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I have a sequence of non negative variables $X_n$ such as: $$E(X_n|C_n)=\frac{C_n}{n^2}$$

where $C_n$ is a sequence of random variables converging almost surely to $1$.

Can I conclude $X_n$ tends to 0 almost surely?

Note: you can replace $\frac{1}{n^2}$ by any sequence with finite sum. The question remains essentially the same and the answer provided by Jason works just the same (see the Borel-Cantelli argument).

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Yes, $X_{n} \to 0$ almost surely. The argument I have is a little convoluted, so bear with me.

First, consider the events $F_{k} = \bigcup_{n \geq k} \{ C_{n} > 2 \}$. By the almost sure convergence of the $C_{n}$ it follows that $P( \bigcap_{k} F_{k} ) = 0$, and since $F_{1} \supseteq F_{2} \supseteq \cdots$ we have $P(F_{k}) \to 0$. So it suffices to show that $X_{n} \to 0$ a.s. within $F_{k}^{c}$, for any $k$.

Now fix a $k$ and an $\varepsilon > 0$. Using the notation $E[X; A]$ to represent $E[X 1_{A}]$, we have for $n \geq k$ \begin{equation} E[X_{n} ; F_{k}^{c}] \leq E[X_{n} ; C_{n} \leq 2] = E[ E(X_{n} | C_{n}) ; C_{n} \leq 2] = E[ C_{n} / n^{2} ; C_{n} \leq 2 ] \leq 2 / n^{2}. \end{equation} This is kind of the key part. (Note, too, that we used the nonnegativity of $X_{n}$ in the first step, to pass from $F_{k}^{c}$ to the larger event $C_{n} \leq 2$.) From here we just need some fairly run-of-the-mill measure theoretic arguments.

The bound above, together with the nonnegativity of $X_{n}$, implies that $P(F_{k}^{c} \cap \{ X_{n} > \varepsilon \}) \leq \frac{2}{n^{2} \varepsilon}$ (for $n \geq k$), so that \begin{equation} \sum_{n \geq k} P(F_{k}^{c} \cap \{ X_{n} > \varepsilon \}) < \infty. \end{equation}

By the Borel-Cantelli Lemma we can now say that the event \begin{equation} F_{k}^{c} \cap \{ X_{n} > \varepsilon \, \text{for infinitely many $n$} \} \end{equation} has probability zero. Since $\varepsilon$ was arbitary, this gets us $X_{n} \to 0$ a.s. on $F_{k}^{c}$.

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  • $\begingroup$ This could be very slightly altered to show that for any exponent $\alpha$ on $n$ such that $\alpha > 1$, $X_n \to 0 \space \text{a.s.}$, it seems to me. $\endgroup$ – jbowman Mar 10 '18 at 2:36
  • $\begingroup$ Thanks a lot. By $E(X;A)$ you mean $E(X|A)$ or $E(X1_A)$ ? $\endgroup$ – Benoit Sanchez Mar 10 '18 at 13:19
  • $\begingroup$ You mean $E(X1_A)$ :-) Maybe you should mention it. Everything seems correct to me, great! Honestly, I don't think there is a simpler proof. $\endgroup$ – Benoit Sanchez Mar 10 '18 at 14:38
  • $\begingroup$ Right, Benoit, I meant $E(X 1_{A})$. I'll make an edit to clarify that. $\endgroup$ – Jason Mar 10 '18 at 23:23
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Set $Z_n=X_n/C_n$. Then $E[Z_n]=1/n^2$ and $Z_n\ge 0$. By Markov's inequality, $P(Z_n>\epsilon)\le E[Z_n]/\epsilon = 1/(n^2\epsilon)$ which has finite sum, so by Borel Cantelli, $P(Z_n>\epsilon \text{ infinitely often})=0$ and $Z_n\to 0$ almost surely.

If $Z_n\to0$ almost surely and $C_n\to 1$ almost surely then $X_n=Z_nC_n\to 0$ almost surely.

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