Example 6.3 Suppose that we flip a coin until a tail first appears and if the number of tosses equals $k$, then we are paid $2^k$ dollars. What is $E[2^k]$?
I believe $E[2^k] = \sum_{k=1}^{\infty} 2^k * \frac{1}{2} ^k$, but what would $E[U(2^k)]$? where U is exponential utility.
Would it be $E[U(2^k)] = -e ^{-\gamma \sum_{k=1}^{\infty} 2^k * \frac{1}{2} ^k}$? If this is right, is there a way simplify $\sum_{k=1}^{\infty} 2^k * \frac{1}{2} ^k$ so that we only have $\gamma$ in our answer?
The expected utility should be calculated as
$$E[U(w)]=\sum_{k=1}^\infty p_kU(w_k)$$
where $p_k$ is the probability of wealth $w_k$ and $U$ is the utility function over wealth. In your example, $w_k=2^k$ and $p_k=\frac{1}{2^k}$. For the exponential case, $U(w)=-e^{-\gamma w}$.
Your expression is
$$-e ^{-\gamma \sum_{k=1}^{\infty} 2^k * \frac{1}{2} ^k}=\Pi_{k=1}^\infty\left[-e^{-\gamma 2^k\frac{1}{2^k}}\right]=\Pi_{k=1}^\infty U(p_kw_k)$$