Suppose $(\xi_i)$ is an iid sequence and $(V_i)$ is a sequence such that $V_n$ converges almost surely to zero. Then $\xi_nV_n$ converges almost surely to zero or in probability? How can I prove this?

I think I can show convergence in probability as $P(\omega:|\xi_n(\omega)|V_n(\omega)|<\varepsilon)$ gets bigger as we increase $n$ due to the fact that $V_n(\omega)$ gets closer to zero for all $\omega$. Does this make sense?

But for a.s. convergence I'm not sure if it is possible to show.

  • $\begingroup$ I guess you mean $\xi_n V_n$ instead of $x_n V_n$, right? And what have you tried? Please add some more thoughts on the problem. $\endgroup$
    – saz
    Sep 21 '14 at 11:31
  • $\begingroup$ No, this doesn't make sense. Just think of the following sequence: e.g. $V_n(\omega) = \frac{1}{n}$ and $\xi_n(\omega)=n$. Then $V_n(\omega) \to 0$ but $\xi_n(\omega) V_n(\omega)$ does not converge to $0$. $\endgroup$
    – saz
    Sep 21 '14 at 12:34
  • 1
    $\begingroup$ But $(\xi)$ is not iid then $\endgroup$
    – radIQ
    Sep 21 '14 at 12:37
  • $\begingroup$ Well, but for some fixed $\omega$ we could have $\xi_n(\omega)=n$, right? My point is simply that it is not that obvious that $V_n \to 0$ implies that the probability gets bigger as $n$ increases. $\endgroup$
    – saz
    Sep 21 '14 at 12:38
  • $\begingroup$ Right that can be the case, but $P(\omega:|\xi_n(\omega|<\varepsilon)$ can't change with $n$ right? Maybe your example shows that we can't have a.s. convergence? $\endgroup$
    – radIQ
    Sep 21 '14 at 12:45

Convergence in probability: Fix $K>0$. It is not difficult to see that

$$\begin{align*} \mathbb{P}(|\xi_n V_n| \geq \varepsilon) &\leq \mathbb{P}\left( |V_n| \geq \frac{\varepsilon}{K} \right)+ \mathbb{P}(|\xi_n| \geq K). \end{align*}$$

Since $V_n \to 0$ almost surely, we get

$$\limsup_{\varepsilon \to 0} \mathbb{P}(|\xi_n V_n| \geq \varepsilon) \leq \mathbb{P}(|\xi_1| \geq K).$$

Assuming that $\mathbb{P}(|\xi_1| = \infty)=0$, this shows that $\xi_n V_n \to 0$ in probability.

Convergence almost surely: The sequence does in general not converge almost surely. Set $V_n:=n^{-1}$, then $V_n \to 0$ almost surely and $$\mathbb{P}(|\xi_n V_n| \geq \varepsilon) = \mathbb{P}\left(|\xi_n| \geq n \varepsilon \right) = \mathbb{P} \left( |\xi_1| \geq n \varepsilon \right).$$

If we choose the distribution of $\xi_1$ such that $\mathbb{P}(|\xi_1| \geq x) \approx \frac{1}{x}$ for $x$ sufficiently large (e.g. Cauchy distribution), then we find

$$\sum_{n \geq 1} \mathbb{P}(|\xi_n V_n| \geq \varepsilon) = \infty$$

and therefore, by the Borel-Cantelli theorem, $|\xi_n(\omega) V_n(\omega)| \geq \varepsilon$ infinitley often (note that since $V_n$ is deterministic, the sets $\{\omega; |\xi_n(\omega) V_n(\omega)| \geq \varepsilon\}$ are independent).

Note that $\xi_n V_n$ converges almost surely to $0$ e.g. if $\xi_1$ is bounded or $V_n = 0$ for $n$ sufficiently large.


When it comes to convergence in probability it is to be shown that $P\left\{ \left|\xi_{n}V_{n}\right|\geq\epsilon\right\} $ converges to $0$ for each $\epsilon>0$. Suppose not. Without essential loss of generality we could say that $P\left(\left|\xi_{n}V_{n}\right|\geq1\right)\geq\frac{1}{2}$ for each $n$.

For a fixed positive integer $k$ we have $\left\{ \left|\xi_{n}V_{n}\right|\geq1\right\} \subseteq\left\{ \left|\xi_{n}\right|\geq k\right\} \cup\left\{ \left|V_{n}\right|>k^{-1}\right\} $ so that $P\left\{ \left|\xi_{n}\right|\geq k\right\} +P\left\{ \left|V_{n}\right|>k^{-1}\right\} \geq P\left\{ \left|\xi_{n}V_{n}\right|\geq1\right\} \geq\frac{1}{2}$. Denoting $P\left\{ \left|\xi_{n}\right|\geq k\right\} $ by $p_{k}$ and letting $n\rightarrow\infty$ leads to $p_{k}\geq\frac{1}{2}$. This cannot be true for every $k$ so a contradiction has been attained.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.