Certainty equivalent: U(C)=E[U(x)] -> find C (certainty equivalent)
4 Utility Theory. The following game is proposed to you. A fair coin will be tossed repeatedly until a tail appears (the first tail will end the game). The pot starts at £1 and is doubled every time a head appears. You win whatever is in the pot when the game ends. Thus, you win £1 if a tail appears on the first toss, £2 if on the second, £4 if on the third, £8 if on the fourth, etc. In short, you win £2-1 if the coin is tossed k times until the first tail appears. A reasonable way to analyse the value of this game is by calculating the certainty equivalent of the game’s random payoff. Calculate the expected utility and the certainty equivalent of the game in the following situations:
a risk-neutral utility function: u(x) = x (hint: note that the probability of the first tail appearing in the k-th toss is )
b power utility function: u(x) = bx with b = -1;
c logarithmic utility function: u(x) = log2 (x) (hint: use the formula Dk-1 (k-1)/2 = 1).
The three parts carry, respectively, 40%, 30%, and 30% of the marks.
The Correct Answer and Explanation is:
$$
U(C) = E[U(X)]
$$
where $X$ is the random payoff of the game.
Game description:
- The game ends the first time a tail appears.
- Probability that the first tail appears on toss $k$ is:
$$
P(k) = \left(\frac{1}{2}\right)^k
$$
(because the first $k-1$ tosses are heads and the $k$-th toss is tail, probability $(1/2)^{k-1} \times (1/2) = (1/2)^k$).
- The payoff if the first tail appears on toss $k$ is:
$$
X_k = 2^{k-1}
$$
Part (a) Risk-neutral utility $u(x) = x$
Step 1: Calculate expected payoff (expected utility)
$$
E[X] = \sum_{k=1}^{\infty} P(k) \cdot X_k = \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^k \cdot 2^{k-1} = \sum_{k=1}^{\infty} \frac{2^{k-1}}{2^k} = \sum_{k=1}^\infty \frac{1}{2} = \infty
$$
Since $\frac{2^{k-1}}{2^k} = \frac{1}{2}$, the infinite sum of $\frac{1}{2}$ diverges to infinity.
Thus,
$$
E[X] = \infty
$$
Step 2: Certainty equivalent
Since $u(x) = x$, $U(C) = C$. The certainty equivalent $C$ satisfies:
$$
C = E[X] = \infty
$$
So the certainty equivalent is infinite under risk-neutral utility, meaning the player values the game infinitely high because the expected payoff diverges.
Part (b) Power utility $u(x) = x^b$ with $b = -1$
Here,
$$
u(x) = x^{-1} = \frac{1}{x}
$$
Step 1: Expected utility
$$
E[U(X)] = \sum_{k=1}^\infty P(k) u(X_k) = \sum_{k=1}^\infty \left(\frac{1}{2}\right)^k \cdot \frac{1}{2^{k-1}} = \sum_{k=1}^\infty \frac{1}{2^k} \cdot 2^{-(k-1)} = \sum_{k=1}^\infty \frac{1}{2^k} \cdot \frac{1}{2^{k-1}} = \sum_{k=1}^\infty \frac{1}{2^{2k-1}}
$$
Simplify:
$$
E[U(X)] = \sum_{k=1}^\infty \frac{1}{2^{2k-1}} = \frac{1}{2} \sum_{k=1}^\infty \left(\frac{1}{4}\right)^k = \frac{1}{2} \cdot \frac{\frac{1}{4}}{1 – \frac{1}{4}} = \frac{1}{2} \cdot \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}
$$
Step 2: Find certainty equivalent $C$
We want $U(C) = E[U(X)]$, so
$$
C^{-1} = \frac{1}{6} \implies C = 6
$$
Part (c) Logarithmic utility $u(x) = \log_2(x)$
Step 1: Calculate expected utility
$$
E[U(X)] = \sum_{k=1}^\infty P(k) \cdot \log_2(X_k) = \sum_{k=1}^\infty \left(\frac{1}{2}\right)^k \cdot \log_2(2^{k-1}) = \sum_{k=1}^\infty \left(\frac{1}{2}\right)^k (k-1)
$$
Use the hint that:
$$
\sum_{k=1}^\infty (k-1)\left(\frac{1}{2}\right)^k = 1
$$
Therefore,
$$
E[U(X)] = 1
$$
Step 2: Find certainty equivalent $C$
$$
U(C) = \log_2(C) = 1 \implies C = 2^1 = 2
$$
Summary:
| Utility Function | Expected Utility $E[U(X)]$ | Certainty Equivalent $C$ |
|---|---|---|
| Risk-neutral $u(x) = x$ | $\infty$ | $\infty$ |
| Power $u(x) = x^{-1}$ | $\frac{1}{6}$ | 6 |
| Logarithmic $u(x) = \log_2(x)$ | 1 |
This problem explores the valuation of a St. Petersburg-type game under different utility functions, focusing on the certainty equivalent $C$. The game payoff doubles each time the coin lands heads before the first tail, leading to payoffs $X_k = 2^{k-1}$ with probabilities $P(k) = (1/2)^k$.
Under a risk-neutral utility $u(x) = x$, the expected payoff is
$$
E[X] = \sum_{k=1}^\infty \left(\frac{1}{2}\right)^k 2^{k-1} = \sum_{k=1}^\infty \frac{1}{2} = \infty
$$
The expected value diverges because the doubling payoff outweighs the rapidly decreasing probability. Thus, the certainty equivalent, which satisfies $U(C) = E[U(X)]$, is infinite. This reflects the classical paradox where the expected value of the game is unbounded, and a risk-neutral player values the game infinitely.
For the power utility function $u(x) = x^{-1}$, which represents risk aversion, the expected utility is finite:
$$
E[u(X)] = \frac{1}{6}
$$
Solving for $C$ yields $C = 6$. This finite certainty equivalent means the player values the game at a fixed amount (£6) reflecting their aversion to the game’s large variance and infinite theoretical expectation.
Finally, with the logarithmic utility $u(x) = \log_2(x)$, which models moderate risk aversion, the expected utility sums to 1. The certainty equivalent $C$ satisfies $\log_2(C) = 1$, so $C=2$. This is lower than the power utility certainty equivalent, indicating more risk aversion: the player values the game less.
In conclusion, the risk preference modeled by the utility function critically impacts the valuation of a stochastic payoff. Risk-neutral players focus on expected value and get infinite certainty equivalents, while risk-averse players discount the value, producing finite and more realistic certainty equivalents.
