← 7. Choice under Uncertainty

Exercises

Problems to check your understanding of this chapter. The solutions are folded away — try them yourself first.

Easy

1EasyCompute

A consumer with u(x)=xu(x) = \sqrt{x} evaluates a lottery paying 144144 with probability 1/21/2 and 3636 with probability 1/21/2. Find the certainty equivalent and the risk premium.

c solves u(c)=U(L),premium=E[L]−cc \text{ solves } u(c) = U(L), \qquad \text{premium} = \mathbb{E}[L] - c
The 144144/3636 lottery with u(x)=xu(x) = \sqrt{x} — certainty equivalent 8181.
Show solution

  1. The expected value is E[L]=12⋅144+12⋅36=90\mathbb{E}[L] = \frac{1}{2} \cdot 144 + \frac{1}{2} \cdot 36 = 90. Expected utility is U(L)=12144+1236=12(12+6)=9U(L) = \frac{1}{2}\sqrt{144} + \frac{1}{2}\sqrt{36} = \frac{1}{2}(12 + 6) = 9.

  2. The certainty equivalent solves c=9\sqrt{c} = 9, so c=81c = 81. The risk premium is 90−81=990 - 81 = 9 — the consumer accepts a sure amount 9 below the expected value to shed the lottery.

Answer

Certainty equivalent c=81c = 81, risk premium =9= 9

Medium

2Medium

Show that in CRRA utility u(x)=x1−γ/(1−γ)u(x) = x^{1-\gamma}/(1-\gamma), the (normalized) utility converges to log utility as γ→1\gamma \to 1: lim⁡γ→1x1−γ−11−γ=ln⁡x\lim_{\gamma \to 1} \frac{x^{1-\gamma} - 1}{1-\gamma} = \ln x.

lim⁡γ→1x1−γ−11−γ=ln⁡x\lim_{\gamma \to 1} \frac{x^{1-\gamma} - 1}{1-\gamma} = \ln x
γ=1\gamma = 1 — the log-utility limit.
Show solution

  1. As γ→1\gamma \to 1, numerator and denominator both go to 00 — a 0/00/0 form. Apply l'Hôpital's rule in γ\gamma: the numerator's derivative is ddγ(x1−γ−1)=−x1−γln⁡x→−ln⁡x\frac{d}{d\gamma}(x^{1-\gamma} - 1) = -x^{1-\gamma}\ln x \to -\ln x, the denominator's is −1-1.

  2. The limit is therefore (−ln⁡x)/(−1)=ln⁡x(-\ln x)/(-1) = \ln x. For fixed γ\gamma, x1−γ/(1−γ)x^{1-\gamma}/(1-\gamma) and (x1−γ−1)/(1−γ)(x^{1-\gamma}-1)/(1-\gamma) differ only by an affine transformation in xx, hence represent the same preferences — which is why CRRA at γ=1\gamma = 1 is called log utility.

3Medium

A consumer with W=100W = 100, loss L=60L = 60 (probability π=0.3\pi = 0.3), and u(x)=xu(x) = \sqrt{x} buys β\beta coverage at premium rate pp per dollar. Differentiate expected utility with respect to β\beta to derive the first-order condition and solve for β∗(p)\beta^*(p). Check the value at p=0.3p = 0.3.

EU(β)=(1−π)W−pβ+πW−L+(1−p)βEU(\beta) = (1-\pi)\sqrt{W - p\beta} + \pi\sqrt{W - L + (1-p)\beta}
d EUdβ=−(1−π)p2W1+π(1−p)2W2=0\frac{d\,EU}{d\beta} = -\frac{(1-\pi)p}{2\sqrt{W_1}} + \frac{\pi(1-p)}{2\sqrt{W_2}} = 0
Fair premium — full insurance β∗=60\beta^* = 60 on the 45-degree line.
Show solution

  1. By the chain rule, d EUdβ=(1−π)−p2W1+π1−p2W2\frac{d\,EU}{d\beta} = (1-\pi)\frac{-p}{2\sqrt{W_1}} + \pi\frac{1-p}{2\sqrt{W_2}}. Setting it to zero gives (1−π)pW1=π(1−p)W2\frac{(1-\pi)p}{\sqrt{W_1}} = \frac{\pi(1-p)}{\sqrt{W_2}} — marginal cost equals marginal benefit of the last dollar of coverage.

  2. Plugging in numbers: 0.7 p100−pβ=0.3 (1−p)40+(1−p)β\frac{0.7\,p}{\sqrt{100 - p\beta}} = \frac{0.3\,(1-p)}{\sqrt{40 + (1-p)\beta}}. Squaring both sides gives a quadratic in β\beta, but at p=0.3p = 0.3 it collapses to 0.21W1=0.21W2\frac{0.21}{\sqrt{W_1}} = \frac{0.21}{\sqrt{W_2}}, so W1=W2W_1 = W_2 — i.e. 100−0.3β=40+0.7β100 - 0.3\beta = 40 + 0.7\beta, giving β∗=60\beta^* = 60. Full insurance is optimal at the fair premium.

Hard

4Hard

Two CRRA consumers A (γA=0.5\gamma_A = 0.5) and B (γB=2\gamma_B = 2) evaluate the 50-50 lottery (150,50)(150, 50). Compute each certainty equivalent and risk premium, and explain via the Arrow-Pratt measure who is more risk averse.

c(γ)=[12⋅1501−γ+12⋅501−γ]11−γc(\gamma) = \left[\tfrac{1}{2} \cdot 150^{1-\gamma} + \tfrac{1}{2} \cdot 50^{1-\gamma}\right]^{\frac{1}{1-\gamma}}
rA(x)=γxr_A(x) = \frac{\gamma}{x}
γ=2\gamma = 2 — certainty equivalent 75.075.0, risk premium 25.025.0.
Show solution

  1. With γA=0.5\gamma_A = 0.5, cA=[12150+1250]2≈9.6592≈93.30c_A = \left[\frac{1}{2}\sqrt{150} + \frac{1}{2}\sqrt{50}\right]^2 \approx 9.659^2 \approx 93.30. The risk premium is 100−93.30=6.70100 - 93.30 = 6.70.

  2. With γB=2\gamma_B = 2, 1−γ=−11-\gamma = -1, so cB=[12⋅150−1+12⋅50−1]−1=10.5/150+0.5/50=10.01333=75.0c_B = \left[\frac{1}{2} \cdot 150^{-1} + \frac{1}{2} \cdot 50^{-1}\right]^{-1} = \frac{1}{0.5/150 + 0.5/50} = \frac{1}{0.01333} = 75.0. The risk premium 100−75=25.0100 - 75 = 25.0 exceeds A's.

  3. The Arrow-Pratt measure is rA(x)=γ/xr_A(x) = \gamma/x, so at x=100x = 100, rAA=0.005r_A^A = 0.005 and rBA=0.02r_B^A = 0.02 — B is more risk averse at every wealth level, which shows up as a lower certainty equivalent and a larger risk premium. This matches the scene's observation that the CE drifts away from the expected value as γ\gamma rises.