resolving expected utility of st. petersburg paradox with logarithmic utility

Question

St. Petersburg paradox is a game where you toss a fair coin repeatedly and if it lands heads on the $k$th trial you get $2^n$ dollars. Expected utility of game is:

$E(U) = \sum_{k=1}^{\infty}[0.5*0 + 0.5*2^n] = \infty$

the question is how much to pay to enter this game and how to make it so expected utility isn't infinity? some texts say that if initial wealth of person taken into account and money is valued logarithmically, the paradox of infinite expected utility goes away. I don't see this. assume initial wealth of player is $m$, and you pay $e$ amount to enter the game, and any amount of money $d$ is valued as $\log_2(d)$. then the new utility $E'(U)$ from playing the game is:

$E'(U) = \sum_{k=1}^{\infty}\frac{1}{2^{k}}\log_2(m - e + 2^k)$

isn't this still $\infty$, just as in original $E(U)$? the term inside $\log_2$ is still increasing so how does having the $\log_2$ make the paradox go away?

Answer 1

Your initial computation is a little confused: it should read

$$\sum_{k=1}^{\infty} \frac{2^k}{2^k} = \infty.$$

The new computation should read (let's keep things simple)

$$\sum_{k=1}^{\infty} \frac{k}{2^k}$$

and this infinite series, unlike the previous one, now converges to a finite value (in fact it converges to $2$) by one of several convergence tests for infinite series, such as the ratio test. The basic fact here on which tests like the ratio test are built is that geometric series with common ratio between $-1$ and $1$ converge.

However, this does not resolve the St. Petersburg paradox. You can bring back the paradox in full force by increasing the payoff after $k$ heads to $2^{2^k}$ dollars. In order to prevent arbitrary increases in payoff from making the expected utility infinite like this you are forced to assume something much more drastic, which is that utility is bounded.

Answer 2

Let's ignore the constants $m$ and $e$ for now, to simplify the expression

The original paradox assumes that utility is directly (linearly) proportional to the money $d$ as $U(d) = d$ (up to a constant). This gives

$$E(U) = \sum\limits_{k=1}^{\infty} \frac{1}{2^k} \cdot 2^k = 1 + 1 + 1 + \cdots = \infty$$

The main problem is that we don't value money like this; if we imagine receiving $10$ trillion dollars one at a time, the $10$ trillionth dollar is much less valuable than the first one we get.

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While our actual utility functions are very unlikely to be precisely logarithmic, they probably look something like it. If $U(d) = \log_2(d)$, then we get

$$E'(U) = \sum\limits_{k=1}^{\infty} \frac{1}{2^k} \cdot \log_2{2^k}$$

but $\log_2{2^k}$ is just $k$, so

$$E'(U) = \sum \limits_{k=1}^{\infty} \frac{k}{2^k}$$

This is an arithmetico-geometric series, and is convergent.

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To evaluate the sum, we can do a little trick:

$$E'(U) = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \cdots$$

$$\frac{E'(U)}{2} = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \cdots $$

The difference of the above two equations is

$$E'(U) - \frac{E'(U)}{2} = \frac{E'(U)}{2} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \sum\limits_{k=1}^{\infty} \frac{1}{2^k} = 1$$

So

$$E'(U) = 2$$

which is considerably less than infinity :)