An investor can choose to put his money in a savings account ($S$) with a return of $\mu_S=1.03$. Since the bank guarantees this return the investment is risk-free, hence $\sigma_S = 0$. Furthermore, the investor has the possibility to invest in a stock ($A$) with expected return $\mu_A=1.05$ and risk $\sigma_A=0.1$. We summarize this information as $(\mu_S, \sigma_S)= (1.03;0)$ and $(\mu_A, \sigma_A) = (1.05; 0.1)$. The utility function of the investor is $U(\mu,\sigma)=\mu-2\sigma^2, \ (\sigma \geq 0)$. The investor puts a fraction $w_1$ in the savings account and he invests a fraction $w_2$ in the stock. Since we assume he invests the full amount available for investment, we obtain the following restriction for the fractions $w_1$ and $w_2$: $w_1+w_2 = 1$ and $w_1 \geq 0, \ w_2 \geq 0.$ By the use of formulas from statistics (that are outside the scope of this book) the expected return and risk of this portfolio $P=w_1S + w_2A$ can be determined. We provide the result of this calculation:
$$\begin{array}{lll}
\mu&=& 1.03w_1 + 1.05w_2,\\
\sigma &=& 0.1w_2.
\end{array}
$$
Determine the optimal portfolio for this investor.
$(w_1,w_2)=(\frac{1}{20},\frac{19}{20})$
$(w_1,w_2)=(-25\frac{1}{4},26\frac{1}{4})$
$(w_1,w_2)=(\frac{1}{4},\frac{3}{4})$
$(w_1,w_2)=(\frac{1}{2},\frac{1}{2})$
Correct: Since $w_1 = 1 - w_2$ we can rewrite these expressions as
$$\begin{array}{lll}
\mu&=& 1.03 + 0.02w_2,\\
\sigma &=& 0.1w_2.
\end{array}
$$
Since $w_2 =10\sigma$ we can rewrite these two equations into one (linear) equation of $\mu$ and $\sigma$:
\[
\mu = 1.03 +0.2 \sigma.
\]
Hence, the investor should solve the following constrained maximization problem:
$$
\begin{array}{ll}
\mbox{maximize}&U(\mu,\sigma) = \mu - 2\sigma^2\\
\mbox{subject to}&\mu = 1.03 + 0.2 \sigma,\\
\mbox{where} & \sigma \geq 0.
\end{array}
$$
We use the substitution method to solve this problem.
Step 1. The restriction $\mu = 1.03 + 0.2 \sigma$ is already written as a function.
Step 2. We substitute the restriction into the utility function:
$$\begin{array}{lll}
u(\sigma) &=& U(1.03 + 0.02 \sigma , \sigma)\\
& = & 1.03 + 0.2\sigma -2 \sigma^2, \ (\sigma \geq 0).
\end{array}
$$
Step 3. We determine the maximum of $u(\sigma)$.
Hence, we have to solve the equation$u'(\sigma) = 0.2- 4 \sigma = 0$. The only stationary point is $\sigma = 0.05$ and since $u''(\sigma) = -4$ it also holds that $u''(0.05) = -4<0$, which means according to the second-order condition for an extremum that $\sigma = 0.05$ is a maximum location.
Step 4. We determine the maximum (location) of $U(\mu, \sigma)$.
By plugging $\sigma = 0.05$ into the restriction we obtain $\mu = 1.03 + 0.2\cdot 0.05=1.04$, which gives the maximum location $(\mu,\sigma)=(1.04,0.05)$. Now we can also determine the fractions. From $w_2=10\sigma$ it follows that $w_2 =\frac{1}{2}$. Hence, $w_1 =\frac{1}{2}$.
Wrong: $\sigma=\frac{1}{20}$.
Try again.
Wrong: Note that $w1\geq 0$.
See Example 1.
Wrong: This combination of $w_1$ and $w_2$ does not result in the highest utility.
Try again.