What a cautious player would pay

Question

Consider the standard uncapped doubling game: a fair coin is flipped until the first tails on flip $k$ , paying $2^k$ dollars. Suppose you value a payout of $m$ dollars by its natural logarithm $\ln(m)$ , so extra dollars matter less and less as the amount grows. Value the game as the single sure payout whose logarithm equals the game's expected log-payout.

What certainty-equivalent dollar value do you assign to the game?

Accepted Answer

First find the expected value of the logarithm of the payout. The payout is 2^k with probability \left(	frac12ight)^k, and \ln(2^k) = k\ln 2, so E[\ln(	ext{payout})] = \sum{k=1}^{\infty}\left(	frac12ight)^k\, k\ln 2 = \ln 2 \sum{k=1}^{\infty} k\left(	frac12ight)^k. The sum \sum{k\ge 1} k x^k = \dfrac{x}{(1-x)^2} equals 2 at x = 	frac12, so E[\ln(	ext{payout})] = 2\ln 2. The certainty-equivalent payout m satisfies \ln(m) = 2\ln 2, hence m = e^{2\ln 2} = 2^2 = 4. A player who values money logarithmically, so that gains shrink in importance as wealth grows, prices this infinite-expec…

What a cautious player would pay

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