Let $Y_1,Y_2,\dots$ iid r.v. Show that

  1. $\dfrac{|Y_n|}{n}\longrightarrow 0$ in probability.
  2. $\dfrac{|Y_n|}{n}\longrightarrow 0$ a.s. iff $\mathbb E\left[|Y_1|\right]<\infty .$


  1. Let $\varepsilon>0$. Then, $$\mathbb P\left(\frac{|Y_n|}{n}>\varepsilon\right)=\mathbb P(|Y_n|>n\varepsilon)$$ Since $\{|Y_{n+1}|>(n+1)\varepsilon\}\subset \{|Y_n|>n\varepsilon\}$, we have that $$\lim_{n\to \infty }\mathbb P(|Y_n|>n\varepsilon)=\mathbb P(Y_\infty =\infty ),$$ Q1) Is $\mathbb P(Y_\infty =\infty )=0$ ? But what would be the sense of $Y_\infty $ ?

  2. I think I have to use the fact that $$\mathbb E[|Y_i|]=\int_0^\infty \mathbb P\{|Y_i|>y\}dy,$$ then $\displaystyle\int |Y_i|d\mathbb P<\infty,$ but I don't see how. Q2) any idea?

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    $\begingroup$ Re 1, you lost me when asserting that $$\{|Y_{n+1}|>(n+1)\varepsilon\}\subset \{|Y_n|>n\varepsilon\}$$ Why should it be so? $\endgroup$ – Did Nov 2 '16 at 12:49
  • $\begingroup$ @Did: You right, there is no reason. Sorry ! Any idea ? $\endgroup$ – user380364 Nov 2 '16 at 13:06
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    $\begingroup$ Sure: $P(|Y_n|>n\epsilon)=P(|Y_1|>n\epsilon)$ hence $\sum\limits_nP(|Y_n|>n\epsilon)=\sum\limits_nP(|Y_1|>n\epsilon)\leqslant\ldots$ $\endgroup$ – Did Nov 2 '16 at 13:39
  • $\begingroup$ @Did Are you posting only comments to give chance to others for reputation? salutes $\endgroup$ – BCLC Nov 2 '16 at 14:03
  • $\begingroup$ What's $$Y_{\infty}$$? $\endgroup$ – BCLC Nov 2 '16 at 14:12

For Question 1: The set inclusion you mentioned is incorrect and not used in the proof. And I don't know what you mean by $Y_\infty$. But since $Y_n$ are independent identically distributed, we may safely say $\mathbb{P}(|Y_n|>n\varepsilon)=\mathbb{P}(|Y_1|>n\varepsilon)$, which tends to zero obviously.

For Question 2: \begin{align}\mathbb{P}(|Y_n|/n\to0)&=\lim_{\varepsilon\to 0}\mathbb{P}\left(\bigcup_{N=1}^\infty\bigcap_{n\ge N}\{|Y_n|/n<\varepsilon\}\right)\\ &=\lim_{\varepsilon\to 0}\lim_{N\to\infty}\prod_{n\ge N}\mathbb{P}(|Y_n|< n\varepsilon)\\ &=\lim_{\varepsilon\to 0}\lim_{N\to\infty}\prod_{n\ge N}(1-\mathbb{P}(|Y_1|\ge n\varepsilon)) \end{align} So the almost surely convergence of $|Y_n|/n$ is equivalent to the convergence of the infinite product $\prod (1-\mathbb{P}(|Y_1|\ge n\varepsilon))$, the latter known to be equivalent to the convergence of series $\sum\mathbb{P}(|Y_1|\ge n\varepsilon)$. Now note that $$\mathbb{E}(|Y_1|)=\int |Y_1|d\mathbb{P}=\sum_{n=0}^\infty\int1_{n\varepsilon\le|Y_i|<(n+1)\varepsilon} |Y_1|d\mathbb{P}$$ Hence $$\mathbb{E}(|Y_1|)\ge\sum_{n=0}^\infty n\varepsilon\mathbb{P}(n\varepsilon\le|Y_i|<(n+1)\varepsilon)$$ $$\mathbb{E}(|Y_1|)\le\sum_{n=0}^\infty (n+1)\varepsilon\mathbb{P}(n\varepsilon\le|Y_i|<(n+1)\varepsilon)$$ Since $\mathbb{P}(n\varepsilon\le|Y_i|<(n+1)\varepsilon)=\mathbb{P}(|Y_i|\ge n\varepsilon)-\mathbb{P}(|Y_i|\ge(n+1)\varepsilon)$, summation by parts yields the desired result.

  • $\begingroup$ $$Y_{\infty} := \lim Y_n ?$$ $\endgroup$ – BCLC Nov 2 '16 at 14:03
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    $\begingroup$ @BCLC It is not necessarily defined then...There is no evidence that $Y_n$ converges a.s. or in probability :( $\endgroup$ – Cave Johnson Nov 2 '16 at 14:06

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