Seminar Economics of Psychology of Risk and Time

Erin Belles

Assignment 1, Group 7, Deadline: March 11, 2016 TiSEM

Question 1

Part A

Expected utility theory with the utility function: \(u(x) = \frac{x}{10}\) with total wealth = 0.

EV= 0.2 x \(\frac{40}{10}\) + 0.6 x \(\frac{50}{10}\) + 0.2 x \(\frac{30}{10}\) = 4.4

CE: \(\frac{m}{10}\) = 4.4 \(\rightarrow\) m=44 The certainty equivalent is 44

Part B

Rank dependent utility with the utility function: \(u(x) = \frac{x}{10}\) and \(w(p) = p^{2}\) with total wealth = 0.}

Probability of outcome q or better:

q=40, p=(0.2) + (0.6) = 0.8
q=50, p=(0.6)
q=30, p=(0.2) + (0.6) + (0.2) = 1

Probability outcome is strictly better than q:

q=40, p=0.6
q=50, p=0
q=30, p=(0.2) + (0.6) = 0.8

\(w(p)=p^2\)
w(0.8) - w(0.6) = \((0.8)^2\) - \((0.6)^2\)= 0.28
w(0.6) - w(0.0) = \((0.0)^2\) - \((0.0)^2\)= 0.36
w(1) - w(0.8) = \((1)^2\) - \((0.8)^2\) = 0.36

Expected utility using these updated probability weights is:

EU = 0.28 x \(\frac{40}{10}\) + 0.36 x \(\frac{50}{10}\) + 0.36 x \(\frac{30}{10}\) = 4

CE: \(\frac{m}{10}\) = 4 \(\rightarrow\) m = 40 The certainty equivalent is 40.

Question 2

Part A

Show that this choice pattern violates Expected Utility theory. Under expected utility, preferences for prospect A implies:

u(3000) \(>\) 0.8 u(4000) + 0.2 u(0)

\(\frac{1}{4}\)(u(3000) \(>\) 0.8 u(4000) + 0.2 u(0)) \(\rightarrow\) 0.25 u(3000) \(>\) 0.2 u (4000) + 0.05 u(0) +0.75 u(0)

\(\rightarrow\) 0.25 u(3000)+ 0.75 u(0) \(>\) 0.2 u(4000)+ 0.8 u (4000)

Preferring prospect D to C implies:

0.2u(4000) + 0.8 u(0) \(>\) 0.25u(3000) +0.75u(0)

Therefore, under prospect theory an individual cannot prefer A to B and D to C.

Part B

Show that Disappointment theory as presented and parameterized on the slides (i.e., with u(x) = x and θ = 0.0002) can accommodate the observed choice pattern.)

EV(A) = 3000 D(A) = 3000

EV(B) = EU = 3200

D(B)=0.8(4000+0.0002x\((4000-3200)^2\) + 0.2(0-0.0002x\((0-3200)^2\))=2892.8

Hence, D(A) \(>\) D(B)

EV(C) = EU = 750

D(C)=0.25(3000+0.0002x\((3000-750)^2\)) + 0.75(0-0.0002x\((0-750)^2\))=918.75

EV(D) = EU = 800

D(D)=0.2(4000+0.0002x\((4000-800)^2\)) + 0.8(0-0.0002x\((0-800)^2\))=1107.2

Hence, D(D) \(>\) D(C)

Part C

Show that Cumulative Prospect Theory can accommodate the choice pattern.
Using the parametrization by Tversky and Kahnemann:

CPT(A)

x p π u(x) = U(x) πU(x)
3000 1 w+(1)-w+(0)=1 1147.8 1147.8

CPT(B)

x p π u(x) πU(x)
4000 0.8 w+(0.8)-w+(0.2)=0.607 63.245 1478.47

CPT(C)

x p π u(x) πU(x)
3000 0.25 w+(0.25)-w+(0)= 0.2707 1147.8 333.67

CPT(D)

x p π u(x) = U(x) πU(x)
4000 10.20 w+(0.20)-w+(0)=0.261 1478.8 385.53

\(\rightarrow\) CPT(A) \(>\) CPT(B)

\(\rightarrow\) CPT(D) \(>\) CPT(C)

Question 3

Part A

(0.25, €75; 0.25, €50; 0.25, €25; 0.25, €0)}

x p π u(x) = U(x) πU(x)
75 0.25 w+(0.25)-w+(0)=0.29074 44.674 12.989
50 0.25 w+(0.50)-w+(0.25)=0.12990 31.268 4.062
25 0.25 w+(0.75)-w+(0.50)=0.14763 16.990 2.508
0 0.25 w+(1)-w+(0.75)=0.43173 0 0

\(u^+(x)\) = \(x^{0.88}\) if x \(\geq\) 0

\(u^-(x)\) = 2.25 * \(-(-x)^{0.88}\) if x \(<\) 0 (with \(\lambda\) = 2.25)

 

\(w^+(p) = \frac{p^{0.61}}{(p^{0.61}+(1-p)^{0.61})^{1/0.61}}\)

\(w^-(p) = \frac{p^{0.69}}{(p^{0.69}+(1-p)^{0.69})^{1/0.69}}\)

