Strong Law of Large Numbers

With only finite mean we can prove almost sure convergene to mean.

proof)

Let $Y_k = X_k 1(|X_k|\leq k )$. $T_n = Y_1 + \dots + Y_n$.  It is enough to show $T_n/n \rightarrow \mu$ a.s.

$\sum\limits^{\infty}_0 P( |X_k| > k ) \leq \int^{\infty}_0 P( |X_1|>t) dt = E|X_1|<\infty$ Therefore 

$P(X_k \neq Y_k \;i.o)=0$ which means that almost surely $X_k=Y_k$ for all $k\geq k(w)$.

Therefore $\sup_k |S_k(w)-T_k(w) | < \infty$ which leads to $ \frac{S_n-T_n}{n} \rightarrow 0 $ a.s

그리고 이제 $k(n)= [\alpha^n] \quad \alpha >1$을 생각한다. 

$\sum\limits^{\infty}_{n=1} P ( |T_{k(n)} - E T_{k(n)} |  > \epsilon k(n) )\leq \sum\limits^{\infty}_{n=1} \frac{E|T_{k(n)} -ET_{k(n)} |^2}{ (\epsilon k(n))^2 } = \frac{1}{\epsilon^2} \sum^{\infty}_{n=1} \frac{1}{k(n)^2} \sum\limits^{k(n)}_{i=1} Var(Y_i) <\infty$ 

따라서 $P(|T_{k(n)} - E T_{k(n)}| > \epsilon k(n) \; i.o)=0$ 이고 

$$\frac{T_{k(n)} - E T_{k(n)} }{k(n)}\rightarrow 0 \quad \text{a.s}$$

Dominated Convergence theorem 에 의해 $EY_k \rightarrow EX_1$이고 따라서 $T_{k(n)}/k(n) \rightarrow EX_1$ almost surley이다.