Let $\{X_n,Y_n:n\in\mathbb N\}$ be a sequence of independent random variables. Suppose that $X_n$ and $Y_n$ have the same distribution for every $n\in\mathbb N$. Prove that if $X_n+Y_n\rightarrow0$ almost surely, then $X_n\rightarrow0$ almost surely.

I've tried to use Borel-Cantelli Lemma to reduce this to prove that $\sum_nP(|X_n+Y_n|>\epsilon)<\infty$ for any $\epsilon>0$ implies $\sum_nP(|X_n|>\epsilon)<\infty$ for any $\epsilon>0$. I observed that $P(|X_n|>\epsilon)=P(|X_n|>\epsilon,|X_n-Y_n|<\epsilon)+P(|X_n|>\epsilon,|X_n-Y_n|\geqslant\epsilon)$, and that the first term $P(|X_n|>\epsilon,|X_n-Y_n|<\epsilon)\leqslant P(|X_n+Y_n|>\epsilon)$. But I don't konw how to control the second term.

Please feel free to discuss it. I am not sure I am heading a right way: I can't find a proper method to use the condition $X_n\stackrel{d}{=}Y_n$.

  • $\begingroup$ Suppose that $Y_n = -X_n$ and $X_n$ has a symmetric distribution for all $n$. Then, $X_n \stackrel{d}{=} Y_n$ for all $n$, $X_n + Y_n = 0$ for all $n$, but $X_n$ does not necessarily converge to zero. (Pick $\{X_n\}$ to be any arbitrary independent sequence with symmetric marginal distribution). $\endgroup$ – passerby51 Dec 13 '16 at 8:38
  • $\begingroup$ We require that $\{X_n,Y_n:n\in\mathbb N\}$ is independent. $\endgroup$ – Eric Yau Dec 13 '16 at 8:41
  • $\begingroup$ it seems that implicit in the assumption is that $\{X_n,Y_n\}$ pairs are also independent, in which case my example is ruled out. $\endgroup$ – passerby51 Dec 13 '16 at 8:43
  • $\begingroup$ So why not use the assumption that you know the joint distribution of $(X_n,Y_n)$? $\endgroup$ – passerby51 Dec 13 '16 at 8:45
  • $\begingroup$ Hmmm....Yep.....:) $\endgroup$ – Eric Yau Dec 13 '16 at 8:45

I interpret the statement as follows: if $N$ is a positive integer then the sequence $\left(Z_n\right)_{1\leqslant n\leqslant 2N}$ is independent, where $Z_n=X_n$ if $1\leqslant i\leqslant N$ and $Z_n=Y_{n-N}$ if $N+1\leqslant n\leqslant 2N$. In other words, the sequence $\left(X_n\right)_{n\geqslant 1}$ is independent of $\left(Y_n\right)_{n\geqslant 1}$, and the sequence $\left(X_n\right)_{n\geqslant 1}$ is independent as well as $\left(Y_n\right)_{n\geqslant 1}$.

Lemma. If $\left(Z_n\right)_{n\geqslant 1}$ is an independent sequence, then $$Z_n\to 0 \mbox{ a.s.}\Leftrightarrow \left(\forall \varepsilon\gt 0,\sum_{n=1}^{+ \infty}\mathbb P\left(\left|Z_n\right|\gt\varepsilon\right)<+\infty \right).$$

This follows from the Borel-Cantelli lemma.

Note that for any positive $\varepsilon$, $$\left\{\left|X_n\right|\geqslant \varepsilon\right\}\cap\left\{\left|Y_n\right|\leqslant \varepsilon/2\right\} \subset \left\{\left|X_n+Y_n\right|\geqslant \varepsilon/2\right\}$$ hence by the Lemma, the series $\sum_{n=1}^{+\infty}\mathbb P\left(\left\{\left|X_n\right|\geqslant \varepsilon\right\}\cap\left\{\left|Y_n\right|\leqslant \varepsilon/2\right\}\right)$ is convergent for any positive $\varepsilon$. Now, using the independence assumption and the fact that $X_n$ has the same distribution as $Y_n$, we derive that $$\mathbb P\left(\left\{\left|X_n\right|\geqslant \varepsilon\right\}\cap\left\{\left|Y_n\right|\leqslant \varepsilon/2\right\}\right)=\mathbb P\left(\left\{\left|X_n\right|\geqslant \varepsilon\right\}\right)\mathbb P\left(\left\{\left|X_n\right|\leqslant \varepsilon/2\right\}\right)=\mathbb P\left(\left\{\left|X_n\right|\geqslant \varepsilon\right\}\right)\left(1-\mathbb P\left(\left\{\left|X_n\right|\gt \varepsilon/2\right\}\right) \right).$$ Therefore, it suffices to prove that $X_n\to 0$ in probability. Denoting by $\varphi_n$ the characteristic function of $X_n$ (hence that of $Y_n$), we get that $\varphi_n(t)^2\to 1$ for each $t$ hence $\left|\varphi_n(t)\right|^2\to 1$ for each $t$ which proves that $X_n-Y_n\to 0$ in distribution hence in probability. Therefore, $X_n\to 0$ in probability, which finishes the proof.


Let $\mu_n$ be the common distribution of $X_n$ and $Y_n$.

We have \begin{align}P(|X_n+Y_n| \ge \epsilon) &= \int_{-\infty}^\infty P(|X_n + t| \ge \epsilon) \mu_n(dt)\\ &= \int_{-\infty}^\infty P(X_n \ge \epsilon - t) \mu_n(dt) + \int_{-\infty}^\infty P(- X_n \ge \epsilon +t) \mu_n(dt)\end{align}

We have \begin{align} \int_{-\infty}^\infty P(X_n \ge \epsilon - t) \mu_n(dt) &\ge \int_{0}^{\infty} P(X_n \ge \epsilon - t) \mu_n(dt) \\&\ge P(X_n \ge \epsilon)\int_{0}^{\infty} \mu_n(dt)\\ &= P(X_n \ge \epsilon) P(X_n \ge 0) \end{align} Similarly, we have $$\int_{-\infty}^\infty P(- X_n \ge \epsilon +t) \mu_n(dt)\ge P(X_n\le-\epsilon)P(X_n<0).$$ If I haven't made any mistake. Similar manipulations might be helpful.

  • $\begingroup$ Then how to conclude $\sum_{n}P(|X_n|\geqslant\frac\epsilon2)<\infty$? $\endgroup$ – Eric Yau Dec 13 '16 at 9:12
  • $\begingroup$ @EricYau, what I wrote was not enough. But you should still be able to estimate things... I have updated my answer, hopefully more useful this time. $\endgroup$ – passerby51 Dec 13 '16 at 9:20

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