Calculating the expected gain in St. Petersburg Paradox

Question

I'd like to understand why can't we calculate the expected gain in St. Petersburg paradox as follows.

Let $G$ denote our gain and let $R$ denote the number of rounds that game proceeds. Then we have, $$ G = 2^{R-1} .$$

In this case, can't we calculate our expected gain as, $$ E[G] = 2^{E[R] -1} ?$$

If we have a fair coin, we simply have $ E[R] = 2 $ which gives us $E[G] = 2$.

So, it doesn't make sense to pay more than $2$ according to my calculation. What's wrong with my reasoning?

Answer 1

You can't just take the expectations of $G$ and $R$ and expect the formula to hold. The definition of $E[G]$ is $\sum G\cdot P(G)$, which does not converge. As an example, let us consider a modified game with $n$ rounds. If keep succeeding and get to $n$ rounds you stop and are paid $2^{n-1}$. Now the sums are finite. $E[R]$ doesn't change much. It is reduced below $2$ by a tiny amount. If you do the sum, the expected value is just about $0.5n$, increased slightly because of the higher chance of stopping at $n$.

Answer 2

It is not true in general that $\mathbb Ef(X)=f(\mathbb EX)$ as you seem to think.