow{P}0$, for $Z_{n(k)}$ there is a subsequence such that $Z_{n (k (\ell))}\xrightarrow{a.e.} 0$ and since $W$ is real-valued then $Z_{n(k (\ell))} W\xrightarrow{a.e.} 0$ and hence $\{Z_n W\} \xrightarrow {P} 0$.