# Convergence in probability implies convergence almost surely

Let $$(X_n)$$ be a sequence of independent and positive random variables. Let $$S_n = X_1 + \dots + X_n$$. Show that $$S_n$$ converges almost surely if and only if it converges in probability.

The first implication is well-known. But the other one is none trivial as this exercise is an instance of a statement that is generally false.

The way I think I should organize my proof is as follows:

1. Define $$A_k = \{\lvert S_k - S \rvert \geq \epsilon \}$$.
2. Majorize $$\mathbb{P}(\lvert S_k - S \rvert \geq \epsilon)$$ using Markov inequality.
3. Show that the upper-bounding general term series is convergent.
4. Conclude that $$\sum \mathbb{P}(A_k) < \infty$$
5. Conclude by Borel-Cantelli lemma, $$\mathbb{P}(\limsup A_k) \rightarrow_k 0$$.

However, I am stuck at:

$$\mathbb{P}(\lvert S_k - S \rvert \geq \epsilon) \leq \frac{\mathbb{E}[\lvert S_k - S \rvert]}{\epsilon} \cdot$$

So far I haven't used the fact that $$S_n = X_1 + ... + X_n$$ nor that $$(X_n)$$ is a sequence of positive variables.

## 1 Answer

Independence is not required at all. $$(S_n)$$ is an increasing sequence of real numbers, so $$\lim S_n$$ exists a.s. (The limit may be $$\infty$$). But convergence in probability implies convergence a.s. for a subsequence from which it follows that $$\lim S_n$$ is finite a.s..