Let $X_n$ be a sequence of iid RVs with $E|X_n|=+\infty$. Show that for every positive number $A$, $P[|X_n|>nA, \; \mbox{i.o.}]=1$ and $P[|S_n|<nA,\;\mbox{ i.o.}]=1$.

Here, $S_n=\sum_{k=1}^nX_k, k=1,2,\ldots$ and i.o. means "infinitely often," i.e. $P(\limsup_{n\to\infty}A_n)=P(A_n,\; \mbox{i.o.})$.

The only place I've ever seen the i.o. terminology used is the Borel-Cantelli Lemma, so I tried to apply this portion of the lemma:

If {$A_n$} is an independent sequence of events such that $\sum_{n=1}^\infty PA_n=\infty$, then $PA=P(A_n,\;\mbox{i.o.})=1$.

Is there a way to apply that here?

  • $\begingroup$ I think the second statement should be $P(|S_n|>nA,\;\mbox{i.o.}) = 1$ $\endgroup$ – spaceisdarkgreen Jan 21 '17 at 0:29
  • $\begingroup$ @spaceisdarkgreen I've checked two editions of the book and both have the equality the same way. But if that is indeed wrong, how would the one you say is correct follow from the first? $\endgroup$ – user408723 Jan 21 '17 at 1:02
  • $\begingroup$ Well, do you disagree with the other answerer's reasoning? I edited my answer to add the proof of the other statement. $\endgroup$ – spaceisdarkgreen Jan 21 '17 at 1:50
  1. \begin{align*} \infty &= E [ |X_1| ] \\& = \int_0^\infty P(|X_1|>t) dt \\ & \le A+\sum_{k=1}^\infty A P(|X_1|>k A)\\ & =A + A\sum_{k=1}^\infty P(|X_k|>k A). \end{align*} Now use Borel-Cantelli (II) to obtain $P(|X_k|>k A\mbox{ i.o.})=1$.

  2. The second statement is wrong, at least as expressed here (I apologize if I misunderstood). Take $X_1$ be any nonnegative random variable taking values in $[1,\infty)$ with infinite expectation (e.g. density $c x^{-2}$ for $x\ge 1$). Then $S_n \ge \frac{1}{2}n$ for all $n$, so statement fails for $A=\frac 12$ (or any $A\in (0,1)$ for that matter).

  • $\begingroup$ The statement was typed exactly as the book wrote it. I just double-checked. It's supposed to be true but I'm not sure why. $\endgroup$ – user408723 Jan 21 '17 at 0:00

Yes, there is. We can write the inequality $$ \frac{|X_n|}{n} \le 1+\sum_{i=1}^\infty I\left(\frac{|X_n|}{n}\ge i\right)$$ where $I$ is the indicator function. This makes sense if you stare at it for awhile... the RHS just adds one until it hits $|X_n|/n.$

Then take the expected value of both sides to get $$\frac{1}{n}E(|X_n|) \le 1 +\sum_{i=1}^\infty P(|X_n|\ge ni)$$ which implies $$ \sum_{i=1}^\infty P(|X_n|\ge ni) = \infty$$ Since the $X$'s are IID, this means we can replace $X_n$ with $X_i$ and write $$ \sum_{i=1}^\infty P(|X_i|\ge ni) = \infty.$$

So you can apply the Borel-Cantelli lemma and conclude that $P(|X_i| > ni,\;\mbox{i.o})=1$ is true for any positive integer $n$. Since $P(|X_i|>ni)$ is decreasing in $n$, it follow that it holds true for any real $n>0$.


As FnaCool showed, the statement that $P(|S_n|<nA,\;\mbox{i.o.})=1$ for all $A>0$ is wrong. However, it is true that $P(|S_n|>nA,\;\mbox{i.o.})=1.$ This follows from the first statement $P(|X_n|>nA,\;\mbox{i.o.})=1$ $\forall A>0.$

To see how, note the first statement is equivalent to $$\limsup_n \frac{|X_n|}{n} = \infty$$ almost surely, and the second is $$\limsup_n \frac{|S_n|}{n} = \infty$$ almost surely.

To prove the second from the first, observe $$ \frac{|X_n|}{n} = \frac{|S_n-S_{n-1}|}{n} \le \frac{|S_n|}{n}+\frac{|S_{n-1}|}{n}$$ so that $$\limsup_n \frac{|S_n|}{n} \ge \frac{1}{2}\limsup_n \frac{|X_n|}{n} $$

  • $\begingroup$ Thank you for the explanation. Without it the indicator function part would have looked foreign to me. Is there a way to apply it to the second one? $\endgroup$ – user408723 Jan 20 '17 at 23:59
  • $\begingroup$ @user408723 The second statement is wrong, as the other answerer noted. If you reverse the inequality (as I commented) then it's true and it can be proven from the first one. $\endgroup$ – spaceisdarkgreen Jan 21 '17 at 0:33

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