# Utility Theory - How to show that this exponential utility function is wealth-independent?

I have a question on the following exercise from chapter 9 of D. Luenberger, Investment Science, International Edition.

Exercise 2 (Wealth Independence)

Suppose an investor has exponential utility function $U(x) = -e^{-ax}$ and an initial wealth level of W. The investor is faced with an opportunity to invest an amount $w \le W$ and obtain a random payoff $x$.

Show that his evaluation of this incremental investment is independent of W.

I first considered that the utility after the investment is $-e^{-aW}*e^{-a(x-w)}$ and that this is a factor above the original utility of $-e^{-aW}$ which is independent of W, and then that's the solution.

However, this doesn't really take into consideration the randomness of x. For instance, I know of one possible example where if $x$ takes only 2 values with probability of $\frac{1}{2}$ each then it can be shown that $w$, the absolute value amount to be invested, is the exact same for any initial wealth $W$.

So if it's the case that this question requires a similar result for any random payoff $x$, how would I go about doing that if I don't know the probability function of $x$?

Perhaps I should consider $w$ as a proportion of $W$ and then show somehow that this is equal to some constant $\frac{k}{W}$ when $E(U(x))$ is maximized with respect to $x$.

If this is the approach, how do you differentiate $E(U(x))$ with respect to x?

• Shouldn't you maximize expectation with respect to $w$, not $x$? – SRKX Sep 4 '15 at 8:43
• Yes. $W$ is what he needs to maximize - which in practice boils down to choosing the share of wealth invested on the risky asset. – phdstudent Sep 4 '15 at 9:22

The answer is relatively straightforward if you assume that $x$ is normally distributed - $x \sim N(\mu_x,\sigma^2_x)$. If $x$ is normally distributed then maximizing $U(x)=−e^{ax}$ is the same as maximizing a mean variance utility: $U = E(W) - 0.5a Var(W)$ .

Now given that:

$E(W) = s\mu_x + (W-s)$ where $s$ is the amount of money on the risky stock and $W-s$ is the amount not invested - this assumes a risk-free rate of zero.

$Var(W) = s^2 \sigma^2_x$

Take first order conditions to: $U = s\mu_x + (W-s) - 0.5a s^2 \sigma^2_x$ and get:

$\mu_x - a s \sigma^2_x = 0$. So : $s = \frac{\mu_x}{a\sigma^2_x}$

Which does not depend on the initial wealth q.e.d.

Edit: Following the comment below let's show it without assuming normality:

Denoting the amount invested on the risky asset by $\theta$ and the initial level of wealth by $W$, the agent's expected utility is: $U = \int \exp(-a[(W-\theta)Rf + \theta x])f(x)dx = \int \exp(-aWRf) \exp[-a\theta(x-Rf)]f(x)dx = \exp[-aWR_f]\int \exp[-a\theta(x-R_f)]f(x)dx$

So the solution to the problem is independent of initial wealth (just take f.o.c. on equation above and note that $W$ does not show up).

• I think the exercise is looking for a demonstration even without assuming a distribution for $x$. – SRKX Sep 4 '15 at 8:41
• Oh I see. So in the final equation if you were to differentiate, set equal to zero and solve for theta, I can see that W just wouldn't show up, so there's no need to do that calculation at all. Thanks. – James Sep 4 '15 at 15:13
• Hi again. I just realised that there should have been a minus sign in the exponent of the exponential utility function that I've edited in now. Can you confirm that everything works the same with this change? – James Sep 4 '15 at 15:35
• Thank you. Everything works the same. I just forgot to put it. – phdstudent Sep 4 '15 at 18:34