I was trying to solve the following question from Larry Wasserman book named "All of statistics":
Suppose we play a game where we start with c dollars. On each play of
the game you either double or halve your money, with equal probability.
What is your expected fortune after n trials?
I saw in this website that people are telling the correct answer is: $(5/4)^n * c$. But I feel that this answer is suspicious. Here is my solution:
Suppose that $X$ is the random variable corresponding to the game result in a single try:
$X = 1$ if win and $X = -1$ if lose
We define another random variable, like $S$, which is the sum of Xs, after n tries:
S = $\sum_{i=0}^{n}X_i$
the amount of money after n trials is:
$2^S.c \Rightarrow E(money) = 2^{E(S)}.c$
Now, we just need to calculate the expected value of S, $E(S)$:
$E(S) = \sum_{i=0}^{n}S_i.f(S_i)$ where $f(S_i)$ is the pdf of $S$. We know that:
$S \sim Binomial(n, p=0.5)$
So the expected value will be:
$ \begin{split} E(S) &= \sum_{i=0}^{n}\binom{n}{i}0.5^i.0.5^{n-i}.((X=1).i + (X= -1).(n-i)) \\&= \sum_{i=0}^{n}\binom{n}{i}0.5^i.0.5^{n-i}.(i + i - n) \\&= \sum_{i=0}^{n}\binom{n}{i}.(0.5^n).(2i - n) \\&= 0.5^n(2\sum_{i=0}^{n}i\binom{n}{i} - n\sum_{i=0}^{n}\binom{n}{i}) \\&= 0.5^n(2n2^{n-1} - n2^{n}) \\&= 0.5^n(n2^{n} - n2^{n}) \\&= 0 \\ \Rightarrow E(S) &= 0 \Rightarrow 2^{E(S)}c = 2^0c = c \end{split} $
So, all of a sudden, it seems that the expected value of the money after n trials, is the initial value of it. I want to see, where I am wrong in this solution. Because all of the steps seems sound to me. Thanks.
the term (X=1).i+(X=-1). (n-i) should be replaced with (2^i)/(2^n) - (2^(n-i))/(2^n), which corresponds to the probability-weighted outcomes for each i. Because when you win, you double your money (which is equivalent to multiplying by 2), and when you lose, you halve your money (which is equivalent to multiplying by 1/2).
probably try after making this change. Otherwise, it looks good to me