Question

4. Standard deviation and risk. The standard deviation o(X) of a random variable is the square root of the variance that is o(X) = Var(X). It characterizes the "spread" of the random variable X. If a random variable X has expected value p and standard deviation o, then X takes values which are on average at distance o from u. Imagine you have the choice to invest in two stock funds: an American fund with a rate return X and an Asian fund with rate of return Y. The rate of return X is a random variable with mean u= 0.15 and standard deviation o(X) and the rate of return Y is a random variable with mean p = 0.15 and standard deviation (Y) with say o(Y) > 0(X). The larger the standard deviation, the riskier your investment is, since a large standard deviation makes it more likely that the rate of return is far from the u. This can be both good or bad, since the rate of return could be either much larger than the average or much smaller. Given the choice between these two investment show that the least risky investment is actually to diversify and invest a proportion a of your fortune in the American stock fund and a proportion 1-a in the Asian stock fund (assume that X and Y are independent). Determine for which proportion a you risk is minimal.

Answer 1

If we invest a proportion $\alpha$ of the fortune in the American stock which has a rate of return X with standard deviation $\sigma(X)$ and

invest a proportion $1-\alpha$ of the fortune in the Asian stock which has a rate of return Y with standard deviation $\sigma(Y)$

The variance of the return for the portfolio is

$\begin{align*} Var(\alpha X+(1-\alpha)Y)&=\alpha^2Var(X)+(1-\alpha)^2Var(Y)\quad\text{X and Y are independent}\\ &=\alpha^2\sigma^2(X)+(1-\alpha)^2\sigma^2(Y) \end{align*}$

We want to minimize this variance (that is risk) by changing the value of $\alpha$.

The first first order condition for the minimization is to equate the first derivative of the portfolio variance w.r.t $\alpha$ to 0

$\begin{align*} &\frac{\partial }{\partial \alpha}Var(\alpha X+(1-\alpha)Y)=0\\ \implies &\frac{\partial }{\partial \alpha}\left(\alpha^2\sigma^2(X)+(1-\alpha)^2\sigma^2(Y) \right )=0\\ \implies &2\alpha \sigma^2(X)-2(1-\alpha)\sigma^2(Y) =0\\ \implies &\alpha(\sigma^2(X)+\sigma^2(Y))=\sigma^2(Y)\\ \implies &\alpha=\frac{\sigma^2(Y)}{\sigma^2(X)+\sigma^2(Y)} \end{align*}$

ans: The value of $\alpha$ for which the risk is minimal is

$\begin{align*} \alpha=\frac{\sigma^2(Y)}{\sigma^2(X)+\sigma^2(Y)} \end{align*}$