Pricing a Game with Infinite Expected Payoff

Expectation · Medium · Free problem

You play a game where a random number $X$ is drawn uniformly on $(0,1)$, and you are paid

/X$ dollars.

(a) What is the expected payoff of this game?

(b) Despite your answer to (a), how much would you actually be willing to pay to play? What frameworks can you use to arrive at a sensible price when the expected value is infinite?

Hints

  1. Compute $\int_0^1 x^{-1} \, dx$ directly -- what happens near the lower bound?
  2. Think about replacing expected payoff with expected utility. What does log utility do to the
    /x$ singularity?
  3. Consider what changes if $X$ is bounded below by some small $\epsilon > 0$. How does the expected value depend on $\epsilon$?

Worked Solution

How to Think About It: This is a continuous version of the St. Petersburg paradox. The payoff

/X$ blows up as $X \to 0$, so you should suspect the expected value diverges. Before computing anything, think about what "infinite expected value" even means practically. If someone offers you this game, you would not pay a billion dollars for it -- most of the time $X$ lands around 0.5, giving a payoff around 2. The infinite expectation is driven entirely by the tiny-probability, enormous-payoff tail near $X = 0$. So the real question is not "what is the expected value?" but "what is a sensible way to price something when the EV criterion breaks down?" That is what the interviewer actually cares about.

Quick Estimate: Half the time $X > 0.5$, giving payoff below 2. A quarter of the time $X \in (0.25, 0.5)$, giving payoff between 2 and 4. So the median payoff is

/0.5 = 2$, and about 75% of the time you get less than 4. A reasonable person would value this somewhere around $\
$--$\$5$, despite the infinite EV. The tail is extreme but vanishingly rare.

Approach: Compute the expected value directly (Part a), then resolve the paradox with several finite-price frameworks, being careful that "willingness to pay" is only well-defined once you fix a utility function AND the wealth/entry-fee formulation (Part b).

Formal Solution:

*Part (a): Expected payoff.* $E[1/X] = \int_0^1 \frac{1}{x}\, dx = \lim_{\epsilon \to 0^+} [\ln x]_{\epsilon}^{1} = \lim_{\epsilon \to 0^+} (0 - \ln \epsilon) = +\infty.$ The expected payoff is infinite. In a pure expected-value framework you would "rationally" pay any finite amount -- which is exactly why EV is the wrong criterion here.

*Part (b): Pricing when the EV is infinite.* There is no single "actual price": the answer depends on the preferences and the formulation you assume. The useful frameworks:

1. Expected utility (standalone certainty equivalent). Treat the prize itself as terminal wealth and replace EV with expected utility. For a CRRA utility $U(w)=\dfrac{w^{1-\gamma}}{1-\gamma}$, the coefficient of relative risk aversion is the constant $\gamma = -\,wU''(w)/U'(w)$.

- *Log utility* $U(w)=\ln w$ corresponds to $\gamma = 1$: $E[\ln(1/X)] = E[-\ln X] = \int_0^1 (-\ln x)\, dx = \big[x - x\ln x\big]_0^1 = 1, \qquad \text{CE} = e^{1} = e \approx \ .72.$ - *Square-root utility* $U(w)=\sqrt{w}=w^{1/2}$ corresponds to

-\gamma = \tfrac12$, i.e. $\gamma = \tfrac12$: $E[\sqrt{1/X}] = E[X^{-1/2}] = \int_0^1 x^{-1/2}\, dx = \big[2\sqrt{x}\big]_0^1 = 2, \qquad \text{CE} = 2^{2} = \$4.$

The key correction: $\sqrt{w}$ is less risk-averse than $\log w$ (its CRRA coefficient $\tfrac12 < 1$), so it penalizes the payoff's dispersion less and is willing to pay more for the heavy upside. Hence its certainty equivalent is higher ($\$4 > \

.72$), not lower. The general rule "more risk aversion $\Rightarrow$ lower price" is correct, but it runs in the opposite direction from the $\log \to \sqrt{}$ comparison: moving from $\log$ to $\sqrt{}$ *reduces* risk aversion and *raises* the price.

2. Wealth-dependent pricing (the formulation matters). The standalone CE above implicitly assumes the prize is your whole wealth. More carefully, with initial wealth $W$ and an up-front entry fee $p$, the willingness-to-pay solves the indifference condition $E\big[\,U(W - p + 1/X)\,\big] = U(W).$ This $p$ is finite and depends on $W$. For log utility, for example, $p \approx \ .00$ at $W=1$, $\approx \$3.17$ at $W=5$, and $\approx \$5.72$ at $W=100$: a larger wealth buffer lets you absorb the variance and pay more. So the price is not a single number -- it is a function of your preferences and your wealth.

3. Truncation (finite precision). Any real random-number generator has finite precision $\epsilon$, so $X \ge \epsilon$ and $E[\,1/X \mid X \ge \epsilon\,] = \frac{1}{1-\epsilon}\int_\epsilon^1 \frac{dx}{x} = \frac{-\ln \epsilon}{1-\epsilon} \approx -\ln \epsilon.$ For $\epsilon = 10^{-6}$, $E \approx 13.8$; for double precision $\epsilon = 10^{-15}$, $E \approx 34.5$. The implied price grows only logarithmically in precision.

4. Median / robust summary. The median of $X$ is $0.5$, so the median payoff is

/0.5 = \
$ -- a finite, tail-robust summary (not the mean).

5. Kelly / growth-rate. A Kelly bettor maximizes long-run log-growth $g(p) = E\big[\ln(1 + (1/X - p)/W)\big]$, which is just the wealth-based log-utility criterion above; it prescribes a finite stake (small relative to $W$), again giving a finite effective price.

Answer: (a) $E[1/X] = +\infty$. (b) There is no unique price without specifying preferences and the wealth/fee formulation. Standalone certainty equivalents: log utility ($\gamma=1$) gives $e \approx \ .72$; square-root utility ($\gamma=\tfrac12$, which is *less* risk-averse than log) gives $\$4 > \ .72$. The general indifference price solves $E[U(W - p + 1/X)] = U(W)$ and depends on wealth $W$; truncation at precision $\epsilon$ gives $-\ln\epsilon/(1-\epsilon)$; the median payoff is $\ $. The lesson: with a tail this heavy, expected value alone is not a usable decision criterion -- you need a risk-preference or robustness framework, and the resulting price is preference- and wealth-dependent.

Intuition

This problem is a continuous cousin of the classical St. Petersburg paradox, and it illustrates a fundamental tension in quantitative finance: expected value is not always the right objective. The payoff

/X$ has a fat tail near $X = 0$ that is just heavy enough to make the integral diverge logarithmically. But logarithmic divergence is about the mildest form of infinity -- truncate even slightly and the answer becomes modest. That is why the "paradox" feels less paradoxical than it sounds: the game is not practically dangerous, it is just theoretically awkward.

In real trading, you encounter this tension constantly. Any position with unbounded upside (e.g., a short put, a carry trade in emerging markets) has a version of this problem lurking in the tail. The standard fixes -- concave utility, position sizing via Kelly, VaR/CVaR constraints -- all do the same conceptual work: they penalize tail outcomes more than linear expectation does. An interviewer asking this question wants to hear that you understand when EV is a useful guide and when it breaks down, and that you have practical tools (not just philosophical hand-waving) to handle the breakdown.

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