What is the maximum expected Sharpe ratio by combining two assets into a portfolio?

Question

Given two assets that have expected excess returns of 7 and 4. Also, given their expected co-variance matrix $$ \begin{bmatrix} 2 & 1 \\ 1 & 1 \\ \end{bmatrix} $$

What is the maximum expected Sharpe ratio that you can achieve by combining two assets into a portfolio?

I appreciate the elaborated solution with the explanation. Thanks

Answer 1

Let A and B be the two assets. From the Covariance Matrix, you get

$\sigma_A^2 = 2$, $\sigma_B^2 = 2$ $Cov(A,B) = 1$

Expected Excess Returns, $R_A = 7$, $R_B = 4$

Let A and B be have weights $w_A$, $w_B$ in the portfolio.

Now Variance of the Portfolio $Var(P) = w_A^2\sigma_A^2+w_B^2\sigma_B^2+2w_Aw_BCov(A,B)$

$\sigma_P = \sqrt{Var(P)}$

$R_P = w_{A}R_A + w_{B}R_B$

$$Sharpe Ratio= \frac{R_P}{\sigma_P}$$

USing the above information set up a solver to find $w_A$ and $w_B$. The below image illustrates and find the optimal solution that will maximize sharpe ratio.

Answer 2

You can do by calculus also:

Take down all the definitions and let $w_A = w$ and $w_B = 1-w$

Then $Var(P) = 1+w^2$

$\sigma_P = \sqrt{1+w^2}$

$R_P = 7w + 4(1-w) = 3w+4$

Now

Sharpe Ratio $S = \frac{3w+4}{\sqrt{1+w^2}}$

If you take $\frac{dS}{dw}$ and set it to 0,

and find the above derivative using quotient rule, you will get

$w = 0.75$ and hence $1-w = 0.25$

Thus$ w_A = 0.75 $ and $w_B = 0.25$, Substitute back in S you will get the maximum Sharpe Ratio to be equal to $5.00$