# Individual invest on each asset

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?

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:

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