Here's the scenario: Suppose we play a game where we start with \$1. On each round of the game you either double or half your money, with equal probability. Fix some $n > 1$, and let $X$ denote the random variable whose value corresponds to the money you have after n rounds. Compute $\mathbb{E}[X]$.
So I started out by looking at what the expected value would be in the first round: $$ (2*.5)+(.5*.5)=5/4 $$ The 2 is one of the two possible outcomes of the game(in this case it's when the money gets doubled) and the .5 just represents the probability of such an outcome occurring. I did the same thing for the other expression in parentheses.
I repeated this for the second and third rounds and got (25/16) and (125/64), respectively. So, from this information, it's pretty easy to see that for n rounds of playing this game, the expected value would just be $(5/4)^n$.
The problem that I am having is that I do not know how to formally prove that this is the case for n rounds. It is intuitive when you look at it, but that's not enough. My teacher said there is a pretty simple inductive proof for the problem, but I do not know how to do it. Any help?
Let $X_n$ be the amount of money you have after $n$ rounds. The conditional distribution $X_{n+1} \mid X_n$ takes values $2X_n$ and $X_n/2$ with equal probability (half). So, $\mathbb{E}[X_{n+1} \mid X_n] = \frac{1}{2} (2X_n) + \frac{1}{2} (X_n/2) = \frac{5}{4} X_n$, and tower rule gives $$\mathbb{E}[X_{n+1}] = \mathbb{E}[\mathbb{E}[X_{n+1} \mid X_n]] = \frac{5}{4} \mathbb{E}[X_n].$$ From the base case $X_0=1$, you get your answer.