<p>The individual will invest the whole <span class="math-container">$\\\$ 10,000$</span>, which makes sense, since in the worst case scenario he still gets a positive return. He has to decide which fraction <span class="math-container">$\alpha$</span> to invest in asset <span class="math-container">$A$</span> and which fraction <span class="math-container">$1-\alpha$</span> to invest in asset <span class="math-container">$B$</span>.</p>
<p>Note that if returns are the same for both assets, the fraction invested on each is irrelevant. Then, his problem can be written as follows,</p>
<p><span class="math-container">$$
\begin{aligned}
\max _{\alpha} B= &amp; \frac{1}{10} \sqrt{1.2 \times 10,000}+\frac{1}{10} \sqrt{1.05 \times 10,000} \\
&amp; +\frac{4}{10} \sqrt{1.2 \times \alpha 10,000+1.05 \times(1-\alpha) 10,000} \\
&amp; +\frac{4}{10} \sqrt{1.05 \times \alpha 10,000+1.2 \times(1-\alpha) 10,000}
\end{aligned}
$$</span></p>
<p>Let's simplify the problem first.</p>
<p>Since there are two situations in which <span class="math-container">$\alpha$</span> does not affect the outcome, we can rewrite the problem as follows,</p>
<p><span class="math-container">$$
\begin{aligned}
\max _{\alpha} \tilde{B}= &amp; \frac{4}{10} \sqrt{1.2 \times \alpha 10,000+1.05 \times(1-\alpha) 10,000} \\
&amp; +\frac{4}{10} \sqrt{1.05 \times \alpha 10,000+1.2 \times(1-\alpha) 10,000} \\
\max _{\alpha} B= &amp; \frac{4 \sqrt{10,000}}{10}[\sqrt{1.2 \alpha+1.05(1-\alpha)}+\sqrt{1.05 \alpha+1.2(1-\alpha)}]
\end{aligned}
$$</span></p>
<p>Then, we can just solve,</p>
<p><span class="math-container">$$
\max _{\alpha} B=\sqrt{1.05+0.15 \alpha}+\sqrt{1.2-0.15 \alpha}
$$</span></p>
<p>The first order condition is given by:</p>
<p><span class="math-container">$$
\frac{\partial B}{\partial \alpha}=0.15(1.05+0.15 \alpha)^{-\frac{1}{2}}-0.15(1.2-0.15 \alpha)^{-\frac{1}{2}}=0
$$</span></p>
<p>Then,</p>
<p><span class="math-container">$$
\begin{aligned}
(1.05+0.15 \alpha)^{-\frac{1}{2}} &amp; =(1.2-0.15 \alpha)^{-\frac{1}{2}} \\
1.05+0.15 \alpha &amp; =1.2-0.15 \alpha \\
0.3 \alpha &amp; =0.15
\end{aligned}
$$</span></p>
<p>Then, the <span class="math-container">$\alpha=0.5$</span>.</p>
<p>This is not surprising, given that the assets are symmetric. Let us finally verify that the second order condition holds:</p>
<p><span class="math-container">$$
\frac{\partial^{2} B}{\partial \alpha^{2}}=-\frac{1}{2} 0.15^{2}(1.05+0.15 \alpha)^{-\frac{3}{2}}-\frac{1}{2}(-0.15)^{2}(1.2-0.15 \alpha)^{-\frac{3}{2}}&lt;0
$$</span></p>

CC BY-SA 4.0
2024-06-27T13:09:49.353Z
2024-06-27T13:09:49.353Z
Lorenzo Castagno
<p>Let's say that an individual has <span class="math-container">$\\\$ 10,000$</span> to invest in a combination of two risky assets. Each asset can lead to a high return or a low one. However, returns are not independent. Their joint probability distribution is provided in the following table.</p>
<div class="s-table-container"><table class="s-table">
<thead>
<tr>
<th></th>
<th>A</th>
<th>A</th>
<th>A</th>
</tr>
</thead>
<tbody>
<tr>
<td>B</td>
<td></td>
<td>20%</td>
<td>5%</td>
</tr>
<tr>
<td>B</td>
<td>5%</td>
<td>0.4</td>
<td>0.1</td>
</tr>
<tr>
<td>B</td>
<td>20%</td>
<td>0.1</td>
<td>0.4</td>
</tr>
</tbody>
</table></div>
<p>This means, for example, that the probability that the returns from assets <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are <span class="math-container">$5 \%$</span> and <span class="math-container">$20 \%$</span> respectively is 0.4 . Let the individual's Bernoulli utility function be given by <span class="math-container">$u(x)=\sqrt{x}$</span>. In such a case, how much will this individual invest on each asset?</p>

