2
$\begingroup$

I practice with some excercises about the Markowitz theory. If we have a portfolio with two stocks A and B, with given return $r_A$ and $r_B$, the expected return can be computed as: $r_P= w_A \cdot r_A + (1-w_A) \cdot r_B$, where $w_A$ is the weight invested in stock A.

For a given risk free rate ($r_f$), I now have to compute the return of the one fund of the one fund theorem. What I am exactly supposed to do here? I know about the one fund theorem, but I don't know what to do. Any hints?

$\endgroup$
1
$\begingroup$

The Sharpe ratio tells us the amount of excess return we get for taking on each additional unit of portfolio standard deviation. $$\frac{\mu_p - r_f}{\sigma_p }$$

We are looking for the combination of the two risky assets with the highest Sharpe ratio ($P^*$). Once we do that, we can take linear combinations of that portfolio and the risk-less asset and form the Capital-Market Line. This is usually solved for numerically rather than analytically but it is possible to do so analytically, particularly in the two asset case.

A portfolio $p$ has an expected return of: $$\mu_p(W_A) = w_A \cdot \mu_A + (1 - w_A) \cdot \mu_B $$

and a standard deviation of: $$\sigma_p(W_A) = \sqrt(w^2_A \cdot \sigma^2_A + (1 - w_A)^2 \cdot \sigma^2_B + 2(1-W_A)W_A \sigma_{AB}) $$ where $\sigma^2_A$ is the variance of asset $A$, $\sigma^2_B$ is the variance of asset $B$, and $\sigma_{AB}$ is their covariance. It therefore has a Sharpe Ratio of:

$$\frac{\mu_p(W_A) - r_f}{\sigma_p(W_A) } = \frac{w_A \cdot \mu_A + (1 - w_A) \cdot \mu_B}{+\sqrt(w^2_A \cdot \sigma^2_A + (1 - w_A)^2 \cdot \sigma^2_B + 2(1-W_A)W_A \sigma_{AB})}$$

To maximize this you'll want to solve: $$ \frac{d}{dW_A} \frac{\mu_p(W_A) - r_f}{\sigma_p(W_A) } = 0 $$ $\Rightarrow W^{*}_A$ s.t. $P(W^{*}_A)=P^*$ and check that the second order condition is met: $$ \frac{d^2}{dW^2_A} \frac{\mu_p(W_A) - r_f}{\sigma_p(W_A) } < 0$$

The algebra is a bit hairy but there is nothing tricky from here on out.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.