(strong law of large numbers) We played a game in a casino. $X_i$ the money we won or lost the i-th time....

Question

>We played a game in a casino. $X_i$ the money we won or lost the i-th time. Each time that we win, we take 1 dollar. When we lost, we lost 1 dollar. If p is the probability of winning and q the probability of losing, use the strong law of large numbers to prove that the average of which we win $\sum_{i=1}^{n}X_i/n$ after n games, goes to infinity if p $\gt$ q, and to minus infinity if q $\gt$ p or to zero if p = q when we play infinite times (n goes to infinity).

Not sure if i understand.But what did I think.

$X_n = \begin{cases} +1, & p \\[2ex] -1, & q \end{cases}$

If $p = q = p - q = 0 \to \lim_{n\to\infty} \sum_{i=1}^n\frac{X_n}{n} = \lim_{n\to\infty} \sum_{i=1}^n\frac{0}{n}= 0 $

Now for the cases $p \gt q$ and $p \lt q$.

Now for the cases $p \gt q$ and $p \lt q$, I understand that the sum is positive ($p \gt q$) and negative ($p \lt q$), but does it go to plus infinity and minus infinity?

Answer 1

Alright, so my interpretation of this question is simply as follows:

You play a gambling game where there is a probability of p that you win 1 dollar and q that you lose 1 dollar (clearly p+q=1). If you play n times, show that if p>q then your expected earnings as n goes to infinity is infinite, if p<q then your expected loss as n goes to infinity is infinite, and if p=q then you're expected not to gain or lose anything.

If this is the correct interpretation, then the proof is very obvious: every game, you are expected to earn p*(1)+q*(-1)=p-q dollars. Therefore by playing an infinite amount of games you are expected to make infinity*(p-q) dollars. If p>q then the limit is clearly infinity, if p<q then the limit is clearly negative infinity. If p=q note we get the indeterminate form infinity*0, so we can instead take the bottom approach instead:

For all finite n, the expected earnings/losings is 0 dollars. Therefore, since we can choose arbitrarily large N such that the amount of earnings is 0 for all n>N, by the definition of a limit, we know the answer is 0.

Answer 2

We can roughly say that from $\lim_{n \to \infty}{P(|\frac{S_n}{n}-EX|\geq \epsilon)=0}$ where $S_n=X_1+...+X_n$, all of the random variables $X_i$ have the same distribution $X$ we can conclude that $\frac{S_n}{n} \approx EX=p-q>0$ almost surely from where we can derive that $S_n \approx n(p-q) \to +\infty$ as $n \to +\infty$ (If $p>q$ and $p$ is probability of winning 1 dollar. It's completely the same for the opposite case.