<p>Since we are suspecting a corner-solution, it is better to write the problem explicitly with its constraint. Even better, use the Fritz John (<strong>FJ</strong>) conditions rather than the Karush-Kuhn-Tucker (<strong>KKT</strong>) ones. We will mention the differences as we go along.  </p>

<p>$$\max_{\alpha} \int u[w+\alpha(z-1)] dF(z),\;\; \text{s.t.}\;\; w-\alpha  \geq 0$$</p>

<p>The lagrangean under the Firtz John formulation is</p>

<p>$$L_{FJ} = \lambda_0\int u[w+\alpha(z-1)] dF(z)\; + \;  \lambda_1(w-\alpha)$$</p>

<p>The new element is the multiplier on the objective function, $\lambda_0$. Without loss of generality we can specify </p>

<p>$$\lambda_0 \in \{0,1\},\;\; \lambda_0 + \lambda_1 \neq 0$$</p>

<p>What is that we gain here, compared to the much more widely known and used <strong>KKT</strong>-conditions?
If a solution necessitates that $\lambda_0 =1$, we obtain the <strong>KKT</strong> conditions <em><strong>with the constraint qualification satisfied</strong></em>. If a solution necessitates that $\lambda_0 =0$, it reflects, among other special cases, the case where the constraint qualification fails to hold.  </p>

<blockquote>
  <p><em>(A standard example is the case where the feasible set for $\alpha$
  has been reduced to a single point due to the constraints imposed.
  Then we will find that the only solution dictates that $\lambda_0=0$,
  which has an intuitive explanation: if $\alpha$ can take one and only
  one value due to the constraints, then the objective function "plays
  no role" in the determination of $\alpha$ and so it gets a zero
  multiplier).</em></p>
</blockquote>

<p>Back to our problem. The first order condition is</p>

<p>$$\frac {\partial L_{FJ}}{\partial \alpha} = \lambda_0\int u'[w+\alpha(z-1)]\cdot (z-1) dF(z) - \lambda_1 \leq 0$$</p>

<p>(note the "lower or equal to zero" which is the case when optimizing under inequality constraints, rather than just "equal").</p>

<p>First, we establish that $\alpha^* &gt;0$. Due to $u'&gt;0$ and the (strict) convexity of the utility function,$u''&gt;0$, and the assumption that $E(z) &gt;1$ we have (using Jensen's Inequality)</p>

<p>$$E[u(w+\alpha(z-1))] &gt; u(w+\alpha(E(z)-1))] &gt; u(w + 0\cdot(E(z)-1))=u(w)$$</p>

<p>We turn now to examine cases.  Since the two multipliers cannot be both zero, and $\lambda_0$ takes only two values, there are three possible combinations.</p>

<p><strong>Examine the case $\lambda_0 =1$.</strong><br>
Then $\lambda_1$ can in principle be zero or positive. Examine the case where $\lambda_1 =0$, i.e. the constraint is not binding which implies that $\alpha^* &lt; w$.  With this candidate pair of multipliers, $\{\lambda_0=1,\lambda_1=0\}$, the first order condition would become</p>

<p>$$\int u'[w+\alpha(z-1)]\cdot (z-1) dF(z) \leq 0 \Rightarrow E(zu') - E(u') \leq 0$$</p>

<p>Since $u'&gt;0 \Rightarrow E(u') &gt; 0$. Also, since $E(z)&gt;1$ we have that</p>

<p>$$E(u')&lt; E(u')E(z) \Rightarrow E(zu') &lt; E(u')E(z) \Rightarrow \text{Cov}(z, u')&lt;0$$</p>

<p>But this cannot hold,  because, since $\alpha^*&gt;0$ and $u''&gt;0$, we have that $u'$ will be strictly increasing in $z$. So the covariance of $z$ and $u'$ cannot be negative. But then the pair of multiplier values $\{\lambda_0=1,\lambda_1=0\}$ cannot be a solution, and this happens due to the assumption $u''&gt;0$.  </p>

<p>We are left with the cases
$\{\lambda_0=1,\lambda_1&gt;0\}$, or $\{\lambda_0=0,\lambda_1&gt;0\}$. In both these cases, $\lambda_1 &gt;0$ i.e. the constraint is binding , i.e. we will have $\alpha^* = w$. <strong>QED.</strong></p>

<blockquote>
  <p><strong>References</strong></p>
</blockquote>

<p>The Fritz John conditions have been stated in
"<strong>F. JOHN. Extremum problems with inequalities as side conditions. In “Studies and Essays, Courant Anniversary Volume”</strong> (K. O. Friedrichs, O. E. Neugebauer and J. J. Stoker, eds.), pp. 187-204. Wiley (Interscience), New York, 1948"</p>

<p>and have been generalized in</p>

<p>"<strong>Mangasarian, O. L., &amp; Fromovitz, S. (1967). The Fritz John necessary optimality conditions in the presence of equality and inequality constraints</strong>. Journal of Mathematical Analysis and Applications, 17(1), 37-47.</p>

CC BY-SA 3.0
2014-12-05T14:22:28.503Z
2014-12-25T02:40:29.690Z
2014-12-25T02:40:29.690Z
Alecos Papadopoulos
<p>Take the standard portfolio choice problem as presented in MWG (p.188-189), but with a risk loving decision maker:</p>

<ul>
<li>with initial wealth  $w$</li>
<li>invests an amount $\alpha$ in a risky asset with a random gross rate of return $z$, where $z$ is distributed according to cdf $F$, and $\int zdF(z)&gt;1$</li>
</ul>

<p>The decision maker solves the following problem
$$
  \max_{\alpha\in[0,w]} \int u(w-\alpha +\alpha z) dF(z),
$$
where $u'&gt;0$, $u''&gt;0$. </p>

