Maximization of CARA utility function: unique solution with an unbounded parameter?

Question

An investor at time $t_0$ can invest his wealth $w_0$ in a risky asset $x$ for an amount $a$ and the remain part in the riskless asset $w_0-a$.

At the end of the period $t_1$, the investor will obtain the wealth $w$: $$ w = a(1+x)+(w_0-a)(1+r_f) =a(x-r_f)+w_0(1+r_f). $$ Using a CARA (exponential negative) utility function we have, $$ U(w)=-e^{-\lambda w}=-e^{-\lambda a(x-r_f)+w_0(1+r_f)} $$ where $\lambda$ is an exogenous parameter for the risk coefficient aversion. Then taking its expectation, it is clear that its maximization it does not depend on $w_0$ that is a fixed quantity but from $a$, $$ \max_a\textrm{ }E[U(w)]=E[-e^{-\lambda a(x-r_f)}\times e^{w_0(1+r_f)}] $$ where $e^{w_0(1+r_f)}$ is a fixed quantity $\tilde{q}$, $$ \max_a\textrm{ }E[U(w)]=E[-e^{-\lambda a(x-r_f)}\times \tilde{q}] $$ Here is my problem, $$ \max_a\textrm{ }E[U(w)]=E[-e^{-\lambda a(x-r_f)}] $$ If $a$ is unbounded this expected function has no point of maxima. It goes to infinity. So $a$ must be bounded and depends from the budget constraint of the initial wealth $w_0$. The professor by mail told me that is a well know result in finance and $a$ exists as an unique optimal solution. Where am I wrong?

Answer 1

This is the canonical Arrow-Pratt "portfolio" model. Couple of points on terminology:

1. For a function $u$, we define the risk aversion function by $r_u(x):=-\frac{u''(x)}{u'(x)}$. In your utility function, $r_u(x) = \lambda$; hence, it is a constant absolute risk aversion utility and $\lambda$ is the "coefficient of risk aversion," not the "risk coefficient aversion".

2. The two points in time, $t_0,t_1$ can be seen as "beginning of period" and "end of period", where "period" is here the time interval $[t_0,t_1]$. This may be important: you don't need a dynamic approach as was suggested by some people. The guy in your problem allocates $a$ to the risky asset and $w_0-a$ to the riskless, over a time interval included between $t_0$ and $t_1$.

3. Your problem is the basic, canonical portfolio choice model with utility over final wealth. The guy in your problem just consumes what he has in the end of the time period. This is also important to bear in mind.

4. It's "negative exponential", not "exponential negative".

5. Rewrite $w(a)$ for final wealth (end of period, or at $t_1$); it depends on $a$, i.e. the part of $w_0$ that is invested in the risky asset. Your problem is:

$$\max_a \;E[U(w(a)] = \max_a \;E[-e^{-\lambda(x-r_f)}]$$

Let $\chi = x-r_f$, i.e. the excess return of the risky asset (relative to the risk-free). Denote its distribution function by $dF(\chi)$ and hence

$$ \max_a \;E[-e^{-\lambda\chi}] = \max_a \;\int -e^{-\lambda z} dF(z)$$

Let $a^* = \arg\max_a \;E[U(w(a))]$. The following condition should hold in order for the (interior) optimum $a^*$ of this function to be bounded (note the redundancy in what I wrote just now):

Assumption (I) The values of the excess return random variable $\chi = x - r_f$ alternate in sign, i.e. $\chi$ takes values $\underline{\chi}\leq 0 \leq \overline{\chi}$ with positive probability.

If $\chi$ was positive almost surely, then $a$ is unbounded precisely because the objective is unbounded, as you very well understood from the beginning. Hence, Assumption (I) should be retained.

Trust your intuition - your professor is wrong.

Addendum:

if $a^*\rightarrow \infty$, i.e. if the optimal solution is unbounded, then the derivative of the expected utility evaluated at the optimal solution is zero - and since this doesn't make any sense, you have to rephrase it as

$$ \lim_{a\rightarrow\infty} E\left[\frac{d}{da}U(w(a)) \right] = 0 $$

Now $U$ is concave. Hence, in order for $a^* \rightarrow \infty$ not to be a critical point, you have to have

$$ \lim_{a\rightarrow\infty} E\left[\frac{d}{da}U(w(a)) \right] <0 $$

and not positive. Replace the parametric form, take the derivative, and you will find a (strict) inequality relating the distribution function and the marginal utility at the limits.

And since this is supposed to be a hint and not a homework helpdesk, I have to stop here :) Already, you were right in your original answer, but you have to prove it as well.

Answer 2

This looks like a general equilibrium model in Economics. It should be described in most of microeconomics textbooks (e.g. this). Yes, you need a budget constraint here for$\ a$, otherwise your optimization problem makes no sense. Moreover, the household prefers consumption today to consumption tomorrow and, hence, you may want to enhance your model by discounting next period’s utility. If you want to derive an equilibrium (along with household consumption problem) you also need to consider firm's profit maximization problem.

Edit: Can you point out a reference where you got this model from? In your framework investor is trying to maximize his wealth. Naturally he invests everything (using leverage) to get maximum return. If you want to run unconstrained optimization, I think that you target function should look a bit different, as you are solving multi-period optimization problem and investor want to maximize his total utility. Usually in economic theory they consider 2 period optimization problem like: $\ \mathop {\arg \max }\limits_a E[U({C_0}) + \frac{1}{{1 + DF }}U({C_1}))]\ $, where$\ C$ is his wealth/consumption,$\ DF$ is discount factor (investor prefers wealth today, rather than tomorrow).

Answer 3

First, your statement that your utility function goes to infinity is wrong. It's minus exponenta. You can think of it as a minimum of $e^{f(x)}$ which is bounded below by zero whatever $f(x)$ is. In other words, your utility function is bounded above by 0.

Second, maximizing expected value, you need to calculate it before deploying maximization techniques.

As an example, assume that $x-r_f$ is distributed as $N(0,1)$. Then $y:=-\lambda a (x-r_f)$ is distributed as $N(-a \lambda,a^2\lambda^2)$. Then, $e^y$ follows the lognormal distribution which mean we can look up, i.e. $E(e^y)=e^{-a\lambda+\frac{a^2\lambda^2}{2}}$

Thus, we can rewrite initial maximization program as $\min_a \; [e^{-a\lambda+\frac{a^2\lambda^2}{2}}]$. Minimum is achieved at $a=\frac{1}{\lambda}$. If you face troubles calculating it, write it in the comment and I'll write down the steps