Law of large numbers

The weak law of large numbers:

A sequence X₁, X₂, ... of independent identically distributed random variables with finite mean m and finite variance is given. Then the sequence S_n = (X₁ + X₂ + ...+ X_n)/n, for n = 1, 2, ... satisfies the following:
For any e > 0

P(|S_n - m| > e) → 0 as n → ∞.

The strong law of large numbers:

Under the same conditions as above S_n satisfies

P(S_n → m as n → ∞) = 1.


The two theorems look very similar, and sort of say that as you perform more and more experiments the mean of the sample will get closer to the expectation. They say it in different ways though, and getting your head round the difference can be tricky.
Roughly speaking SLLN says something about the behaviour of the particular sequences S₁, S₂, ... once we have decided the values for X₁, X₂, ..., while WLLN rather tells you something about the the probability distribution of S_n at different values of n.

The difference may be easier to understand if I give an example of a random sequence T₁, T₂, ... that satisfies the result of WLLN but not SLLN.
Let Y₁, Y₂, ... be sequence of independent random variables, such that Y_k is uniformly distributed on the set {2^k, 2^k+1, ..., 2^k+1-1}. Define T_n = 1 if n = Y_k for some k and T_n = 0 otherwise.
For T_n thus defined we see that

P(T_n = 0) = 2^-k

for 2^k ≤ n < 2^k+1, but that T_n always contains an infinite number of both ones and zeroes. Thus T_n satisfies the result of WLLN (with m = 0), but not the result of SLLN.
It follows that SLLN cannot be derived from WLLN. A shame really, since WLLN can be proved easily using Chebyshev's inequality, while SLLN is the result that we are really interested in.

It is, however, possible to derive WLLN from SLLN, which we might as well do while we're at it.
Suppose that a random sequence S_n satisfies the conditions of SLLN (with m = 0 for simplicity).
Given e > 0 let E_N be the event that |S_n| < e for n ≥ N. We note that E_N ⊆ E_N+1 for all N and UNION(N = 0, ∞)(E_N) ⊇ {S_n -> 0 as n -> ∞}, so lim_N→∞P(E_N) ≥ P(S_n → 0 as n → ∞) = 1 (this is a basic result of measure theory). Also {|S_n| < e} ⊇ E_n for all n, so P(|S_n| < e) ≥ P(E_n).
Hence

lim_n→∞P(|S_n| < e) ≥ lim_n→∞P(E_n) ≥ 1

So S_n satisfies the result of WLLN with m = 0.

Thus we have shown that (unsurprsingly) the strong law of large numbers is a stronger result than the weak law of large numbers. (In the language of measure theory we have shown that if X_n → X almost surely then X_n → X in probability, but that the converse does not necessarily hold.)