I truly don't know how to name this question properly, and I will try my best to be more specific. And here is my problem
Suppose a person has 2 choices. For choice A, he gets $Z$ guaranteed dollars; for choice B, Flip a fair coin repeatedly until the time T when the first tail appears. The payout is then $P(T) = 2^{T-1}$. Please note that $T-1$ is the number of heads but not number of flips. The Utility function of this person is $U(x) = x^c$. So for what value of $Z$ and $c$, will $E[U(B)] \gt E[U(A)]$ ?
Below is my attempt
The utility of get Z dollars is $E[U(Z)] = z^c$
Because it is fair coin, pmf of T is $f_T(t) = p(1-p)^{t-1}$, as here p = 0.5, $f_T(t) = (\frac{1}{2})^t$
$E[U(P(T))] = \sum_{t = 1}^{\infty} f_T(t) U(P(T = t)) = \sum_{t = 1}^{\infty} (\frac{1}{2})^t (2^{t-1})^c$
$E[U(P(T))] = \sum_{t = 1}^{\infty} 2^{tc - t - c}$
However, I am stuck here for a long time. Could someone tell me what to do next? It seems I should now compare the 2 expectations, but I am unsure of that. Any help or hint is welcome.
If $c \ge 1$ then the terms in the sum you obtained for case $B$ monotonically increase or remain constant. Hence the sum diverges, and $E(B)$ is infinitely large. Therefore one should always prefer $B$ over $A$ in this case.
If $c < 1$ then the sum you obtained for case $B$ can be evaluated analytically, and the result is: $$E(B) = \frac {1/2} {1 - (1/2)^{1-c}}$$
You should then compare this expectation value with $E(A) = Z^c$ and choose the larger of the two.