Borel-Cantelli Lemma

Many interesting theorem in probability utilizes measure theory, as it often provides some form of one-line argument that allows the proof to go through. Among one of them is the law of large numbers, which informally states that the average of iid samples approaches the mean of the distribution as the sample size grows large.

We would like to be more precise about what the words "approaches" mean. Clearly we cannot replace the word "approaches" with "converges to", as one could be unlucky and consistently draw below or above the mean so that the sample average may never even be close to the true mean. Despite of this, one thing that we do know is that consistently drawing below or above the mean, while not impossible, becomes increasingly unlikely as the sample grows large. This naturally leads to the idea of convergence in probability.

Convergence in probability

A sequence of random variables ${\textstyle X_{n}}$ converges in probability to ${\textstyle X}$ if for any ${\textstyle \epsilon >0}$,

$$\lim _{n\to \infty }\mathbb {P} (|X_{n}-X|>\epsilon )=0.$$

Notice that this is simply convergence in measure in probability space. Equipped with this definition, we can now state the weak law of large numbers: the sample average converges in probability to the mean. The proof technique of this result is basic but unrelated to measure theory, so we will skip it.

Almost sure convergence

A sequence of random variables ${\textstyle X_{n}}$ converges almost surely to ${\textstyle X}$ if

$$\mathbb {P} (\lim _{n\to \infty }X_{n}=X)=1.$$

Note that this is the almost everywhere convergence in probability space, which is a stronger version of convergence as in a finite measure space, convergence almost everywhere implies convergence in measure.

The strong law of large number carries this stronger notion of converges and states: the sample average converges almost surely to the mean. Before the proof, we first introduce a lemma.

Borel Catelli Lemma

Let ${\textstyle (\Omega ,B,\mathbb {P} )}$ be a probability space, and ${\textstyle \{E_{n}\}}$ any sequence of events in ${\textstyle B}$. If ${\textstyle \sum _{i=i}^{\infty }\mathbb {P} (E_{i})<\infty }$, then ${\textstyle \mathbb {P} (\cap _{n=1}^{\infty }\cup _{k\geq n}E_{k})=0}$.

On the other hand, if ${\textstyle \sum _{i=1}^{\infty }\mathbb {P} (E_{i})=\infty }$, then ${\textstyle \mathbb {P} (\cap _{n=1}^{\infty }\cup _{k\geq n}E_{k})=1}$.

Proof.

Fix ${\textstyle \epsilon >0}$, there exists ${\textstyle n}$ sufficiently large such that ${\textstyle \sum _{i=n}^{\infty }\mathbb {P} (E_{i})<\epsilon }$. By union bound we obtain ${\textstyle \mathbb {P} (\cup _{k\geq n}E_{k})<\epsilon }$. So ${\textstyle \lim _{n\to \infty }\mathbb {P} (\cup _{k\geq n}E_{k})=0}$. By continuity of measure , we have ${\textstyle \mathbb {P} (\cap _{n=1}^{\infty }\cup _{k\geq n}E_{k})=\lim _{n\to \infty }\mathbb {P} (\cup _{k\geq n}E_{k})=0}$.

The other statement follows easily from the similar technique.

Strong law of large numbers

To be finished...