# Law of large numbers and a product of random variables

A gambler makes a long sequence of bets against a rich friend. The gambler has initial capital C. On each round, a coin is tossed; if the coin comes up tails, he loses 30% of his current capital, but if the coin comes up heads he instead wins 35% of his current capital.

. . .

(c) Consider log$C_n$. What does the law of large numbers tell us about the behaviour of $C_n$ as $n \rightarrow \infty$? How is this consistent with the behaviour of $\mathbb{E}C_n$?

In parts (a) and (b), I've written $C_n$ as the gambler's capital after n rounds with $C_n$ as a product $CY_1Y_2...Y_n$ where $Y_i$ are i.i.d random variables and have calculated $\mathbb{E}C_n$ as $1.025^nC$.

Looking at (c), I can see that log$C_n =$ log$C +$ log$Y_1 + \space ... \space$ log$Y_n$ and that the law of large numbers tells us that $\frac{1}{n}$(log$Y_1 + \space ... \space$ log$Y_n) \rightarrow \mathbb{E}($log$Y_1$).

I don't know what this tells us about the behaviour of $C_n$ and I don't know how this compares to the behaviour of $\mathbb{E}C_n$. If someone was able to explain this to me I'd really appreciate it.

Well, what is $E(\log(Y))$? I compute $$E(\log(Y)) = \frac{1}{2}(\log(1.35) + \log(.7)) \approx -0.03.$$
Then, from what you said about SLLN, we have, almost surely, $\log(C_n)/n \to -0.03$ which means $\log(C_n)\to -\infty$ which means $C_n\to 0$ almost surely.
On the other hand, you've calculated $E(C_n) =1.025^n C \to \infty.$
Here's how I think about it: imagine an ensemble of millions of gamblers each playing this game starting with $\$ 1$. It is perfectly consistent for the following two things to be true with high probability at some large time$n\$ 1) A very small fraction of the gamblers have more that a penny left. 2) The average wealth of the gamblers is more than 100 trillion dollars.