expected utility and jensen s inequality
TRANSCRIPT
-
7/31/2019 Expected Utility and Jensen s Inequality
1/30
Economic Rationality QuaExpected Utility Theory
The math behind it via Jensens inequality
-
7/31/2019 Expected Utility and Jensen s Inequality
2/30
x is a random variable with realizations xand x 2, x1 < x2.u is the utility function. u is increasing and
strictly concave.p = prob{x = x 1}, 1 p = prob{x = x 2} Model Setup
-
7/31/2019 Expected Utility and Jensen s Inequality
3/30
x is a random variable with realizations xand x 2, x1 < x2. u is the utility function. u is increasing and
strictly concave.p = prob{x = x 1}, 1 p = prob{x = x 2} Model Setup
-
7/31/2019 Expected Utility and Jensen s Inequality
4/30
x is a random variable with realizations xand x 2, x1 < x2.u is the utility function. u is increasing and
strictly concave. p = prob{x = x 1}, 1 p = prob{x = x 2}Model Setup
-
7/31/2019 Expected Utility and Jensen s Inequality
5/30
E(x) = px1 + (1-p)x 2E(u(x)) = pu(x 1) + (1-p)u(x 2)
Expectation of x andExpectation of u(x)
-
7/31/2019 Expected Utility and Jensen s Inequality
6/30
E(x) = px1 + (1-p)x 2 E(u(x)) = pu(x 1) + (1-p)u(x 2)
Expectation of x andExpectation of u(x)
-
7/31/2019 Expected Utility and Jensen s Inequality
7/30
If u is strictly concave and 0 < p < 1, tu(E(x)) > E(u(x)).
Jensens Inequality Derivation for the two-point support case
-
7/31/2019 Expected Utility and Jensen s Inequality
8/30
If u is strictly concave and 0 < p < 1,u(E(x)) > E(u(x)).
Jensens Inequality Derivation for the two-point support case
-
7/31/2019 Expected Utility and Jensen s Inequality
9/30
If u is strictly concave and 0 < p < 1,then u(E(x)) > E(u(x)).
Jensens Inequality Derivation for the two-point support case
-
7/31/2019 Expected Utility and Jensen s Inequality
10/30Looking at it Graphically
-
7/31/2019 Expected Utility and Jensen s Inequality
11/30
Larger D: Base = x 2 - x1 Height = u(x 2) u
Smaller D: Base = E(x) - x1
Height = ?
Base and Height
-
7/31/2019 Expected Utility and Jensen s Inequality
12/30
Larger D: Base = x 2 - x1 Height = u(x 2) u
Smaller D: Base = E(x) - x1
Height = ?
Base and Height
-
7/31/2019 Expected Utility and Jensen s Inequality
13/30
Larger D: Base = x 2 - x1 Height = u(x 2)
Smaller D: Base = E(x) - x 1 Height = ?
E(x) - x1 = px1 + (1 p)x2 x1 = (1 p)(x
A little bit of algebra
-
7/31/2019 Expected Utility and Jensen s Inequality
14/30
Larger D: Base = x 2 - x1 Height = u(x 2)
Smaller D: Base = (1 p)(x2 x1).
Smaller D: Height = (1 p)[u(x2) u(x1)].
Conclusion
-
7/31/2019 Expected Utility and Jensen s Inequality
15/30
u(x1) + (1 p)[u(x2) u(x1)] =
u(x1) - (1 p)u(x 1) + (1 p)u(x2) =
pu(x 1) + (1 p)u(x 2) = E(u(x))
More Algebra
-
7/31/2019 Expected Utility and Jensen s Inequality
16/30
u(x1) + (1 p)[u(x2) u(x1)] =
u(x1) - (1 p)u(x 1) + (1 p)u(x2) =
pu(x 1) + (1 p)u(x 2) = E(u(x))
More Algebra
-
7/31/2019 Expected Utility and Jensen s Inequality
17/30
u(x1) + (1 p)[u(x2) u(x1)] =
u(x1) - (1 p)u(x 1) + (1 p)u(x2) =
pu(x 1) + (1 p)u(x 2) = E(u(x))
More Algebra
-
7/31/2019 Expected Utility and Jensen s Inequality
18/30
Justification of Labeling
-
7/31/2019 Expected Utility and Jensen s Inequality
19/30
Constant Absolute Risk Aversion
Constant Relative Risk Aversion
Quadratic Utility
Three Functional Forms for u
-
7/31/2019 Expected Utility and Jensen s Inequality
20/30
Constant Absolute Risk Aversion
Constant Relative Risk Aversion
Quadratic Utility
Three Functional Forms for u
-
7/31/2019 Expected Utility and Jensen s Inequality
21/30
Constant Absolute Risk Aversion
Constant Relative Risk Aversion
Quadratic Utility
Three Functional Forms for u
-
7/31/2019 Expected Utility and Jensen s Inequality
22/30
u(x) = -ke-gx
+ c; k, g, and c constants; k,
u(x) = gke -gx > 0, u (x) = -g2ke -gx < 0.
-u(x)/u(x) = g.
Constant Absolute Risk Aversion
-
7/31/2019 Expected Utility and Jensen s Inequality
23/30
u(x) = -ke-gx
+ c; k, g, and c constants; k,
u(x) = gke -gx > 0, u (x) = -g2ke -gx < 0.
-u(x)/u(x) = g.
Constant Absolute Risk Aversion
-
7/31/2019 Expected Utility and Jensen s Inequality
24/30
u(x) = -ke-gx
+ c; k, g, and c constants; k,
u(x) = gke -gx > 0, u (x) = -g2ke -gx < 0.
-u(x)/u(x) = g.
Constant Absolute Risk Aversion
-
7/31/2019 Expected Utility and Jensen s Inequality
25/30
u(x) = kxa
+ c; k, a , and c constants; k > 0, 0 0, u (x) = a(a-1) kxa-2 < 0.
-xu(x)/u(x) = 1 - a.
Constant Relative Risk Aversion
-
7/31/2019 Expected Utility and Jensen s Inequality
26/30
u(x) = kxa
+ c; k, a , and c constants; k > 0, 0 0, u (x) = a(a-1) kxa-2 < 0.
-xu(x)/u(x) = 1 - a.
Constant Relative Risk Aversion
-
7/31/2019 Expected Utility and Jensen s Inequality
27/30
u(x) = kxa
+ c; k, a , and c constants; k > 0, 0 0, u (x) = a(a-1) kxa-2 < 0.
-xu(x)/u(x) = 1 - a.
Constant Relative Risk Aversion
-
7/31/2019 Expected Utility and Jensen s Inequality
28/30
-
7/31/2019 Expected Utility and Jensen s Inequality
29/30
u(x) = -ax2
+ bx + c; a, b, and c constants; a, b >
u(x) = -2ax + b > 0 if x < b/2a, u (x) = -2a
u(x) = -2ax + b > 0 if x < b/2a, u (x) = -2a