\(w^+(0) = \frac{0^{0.61}}{(0^{0.61}+(1)^{0.61})^{1/0.61}}\) = 0
\(w^+(0.25) = \frac{0.25^{0.61}}{(0.25^{0.61}+(0.75)^{0.61})^{1/0.61}}\) = 0.290
\(w^+(0.50) = \frac{0.50^{0.61}}{(0.50^{0.61}+(0.50)^{0.61})^{1/0.61}}\) = 0.42064
\(w^+(0.75) = \frac{0.75^{0.61}}{(0.75^{0.61}+(0.25)^{0.61})^{1/0.61}}\) = 0.56827
\(w^+(1) = \frac{1^{0.61}}{(1^{0.61}+(0)^{0.61})^{1/0.61}}\) = 1

0.29074 - 0 = 0.29074
0.42064 - 0.29074 = 0.12990
0.56827 - 0.42064 = 0.14763
1 - 0.56827 = 0.43173

\(u^+(75)\) = \(75^{0.88}\) = 44.674
\(u^+(50)\) = \(50^{0.88}\) = 31.268
\(u^+(25)\) = \(25^{0.88}\) = 16.990
\(u^+(0)\) = \(0^{0.88}\) = 0

CPT = 12.989 + 4.062 + 2.508 + 0 = 19.559

\(CE^{0.88}\) = 19.559 \(\rightarrow\) CE = 29.339

EV = \(\frac{1}{4}(75)+\frac{1}{4}(50)+\frac{1}{4}(25)+\frac{1}{4}(0)=37.5\)
RP = 37.5 - 29.339 = 8.161

Part B

  1. (0.25, €0; 0.25, -€25; 0.25, -€50; 0.25, -€75)
x p π u(x) = U(x) πU(x)
0 0.25 w+(0.25)-w+(0)=0.29074 0 0
-25 0.25 w-(0.50)-w+(0.25)=0.12990 -38.227 -6.591
-50 0.25 w-(0.50)-w+(0.25)=0.14763 -70.352 -11.289
-75 0.25 w-(0.25)-w-(0)=0.29352 -100.516 -29.503

\(w^-(0.75) = \frac{0.75^{0.69}}{(0.75^{0.69}+(0.25)^{0.69})^{1/0.69}}\) = 0.62640 0.62640-0.45399= 0.17241 \(u^-(-75)\) = 2.25 * \(-(--75)^{0.88}\) = -100.516

CPT = 0 + -6.591 + -11.289 + -29.503 = -47.383 \(CE^{0.88}\) = -47.383 \(\rightarrow\) CE = -31.909

EV = \(\frac{1}{4}(0)+\frac{1}{4}(-25)+\frac{1}{4}(-50)+\frac{1}{4}(-75)= -37.50\) RP = -37.50 - -31.909 = -5.591

Question 4

Using the example from the lecture slides, people can prefer E to F and G to H without violating the expectations of an s-shaped utility curve (see the graph on the following slide). The intuition is that losing x hurts more than gaining x brings pleasure. Although people should prefer 0.75 x 3000 to 0.5 x 4500, the possible loss of 6000 with probability of 0.25 hurts too much and will outweigh the extra utility from 0.75 x 3000.

\(w^+(p)=w^-(p)=p^2\)
\(u^+(x)=x^5\)
\(u^-(x)=-2x^5\)

CPT(E) = \(w^-(0.5)u^-(-3000)+w^+(0.5)u^+(4500)\) = \(0.5^2\) x (-2) x \(3000^5\) + \(0.5^2\) x \(4500^5\) = 4.6 x \(10^6\)

CPT(F) = \(w^-(0.25)u^-(-6000)+w^+(0.75)u^+(3000)\) = \(0.25^2\) x (-2) x \(6000^5\) + \(0.75^2\) x \(3000^5\) = -8.35 x \(10^{17}\)

CPT(E) \(>\) CPT(F)

CPT(G) = \(w^-(0.5)u^-(-1500)+w^+(0.5)u^+(4500)\) = \(0.5^2\) x (-2) x \(1500^2\) + \(0.25^2\) x \(4500^5\) = 4.58 x \(10^{17}\)

CPT(H) = \(w^-(0.25)u^-(3000)+w^+(0.75)u^+(3000)\) = \(0.25^2\) x (-2) x \(3000^5\) + \(0.75^2\) x \(3000^5\) = 1.06 x \(10^{17}\)

CPT(G) \(>\) CPT(H)

Marginal Utility is Dependent on Domain

CPT utility curve over domains of losses and gains
CPT utility curve over domains of losses and gains

A second possible explanation that comes to the same conclusion that we do not agree with the statement by Levy and Levy (2002) is the following: Looking at the probability weighting of cumulative prospect theory, shows that small probabilities are overweighted and moderate and large probabilities are underweighted (see the second graph below).

Probability weighting function
Probability weighting function

Why do people prefer E over F? In prospect F there is a small probability of a very large negative outcome. Since it is a small probability and a negative outcome, people are pessimistic according to CPT. Prospect E on the other hand has equal moderate probabilities and is hence preferred over a prospect with a small overweighted outcome of loosing 6000 with probability 0.25.

The same logic holds for prospects G and H. In Prospect H there is a small probability of a very large negative outcome. Since it is a small probability and a negative outcome, people are pessimistic according to CPT and hence prefer prospect G over prospect H.