I am not sure if this is a special case of the gamblers ruin or just a Markov chain problem. A game where we start with M0 amount of money.
With probability 1/2 we win M0/2 and probability 1/2 we lose M0/2. We do this n times, so after the nth time, we gain or lose (Mn−1)/2.
Our total money after n times is, therefore, Mn = Mn−1 ± (Mn−1)/2, with each occurring with probability 1/2.
My first problem is finding the expectation of Mn.
This is what I tried E(Mn)= 1/2 (Mn−1 + (Mn−1)/2)+ 1/2 (Mn−1 - (Mn−1)/2) = Mn-1
This doesn't make sense since Mn-1 also depends on Mn-2 and so on.
I stimulated the problem on Matlab
For any initial values of M0 and n I used, I noticed that It always converges to zero as n gets large. So I guess E(Mn) goes 0? What's the general expression for E(Mn) (My plotted graph is just one realization)
Secondly, we will define Xn = log(Mn/M0), approximate P(Xn > s) fixed s, and large n. We then approx P(M100 > M0/8).
I tried using the central limit theorem for this but I'm stuck. Anything will help, thanks
The expected value of money at nth step is equal to n-1th step and hence equal to n-2th step and so on .. equal to M0 Your calculation seems to be correct. Regarding the expression you mentioned last yes you can use law of large no for finding the probability , at every step the value gets multiplied with 1/2 or with 3/2 with half probability each so at nth step the value is $$M0 * (\frac{1}{2})^{n-r} * (\frac{3}{2})^{r}$$ apply the inequality and apply log on both sides to convert the above multiplication expression into summation expression, the analogy is that a random variable takes values log(3/2) and log(1/2) with half probability each and we are picking n iid random variables whose sum must be greater than or equal to log(1/8)