I play the following game using a coin that lands heads with probability $p$. I start with $X_0$ = $1$ and at each stage I gamble all I have on the toss of the coin. If it lands heads I end up with twice what I started with; if it lands tails I lose everything. All coin tosses are independent.

(1) With $X_n$ denoting how much money I have after the nth toss, find $E[X_{n+1}|X_n = k]$ in terms of k. (2) Find $E[X_n]$ for all $n$.

My attempt at solution:

(1) If we know that $X_n = k$, and the $n+1$ toss will be either H or T with probability $p$ and $1-p$ respectively, that means Xn+1 will be either 2k or 0 respectively. So $$E[X_{n+1}|X_n = k] = (p)(2k) + (1-p)(0) = 2pk$$

(2) Not sure if I have the right thought process but $E[X_n] = 2p(X_{n-1})$ because there's a $p$ probability of doubling $X_{n-1}$ and $1-p$ probability of getting $0$. So using the same logic as in (1), it would be $2p(X_{n-1})$.

I think my reasoning for (2) is correct, but I'm not sure if that's the final answer.


1 Answer 1


Your both answers are correct (ignoring the abuse of notation in the second part that $E[X_n]$ cannot be equal to $2p X_{n-1}$ since $X_{n-1}$ is random, but $E[X_n]$ is not).

For (1), you've already written the correct solution. For (2), we'll just use Law of Iterated (Total) Expectation: $$E[X_n]=E[E[X_n|X_{n-1}]]=E[2pX_{n-1}]=2pE[X_{n-1}]$$ Going towards $n=0$, we have: $$E[X_n]=2pE[X_{n-1}]=(2p)^2E[X_{n-2}]=\ ...\ =(2p)^nE[X_0]=(2p)^n$$

I can suggest another solution for the second part by the way:

In $n$-th toss, you'll have either $2^n$ dollars or $0$ dollars. And, you'll get $2^n$ only if your all tosses are success, i.e. with probability $p^n$. In any other case, i.e. $1-p^n$, you'll get $0$ dollars. So, the expectation will be $p^n2^n+(1-p^n)0=(2p)^n$.


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