Varian problem 11.6
Esperanza has a utility function u (r) which satisfies u0 > 0 and u00 < 0. Her return R is normally
distributed so it has a density function
f (r) = √
1
2πσ 2
e−
(r−µ)2
2σ 2
,
and hence her expected utility is
Z
+∞
EU =
Z
+∞
u (r) √
u (r) f (r) dr =
−∞
−∞
1
2πσ 2
e−
(r−µ)2
2σ 2
dr.
(a) As the function f only depends on µ and σ 2 , EU can also only depend on µ and σ 2 .
(b) Define R2 = R − µ. Then R2 ∼ N 0, σ 2 , and we can rewrite Esperanza’s expected utlity as
+∞
Z
u (r2 + µ) √
EU =
−∞
2
r2
1
2πσ 2
e− 2σ2 dr2 .
We see that µ only appears in the utility function. To see the effect of an increase in µ, Leibniz’s formula
tells us that we can simply diffrentiate the integrand, hence
∂EU
=
∂µ
as u0 > 0 and
+∞
Z
u0 (r2 + µ) √
−∞
2
r2
1
2πσ 2
e− 2σ2 dr2 > 0
r2
2
√ 1
e− 2σ2
2πσ 2
(c) Define now R3 =
> 0.
R−µ
σ .
Then R3 ∼ N (0, 1) and the expected utility becomes
+∞
Z
EU =
−∞
2
r3
1
u [σr3 + µ] √ e− 2 dr3
2π
so
∂EU
=
∂σ
To find the sign of this, notice that
∂EU
∂σ
Z
+∞
−∞
2
√1 e−r3
2π
=
2
r3
1
r3 u0 [σr3 + µ] √ e− 2 dr3 .
2π
2
√1 e−(−r3 )
2π
so the density is symmetrical around 0.
+∞
2
r3
1
[r3 u0 [σr3 + µ] + (−r3 ) u0 [σ (−r3 ) + µ]] √ e− 2 dr3
2π
0
Z +∞
2
r3
1
=
r3 [u0 (σr3 + µ) − u0 (−σr3 + µ)] √ e− 2 dr3 .
2π
0
Z
=
As u is concave, the second derivtive is larger for small values than for large values, so u0 (σr3 + µ) <
u0 (−σr3 + µ). Hence the expression in square brackets is negative and the integral is negative, so the
expected utility declines as σ increases.
1
ECON 4230 / 4235
Fall 2005
Supplementary comments on Varian problem 11.6
Let R be a stochastic variable, normally distributed with mean µ and variance σ2. That is,
the probability density is given by:
⎧⎪ 1 ⎛ r − µ ⎞ 2 ⎫⎪
1
f (r ) =
exp ⎨− ⎜
⎟ ⎬
σ 2π
⎩⎪ 2 ⎝ σ ⎠ ⎭⎪
In addition, assume that the decision maker has constant absolute risk aversion, equal to λ
> 0, that is, the expected utility function is given by
u (r ) = e− λ r = exp ( −λ r )
(1)
The decisions maker’s expected utility of receiving the uncertain reward R then becomes:
⎧⎪ 1 ⎛ r − µ ⎞ 2 ⎫⎪
∞
1
(2)
E Ru ( r ) = ∫
exp ⎨− ⎜
⎟ ⎬ exp ( −λ r ) dr
−∞
σ 2π
⎩⎪ 2 ⎝ σ ⎠ ⎭⎪
The sum of the exponents in equation (2) can be written
1
2
(3)
− 2 ( r − µ ) + 2σ 2 λ r
2σ
By “completing the square” we see that the expression (3) is equal to
{
}
1 ⎛ r − (µ −σ λ ) ⎞
1
⎟ − λµ + σ 2 λ 2
(4) − 2 r − ( µ − σ λ ) + 2σ λµ − σ λ = − ⎜
⎟
2σ
2⎜
2
σ
⎝
⎠
By inserting (4) into (2) and moving terms not containing r outside the integral sign, we
conclude that the decisions maker’s expected utility of receiving R equals
⎧ ⎛ r − µ − σ 2λ ⎞2 ⎫
∞
(
) ⎟ ⎪dr
1
1
⎪ 1
⎛
⎞
E R u ( r ) = exp ⎜ −λµ + σ 2 λ 2 ⎟ ∫
exp ⎨− ⎜
(5)
⎬
⎟ ⎪
2
σ
⎝
⎠ −∞ σ 2π
⎪⎩ 2 ⎜⎝
⎠ ⎭
The integral in (5) must be equal to 1, since it is the integral over the whole range of the
density function of a probability distribution, namely the normal distribution with mean
µ - σ2λ and variance σ2. Therefore, the expression (5) equals
1
⎛ ⎛
1
⎛
⎞
⎞⎞
exp ⎜ −λµ + σ 2 λ 2 ⎟ = exp ⎜ −λ ⎜ µ − σ 2λ ⎟ ⎟
2
2
⎝
⎠
⎠⎠
⎝ ⎝
By (1) we then get
1
⎛
⎞
E Ru ( r ) = u ⎜ µ − σ 2λ ⎟
2
⎝
⎠
Conclusion: The decision maker is indifferent between the uncertain reward R and a
1
certain reward equal to µ − σ 2 λ , that is, the expectation of R minus a risk premium
2
1 2
σ λ . The risk premium is proportional to the coefficient of absolute risk aversion λ,
2
and also proportional to the variance σ2 of the uncertain reward R.
1
{(
2
)
2
2
4
2
}
2
2