Maximize Expected Profit and Minimize Volatility of a Stock Portofio

Question

The following question was given on a sample Quantitative Research Exam:

You are given 1 million dollars to invest in a portfolio consisting of two different stocks: A and B. Stock A has an expected return of 10 percent and a volatility (standard deviation of return) of 10 percent while stock B has an expected return of 15 percent and an expected volatility of 20 percent. The following questions are then given:

Find the investments that

1. Maximize the expected portfolio profit

2. Minimize the portfolio volatility

3. Maximize the Sharpe ratio (the ratio of the expected return to the volatility).

Here are my answers to questions 1 and 2:

1. The expected return of an investment of $x$ percent of 1 million dollars in A and $1-x$ million dollars in $B$ is $x * (0.10) + (1-x)(0.15) = 0.15 - .05*x$. This is obviously maximized if $x=0$, so that the entire 1 million dollars is invested in stock B. This somehow seems too simple to me; am I missing something? Is this correct?

2. According to the formula for variance of two stock's with given variance at this site: https://www.investopedia.com/terms/p/portfolio-variance.asp, the variance of a stock that invested x percent of 1 million in stock A and x percent of 1 million in stock B gives the following formula for the variance:

$$\sigma_{AB}^{2}(x) = x^{2}(0.10)^{2}+(1-x)^{2}(0.20)^{2}=0.05x^{2}-0.08x+0.04$$

Note that $\sigma_{AB}(x)$ is actually the volatility of the portfolio. Since minimizing $\sigma_{AB}(x)$ amounts to minimizing $\sigma_{AB}^{2}$, we can do the following: for simplicity, I will let $f(x)=\sigma_{AB}^{2}(x)$. Then $f'(x)=0.1x-0.08$, so we obtain $f'(x)=0$ if and only if $x=0.08/0.1=0.8$. Therefore, the volatility will be minimized if we invest 800,000 dollars in stock A and 200,000 dollars in stock B. The volatility given will be approximately 0.09 or 9 percent according to wolfram alpha. Is this correct?

3. In this case the Sharpe ratio will be given by the following: $S(x)=\frac{0.15-0.05x}{0.01x^{2}+(0.04)(1-x)^{2}}$. Differentiating gives $S'(x)=\frac{x^{2}-6x+4}{(x^{2}-1.6x+0.8)^{2}}$. The roots are then $x=3-\sqrt{5}$ and $x=3+\sqrt{5}$. Plugging each of these back in, we should recognize a maximum at $x=3-\sqrt{5}$ of $S=\frac{5}{8}(11+5\sqrt{5})$ (I used the assistance of wolfram alpha). $3-\sqrt{5}$ is approximately $0.7639320225$, so we should invest approximately 763,932.02 dollars in stock A and 236,067.98 dollars in stock B give a maximum Sharpe ratio of approximately 13.86. Is this correct?

Please let me know your thoughts; thanks!

Answer 1

Since minimizing () σ A B ( x ) amounts to minimizing 2 σ A B 2 , we can do the following

While this statement is correct, this doesn't mean that you can just ignore the formula you correctly write out for the volatility:

${^2_{}(x)=^2(0.10)^2+(1−)^2(0.20)^2=0.05^2−0.08+0.04}$

Thus:

${_{}(x)=\sqrt{0.05^2−0.08+0.04}}$

For part 3 of the question, when you differentiate the formula for the Sharpe ratio you've left out the square root in the denominator, which leads to the wrong answer- at least your result is different to mine when I included the square root.

Also, you haven't checked whether the stationary points you have found are maxima or minima for any part of the question, which I'd imagine would be important to include.

Sorry for the late answer but hopefully this is helpful for someone.