5
$\begingroup$

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 $X_n\stackrel{d}{=}Y_n$.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.