Chapter 13.2, Problem 1P

Explanation of Solution

a.

Given:

The utility function for asset position $x$ is given by,

$u(x)=\ln x$

Determining whether Am I Risk-Reverse, Risk-Neutral, or Risk-Seeking:

If $u(x)$is differentiable, then it is,

• Risk-Reverse when $u\text{'}\text{'}(x)<0$,
• Risk-Neutral when $u(x)$is a linear function,
• Risk Seeking when $u\text{'}\text{'}(x)>0$

Explanation of Solution

b.

Given:

The utility function for asset position $x$ is given by,

$u(x)=\ln x$

Now, I have an asset of $20,000.

Consider the two lotteries,

L1: With probability 1, the asset becomes $19,000 (Loss of $1,000)

L2: With probability 0.9, the asset becomes $20,000 (Gain of $0)

       With probability 0.1, the asset becomes $10,000 (Loss of $10,000)

Determining which lottery, I prefer:

For a given lottery, $L=(p_1,r_1;p_2,r_2;\mathrm{...};p_n,r_n)$, define the expected utility of the lottery L given as,

                  $E(U\text{ for L) = }{\displaystyle \sum _{i=1}^{i=n}{p}_{i}u}(r_i)$

• Here, $p$ is the probability,
• $u(r_i)$is the utility function,
• $r$is the reward.

It is given that,

$u(x)=\ln x$

Now, find the expected utility of the lottery L₁,

$\begin{array}{l}E(U{\text{ for L}}_{\text{1}}\text{) =1}\times \text{u(19000)}\\ \text{ =1}\times \text{ln(19000)}\\ \text{ =9}\text{.8522}\end{array}$

Similarly, find the expected utility of the lottery L₂,

$\begin{array}{l}E(U{\text{ for L}}_{\text{2}}\text{) =(0}\text{.9}\times \text{u(20000))+(0}\text{.1}\times \text{u(10000))}\\ \text{ =(0}\text{.9}\times \text{ln(20000))+(0}\text{.1}\times \text{ln(10000))}\\ \text{ =9}\text{.8342}\end{array}$

For the two lotteries L₁ and L₂, use preference criteria via expected utility criteria. The criteria are as follows:

• We prefer L₁ over L₂ (L₁p L₂) if $E(U{\text{ for L}}_{\text{1}})>E(U{\text{ for L}}_{\text{2}})$
• We prefer L₂ over L₁ (L₂p L₁) if $E(U{\text{ for L}}_{\text{2}})>E(U{\text{ for L}}_{\text{1}})$
• The lottery L₁ is indifferent to L₂ (L₁i L₂) if $E(U{\text{ for L}}_{\text{1}})=E(U{\text{ for L}}_{\text{2}})$

We have,

$E(U{\text{ for L}}_{\text{1}}\text{) = 9}$