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Let $(X_n)_{n \geq 1}$ be a sequence of non-negative i.i.d random variables.

Show that $$E[X_1]\leq \sum_{n \in N} P(X_1 > n) $$

Further, show that if $E[X_1]=\infty$ then $P(X_n > n, i.o) = 1$ and $\frac{S_n} {n}$ converges almost always to $\infty$

I think the second part follows from the strong law of large numbers but I don't know how to proof the 1st one. I tried using Markov's inequality but it didn't seem to work.

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  • $\begingroup$ What are the possible values of $X_1$? Are they positive integers? nonnegative integers? nonnegative real numbers? real numbers? What exactly is $N$? $\endgroup$ Nov 18, 2019 at 23:21
  • $\begingroup$ @GregMartin N is the natural numbers and the problem says X1 is any random variable (most likely real valued since this is chapter centers on continuous probability). $\endgroup$ Nov 18, 2019 at 23:31
  • $\begingroup$ Actually, $X_1$ is a non-negative random variable $\endgroup$ Nov 18, 2019 at 23:37
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    $\begingroup$ math.stackexchange.com/questions/3399424/… $\endgroup$
    – Robert W.
    Nov 18, 2019 at 23:43
  • $\begingroup$ The first part follows from the answer that @RobertW. linked to. For the other part, note that $\{X_n>n, \mathrm{i.o.}\}$ is a tail event, and so either has probability zero or one. Also see the discussion here: math.stackexchange.com/questions/1644218/… $\endgroup$
    – Math1000
    Nov 19, 2019 at 3:11

1 Answer 1

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The first part was already answered by @RobertW.

The second part directly follows from the second Borel-Cantelli lemma since $ \sum_{n \in N} P(X_n > n) = \sum_{n \in N} P(X_1 > n) \ge E[X_1]=\infty.$

For the third part regarding $\frac{S_n}{n}$, you can use Kronecker's lemma to create a contradiction. Suppose $\frac{S_n} {n}$ is finite for some value $\omega \in \{x_n > n\ \ i.o.\}$. Then we can write $\lim_{n\rightarrow \infty}\sum_{m=1}^n\frac{x_m(\omega)}{n}=s$ where $s$ is a real value. By Kronecker's Lemma, it follows that $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{m=1}^nm\frac{x_m(\omega)}{n}=0.$ But this is a contradiction since $ \frac{1}{n}\sum_{m=1}^nm\frac{x_m(\omega)}{n}\ge \frac{1}{n}n\frac{x_n(\omega)}{n}=\frac{x_n(\omega)}{n} > 1 $ i.o.

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  • $\begingroup$ I don't think this proof is quite complete. The third part seems to prove that we cannot have $S_n/n$ converging to a finite limit with positive probability. But we still need to rule out the possibility that it fails to converge at all. In other words, we have shown $\limsup S_n/n = +\infty$ a.s., but we haven't shown that the liminf is a.s. infinite also. $\endgroup$ Jun 17, 2020 at 13:54

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