<p><strong>Is it true that the optimal solution is $\alpha^*=w$?</strong></p>

<p>Note that since $u$ is convex, the usual first order condition is not sufficient for a maximum. </p>

<p>Intuitively the answer should be yes. But what's the reason, exactly?</p>

<decision-theory><risk>
Portfolio choice of a risk lover
Herr K.
# Portfolio choice of a risk lover

(Asked by: username_0 on 2014-12-05)

Take the standard portfolio choice problem as presented in MWG (p.188-189), but with a risk loving decision maker:

* with initial wealth $w$
* invests an amount $\alpha$ in a risky asset with a random gross rate of return $z$, where $z$ is distributed according to cdf $F$, and $\int zdF(z)>1$

The decision maker solves the following problem
$$
\max\_{\alpha\in[0,w]} \int u(w-\alpha +\alpha z) dF(z),
$$
where $u'>0$, $u''>0$.

**Is it true that the optimal solution is $\alpha^\*=w$?**

Note that since $u$ is convex, the usual first order condition is not sufficient for a maximum.

Intuitively the answer should be yes. But what's the reason, exactly?

**Tags:** decision-theory, risk

---

## 1 answers

### Answer by username_1 on 2014-12-05. Score: 4 | Accepted answer

Since we are suspecting a corner-solution, it is better to write the problem explicitly with its constraint. Even better, use the Fritz John (**FJ**) conditions rather than the Karush-Kuhn-Tucker (**KKT**) ones. We will mention the differences as we go along.

$$\max\_{\alpha} \int u[w+\alpha(z-1)] dF(z),\;\; \text{s.t.}\;\; w-\alpha \geq 0$$

The lagrangean under the Firtz John formulation is

$$L\_{FJ} = \lambda\_0\int u[w+\alpha(z-1)] dF(z)\; + \; \lambda\_1(w-\alpha)$$

The new element is the multiplier on the objective function, $\lambda\_0$. Without loss of generality we can specify

$$\lambda\_0 \in \{0,1\},\;\; \lambda\_0 + \lambda\_1 \neq 0$$

What is that we gain here, compared to the much more widely known and used **KKT**-conditions?
If a solution necessitates that $\lambda\_0 =1$, we obtain the **KKT** conditions ***with the constraint qualification satisfied***. If a solution necessitates that $\lambda\_0 =0$, it reflects, among other special cases, the case where the constraint qualification fails to hold.

> *(A standard example is the case where the feasible set for $\alpha$
> has been reduced to a single point due to the constraints imposed.
> Then we will find that the only solution dictates that $\lambda\_0=0$,
> which has an intuitive explanation: if $\alpha$ can take one and only
> one value due to the constraints, then the objective function "plays
> no role" in the determination of $\alpha$ and so it gets a zero
> multiplier).*

Back to our problem. The first order condition is

$$\frac {\partial L\_{FJ}}{\partial \alpha} = \lambda\_0\int u'[w+\alpha(z-1)]\cdot (z-1) dF(z) - \lambda\_1 \leq 0$$

(note the "lower or equal to zero" which is the case when optimizing under inequality constraints, rather than just "equal").

First, we establish that $\alpha^\* >0$. Due to $u'>0$ and the (strict) convexity of the utility function,$u''>0$, and the assumption that $E(z) >1$ we have (using Jensen's Inequality)

$$E[u(w+\alpha(z-1))] > u(w+\alpha(E(z)-1))] > u(w + 0\cdot(E(z)-1))=u(w)$$

We turn now to examine cases. Since the two multipliers cannot be both zero, and $\lambda\_0$ takes only two values, there are three possible combinations.

**Examine the case $\lambda\_0 =1$.**  
Then $\lambda\_1$ can in principle be zero or positive. Examine the case where $\lambda\_1 =0$, i.e. the constraint is not binding which implies that $\alpha^\* < w$. With this candidate pair of multipliers, $\{\lambda\_0=1,\lambda\_1=0\}$, the first order condition would become

$$\int u'[w+\alpha(z-1)]\cdot (z-1) dF(z) \leq 0 \Rightarrow E(zu') - E(u') \leq 0$$

Since $u'>0 \Rightarrow E(u') > 0$. Also, since $E(z)>1$ we have that

$$E(u')< E(u')E(z) \Rightarrow E(zu') < E(u')E(z) \Rightarrow \text{Cov}(z, u')<0$$

But this cannot hold, because, since $\alpha^\*>0$ and $u''>0$, we have that $u'$ will be strictly increasing in $z$. So the covariance of $z$ and $u'$ cannot be negative. But then the pair of multiplier values $\{\lambda\_0=1,\lambda\_1=0\}$ cannot be a solution, and this happens due to the assumption $u''>0$.

We are left with the cases
$\{\lambda\_0=1,\lambda\_1>0\}$, or $\{\lambda\_0=0,\lambda\_1>0\}$. In both these cases, $\lambda\_1 >0$ i.e. the constraint is binding , i.e. we will have $\alpha^\* = w$. **QED.**

> **References**

The Fritz John conditions have been stated in
"**F. JOHN. Extremum problems with inequalities as side conditions. In “Studies and Essays, Courant Anniversary Volume”** (K. O. Friedrichs, O. E. Neugebauer and J. J. Stoker, eds.), pp. 187-204. Wiley (Interscience), New York, 1948"

and have been generalized in

"**Mangasarian, O. L., & Fromovitz, S. (1967). The Fritz John necessary optimality conditions in the presence of equality and inequality constraints**. Journal of Mathematical Analysis and Applications, 17(1), 37-47.


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