4
$\begingroup$

(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!

$\endgroup$
10
  • $\begingroup$ I have a strong feeling information is missing. $\endgroup$ Oct 2, 2020 at 22:14
  • $\begingroup$ 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. $\endgroup$
    – Mark
    Oct 2, 2020 at 22:16
  • $\begingroup$ @MathQED Well, it is posted as an exercise in the book mentioned (Pg 325, Chapter 6) $\endgroup$
    – af13
    Oct 2, 2020 at 22:20
  • $\begingroup$ @Mark any ideas on how to prove them? $\endgroup$
    – af13
    Oct 2, 2020 at 22:30
  • 1
    $\begingroup$ 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?). $\endgroup$
    – Aphelli
    Oct 3, 2020 at 7:03

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .