Why does the maximum probability of profiting occur when std. deviations of two different stock prices are equal?

Question

I am working through the "Math for Quantitative Finance" course on brilliant.org. The following question was given as an example:

An investor wishes to invest $700.

There are two independent stocks the investor can choose to invest in, both of which are currently trading at the same share price. The daily returns of the first stock are historically normally distributed with a mean of 3% and a standard deviation of 1.5%. The daily returns of the second stock are historically normally distributed with a mean of 4% and a standard deviation of 2%.

How much should the investor choose to invest (in dollars) in the first stock to maximize his probability of having a positive profit over the course of a day?

The solution given is:

Suppose he invests $x$ dollars into the first stock and $y$ dollars into the second stock. He will have the maximum probability of profiting when the two standard deviations are equal, i.e. $1.5x = 2y$. This, together with $x+y = 700$ gives $x = 400$.

I understand how to solve algebraic systems of equations, so I understand why, given these two equations, the answer is that the investor maximizes their probability of profiting when investing $400 into the first stock.

However, I do not understand why it is the case that maximum probability of profiting occurs when the std. deviations are equal. What is the theory behind this?

Answer 1

The probability that a gaussian random variable is positive is maximized by maximizing the quantity $\frac{\mu}{\sigma}$. For an affine combination of two gaussians we have $\mu = x \mu_1 + (1-x) \mu_2$ and $\sigma^2 = x^2 \sigma_1^2 + (1-x)^2 \sigma_2^2$, so we would like to maximize $$\frac{x \mu_1 + (1-x)}{\sqrt{x^2 \sigma_1^2 + (1-x)^2 \sigma_2^2}}$$

Wolfram Alpha gives the derivative of this wrt $x$:

$$\frac{\mu_1 \sigma_2^2 x - \mu_1 \sigma_2^2 + \mu_2 \sigma_1^2x}{( \sigma_1^2x^2+\sigma_2^2(1-x)^2)^{\frac{3}{2}}}$$which if set to $0$, yields $x^* = \frac{\mu_1 \sigma_2^2}{\mu_1 \sigma_2^2 + \mu_2 \sigma_1^2}$. (Of course this is only a valid solution if $0 \leq x^* \leq 1$)

This conveniently works out to $x^* = \frac{4}{7}$ for the numbers in OP's post.

The "strategy" from OP's post seems to hold true whenever $\mu_1 \sigma_2 = \mu_2 \sigma_1$. If you work out the algebra, you'll find that $x^* \sigma_1 = (1-x^*) \sigma_2$ in this case.

Answer 2

Assuming that the historical return and volatility are good estimates for expected return and volatility, the proposed answer is wrong (or at least it doesn't generalize to any values used).

The problem to solve is the following: $$\max_x P(A>0) \\ \text{where }A= x\cdot R_1 + (1-x) R_2 \\ $$

$x$ is the share of your wealth invested in stock 1
$R_i$ is the return of stock $i$
$\mu_i$ is the expected return of stock i, and $\sigma_i$ is its standard deviation

Given that the stocks are independent, we have: $$A\sim N(x \mu_1 + (1-x) \mu_2, x^2 \sigma_1^2 + (1-x)^2 \sigma_2^2 ) $$

If $Z \sim N(0, 1)$, we have: $$ P(A>0) = P(\frac{A - (x \mu_1 + (1-x) \mu_2)}{\sqrt{x^2 \sigma_1^2 + (1-x)^2 \sigma_2^2}}>\frac{-(x \mu_1 + (1-x) \mu_2)}{\sqrt{x^2 \sigma_1^2 + (1-x)^2 \sigma_2^2}}) \\ = P(Z>\frac{-(x \mu_1 + (1-x) \mu_2)}{\sqrt{x^2 \sigma_1^2 + (1-x)^2 \sigma_2^2}})$$

The maximization problem is thus be equivalent to: $$ \max_x \frac{x \mu_1 + (1-x) \mu_2}{\sqrt{x^2 \sigma_1^2 + (1-x)^2 \sigma_2^2}} $$ You can see that the clearly depends on the expected returns. See that expected return has a positive effect on the amount you want to invest in a stock, and volatility (sd) has a negative effect.