<microeconomics><game-theory>
Individual invest on each asset
DanIsabel
# Individual invest on each asset

(Asked by: username_0 on 2024-06-27)

Let's say that an individual has $\\\$ 10,000$ to invest in a combination of two risky assets. Each asset can lead to a high return or a low one. However, returns are not independent. Their joint probability distribution is provided in the following table.

|  | A | A | A |
| --- | --- | --- | --- |
| B |  | 20% | 5% |
| B | 5% | 0.4 | 0.1 |
| B | 20% | 0.1 | 0.4 |

This means, for example, that the probability that the returns from assets $A$ and $B$ are $5 \%$ and $20 \%$ respectively is 0.4 . Let the individual's Bernoulli utility function be given by $u(x)=\sqrt{x}$. In such a case, how much will this individual invest on each asset?

**Tags:** microeconomics, game-theory


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## 1 answers

### Answer by username_1 on 2024-06-27. Score: 1 | Accepted answer

The individual will invest the whole $\\\$ 10,000$, which makes sense, since in the worst case scenario he still gets a positive return. He has to decide which fraction $\alpha$ to invest in asset $A$ and which fraction $1-\alpha$ to invest in asset $B$.

Note that if returns are the same for both assets, the fraction invested on each is irrelevant. Then, his problem can be written as follows,

$$
\begin{aligned}
\max \_{\alpha} B= & \frac{1}{10} \sqrt{1.2 \times 10,000}+\frac{1}{10} \sqrt{1.05 \times 10,000} \\
& +\frac{4}{10} \sqrt{1.2 \times \alpha 10,000+1.05 \times(1-\alpha) 10,000} \\
& +\frac{4}{10} \sqrt{1.05 \times \alpha 10,000+1.2 \times(1-\alpha) 10,000}
\end{aligned}
$$

Let's simplify the problem first.

Since there are two situations in which $\alpha$ does not affect the outcome, we can rewrite the problem as follows,

$$
\begin{aligned}
\max \_{\alpha} \tilde{B}= & \frac{4}{10} \sqrt{1.2 \times \alpha 10,000+1.05 \times(1-\alpha) 10,000} \\
& +\frac{4}{10} \sqrt{1.05 \times \alpha 10,000+1.2 \times(1-\alpha) 10,000} \\
\max \_{\alpha} B= & \frac{4 \sqrt{10,000}}{10}[\sqrt{1.2 \alpha+1.05(1-\alpha)}+\sqrt{1.05 \alpha+1.2(1-\alpha)}]
\end{aligned}
$$

Then, we can just solve,

$$
\max \_{\alpha} B=\sqrt{1.05+0.15 \alpha}+\sqrt{1.2-0.15 \alpha}
$$

The first order condition is given by:

$$
\frac{\partial B}{\partial \alpha}=0.15(1.05+0.15 \alpha)^{-\frac{1}{2}}-0.15(1.2-0.15 \alpha)^{-\frac{1}{2}}=0
$$

Then,

$$
\begin{aligned}
(1.05+0.15 \alpha)^{-\frac{1}{2}} & =(1.2-0.15 \alpha)^{-\frac{1}{2}} \\
1.05+0.15 \alpha & =1.2-0.15 \alpha \\
0.3 \alpha & =0.15
\end{aligned}
$$

Then, the $\alpha=0.5$.

This is not surprising, given that the assets are symmetric. Let us finally verify that the second order condition holds:

$$
\frac{\partial^{2} B}{\partial \alpha^{2}}=-\frac{1}{2} 0.15^{2}(1.05+0.15 \alpha)^{-\frac{3}{2}}-\frac{1}{2}(-0.15)^{2}(1.2-0.15 \alpha)^{-\frac{3}{2}}<0
$$


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