Can you prove a strong law of large numbers? A weak law? (Allan Gut, Probability: A graduate Course) Suppose $X$ and $\{X_k, k\geq1\}$ are independent, identically distributed random variables, such that
$$P(X=n)=\frac{1}{n(n-1)},\;\;\text{for}\;n=2,3,...$$
Can you prove a strong law of large numbers? A weak law?
I have no idea about it. Any hint please. Thanks!
 A: If you prove that $\lim_ {n \to \infty} \dfrac{1}{n^2} \sum_{k=1}^{n} Var(X_{k})=0$ then you prove that $\dfrac{\sum_{k=1}^{n} X_{k}}{n} \rightarrow^{P} E(X)$, in this case the weak law.
You can see that $\forall n \geq 2$:
$E(X)=E(X_k)=\dfrac{1}{n-1}$
$E(X^2)=E(X_k^2)=\dfrac{n}{n-1}$
$Var(X)=Var(X_k)=\dfrac{n}{n-1}-(\dfrac{1}{n-1})^2=\dfrac{n(n-1)-1}{(n-1)^2}$
Then;
$\lim_ {n \to \infty} \dfrac{1}{n^2} \sum_{k=1}^{n} Var(X_{k})=\lim_ {n \to \infty}  \dfrac{1}{n^2}  \sum_{k=1}^{n} \dfrac{n(n-1)-1}{(n-1)^2}=\lim_ {n \to \infty} \dfrac{n(n-1)-1}{n(n-1)^2}= 0$
With this:
$P(|\dfrac{\sum_{k=1}^{n} X_{k}}{n}-E(x)|>\epsilon) \leq \dfrac{Var(\dfrac{\sum_{k=1}^{n} X_{k}}{n})}{\epsilon^2}=\dfrac{\dfrac{1}{n^2} \sum_{k=1}^{n} Var(X_{k})}{\epsilon^2}$
If you tends to infinity , you find that:
$\lim_{n \to \infty} P(|\dfrac{\sum_{k=1}^{n} X_{k}}{n}-E(x)|>\epsilon)=0$
Therefore, $\dfrac{\sum_{k=1}^{n} X_{k}}{n} \rightarrow^{P} E(X)$ and verifies weak law.
This is similar for strong law, you have to use almost sure convergence.
