Show that $P(|X_n| > an \text{ for infinitely many } n) = 1$ for any $a > 0$ with $E[|X_n|] = ∞$

Question

I want to check my solutions for this problem. Can someone help me?

Let $(X_n)_{n\geq1}$ be independent and identically distributed real-valued random variables.

Suppose $E[|X_n|] = ∞$.

a. Show that $P(|X_n| > an \text{ for infinitely many } n) = 1$ for any $a > 0$. Hint: Justify $\int_{0}^{\infty} P(|X_1| > at) dt = ∞$ and then use $P(|X_n| > an) > P(|X_n| > at)$ for $t > n$.

My solutions:

First, let's justify $\int_{0}^{\infty} P(|X_1| > at) dt = ∞$.

We know that $E[|X_1|] = ∞ = \int_{0}^{\infty} P(X_1 \geq t) dt$ for positive $X_1$.

So, we have $E[|X_1|] = ∞ = \int_{0}^{\infty} P(X_1 \geq t) dt = \int_{0}^{\infty} P(|X_1| \geq t) dt$.

We also know that $E[|aX_1|] = aE[|X_1|]$ for positive $a$.

Therefore, $aE[|X_1|] = \infty = a \int_{0}^{\infty} P(X_1 \geq t) dt = \int_{0}^{\infty} P(X_1 \geq at) dt = E[|aX_1|] = \infty$.

Now, we can use the hint that $P(|X_n| > an) > P(|X_n| > at)$, and since $\int_{0}^{\infty} P(X_1 \geq at) dt = \infty$, it implies that $P(X_1 \geq at) = 1$. Therefore, $P(X_1 \geq an) = 1$ follows from $P(|X_n| > an) > P(|X_n| > at)$ for $t > n$.

Are my solutions correct?

Answer 1

There are many mistakes in the last part and you did not even prove what was asked.

Integrate $P(|X_n| > an) \geq P(|X_1| > at)$ from $t=n$ to $t=n+1$ and add. You get $\sum_n P(|X_n| > an) \geq \int_0^{\infty} P(|X_1| \geq at)dt=\infty$. The conclusion now follows by the converse part of Borel-Cantelli Lemma.