A fair coin is tossed until heads turns up for the first time. If heads turns up for the first time on the nth toss you receive $2^n$ dollars.
(a) Show that the expected value of your winnings does not exist (i.e., is given by a divergent sum) for this game. Does this mean that this game is favorable no matter how much you pay to play it?
I think $E(X)=\sum_{i=1}^{\infty} 2^n(1/2)^n = \infty$. I'm not sure how to answer the second part of the problem, I think the answer is no but I can't think of a reason why.
(b) Assume that you only receive $2^{10}$ dollars if any number greater than or equal to ten tosses are required to obtain the first head. Show that your expected value for this modified game is finite and find its value.
$E(X)=\sum_{i=1}^{9} 2^n(1/2)^n + \sum_{i=10}^{\infty} 2^{10} (1/2)^n $
$E(X)= 9 + 2^{10}(\sum_{i=1}^{\infty} (1/2)^n) - \sum_{i=1}^{9} (1/2)^n)=9+2^{10}(1-0.998047)=11$
(d) Now assume that the utility of $n$ dollars is $\sqrt{n}$. Write an expression for the expected utility of the payment, and show that this expression has a finite value. Estimate this value. Repeat this exercise for the case that the utility function is $log(n)$.
I'm really confused about what this question is asking, do I replace $2^n$ in the above with $\sqrt{n}$ and $log(n)$?
Edit:
if I sub in $\sqrt{2}^n$ for $2^n$
$E(X)=\sum_{i=1}^{\infty} \sqrt{2}^n(1/2)^n = \sum_{i=1}^{\infty} (\sqrt{2}/2)^n = \frac{\sqrt{2}/2}{1-\sqrt{2}/2}=2.414$
if I sub in $log(2^n)$
$E(X)=\sum_{i=1}^{\infty} log(2^n)(1/2)^n = log(2)\sum_{i=1}^{\infty}n(1/2)^n$ = 2log(2)$
Is any of the above correct?