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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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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.

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