Is this proof of almost sure convergence of a sequence of random variables correct?

Question

Let $ X_1,X_2,...,X_n $ be independent and identically distributed random variables with $ E|X_i| < \infty $. Show than $ \frac{X_n}{n} \to 0 $ almost surely.

My attempt

$$ \infty \gt E|X_i| = \int_{-\infty}^\infty |x|f(x)dx \ge \int_{-\infty}^\infty xf(x)dx = E(X_i) $$

Since $ E(X_i) \lt \infty $ and $ X_i $ are iid $ \implies \frac{X_n}{n} \to 0 $ almost surely.

My question is with the finite mean and the independent and identical distribution. I see these being used as criteria to prove almost sure convergence in my lecture notes, but my question is, is this a well established theorem, or should I have to prove this? I can't find any proofs of it in my textbook or online. My attempt above seems almost too easy for this exercise.

You claim to have shown $E(X_n)$ is finite but that in itself is not a proof that $\frac{X_n}{n} \to 0$ almost surely. — Henry, Jan 18 '23 at 23:39
You have not given any proof at all. Also, not all r.v.'s have densities. — geetha290krm, Jan 18 '23 at 23:41

score 3 · Answer 1 · edited Jan 19 '23 at 16:52

3

We have $\sum P(|\frac {X_n} n| >\epsilon)=\sum P(\frac {X_1} {\epsilon} >n)<\infty$ because $E|\frac {X_1} {\epsilon}| <\infty$. By Borel-Cantelli Lemma the result follows.

edited Jan 19 '23 at 16:52

Matija

3,526

answered Jan 18 '23 at 23:43

geetha290krm

36,632

Is this proof of almost sure convergence of a sequence of random variables correct?

My attempt

1 Answers1