# Lower and upper bounds for expected value

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.

• What are the possible values of $X_1$? Are they positive integers? nonnegative integers? nonnegative real numbers? real numbers? What exactly is $N$? Nov 18, 2019 at 23:21
• @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). Nov 18, 2019 at 23:31
• Actually, $X_1$ is a non-negative random variable Nov 18, 2019 at 23:37
• math.stackexchange.com/questions/3399424/… Nov 18, 2019 at 23:43
• 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/… Nov 19, 2019 at 3:11

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.
• 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. Jun 17, 2020 at 13:54