# 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, 2015 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. Sep 4, 2015 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, 2015 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. Sep 4, 2015 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? Sep 4, 2015 at 15:35
• Thank you. Everything works the same. I just forgot to put it. Sep 4, 2015 at 18:34