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 a strong feeling information is missing. Oct 2, 2020 at 22:14
• If the random variables are iid and the expectation exists (even if it is infinite, like in this example) then both the strong and weak laws hold. This is a very nontrivial theorem though.
– Mark
Oct 2, 2020 at 22:16
• @MathQED Well, it is posted as an exercise in the book mentioned (Pg 325, Chapter 6)
– af13
Oct 2, 2020 at 22:20
• @Mark any ideas on how to prove them?
– af13
Oct 2, 2020 at 22:30
• Well, you know that almost surely, for all integers $p > 0$, $(X_1+\ldots+X_n) \geq n(1+o(1))p$, thus if $n$ is large enough $(X_1+\ldots+X_n)/n \geq p-1$. So $(X_1+\ldots+X_n)/n$ goes to infinity and we’re done (or aren’t we?). Oct 3, 2020 at 7:03

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.