3
$\begingroup$

Based on the definitions of A, B, C, and D in "An Analytic Derivation Of The Efficient Portfolio Frontier" by Robert Merton (1972), how can I prove the following in a line-by-line derivation?

$cov({x}_{p},{x}_{q}) = {x}_{p} \Omega {x}_{q} = \frac{C}{D}\left [E({R}_{p}) - \frac{A}{C} \right ]\left [E({R}_{q}) - \frac{A}{C} \right ] + \frac{1}{C}$

where the term on the left is the covariance between two given frontier portfolios.

$\endgroup$

1 Answer 1

5
$\begingroup$

Let $\Sigma$ denote the covariance matrix of our asset universe, $\mu$ is the vector of expected returns. Further, $\mathbb{1}$ is a vector of ones. Let's identify the vector of the minimum variance portfolio's asset weights with $w_0$ and that of the tangency portfolio with $w_m$. Let's further identify

$$ \begin{align} a&\equiv \mathbb{1}^T\Sigma^{-1}\mathbb{1}\\ b&\equiv \mathbb{1}^T\Sigma^{-1}\mathbb{\mu}\\ c&\equiv \mathbb{\mu}^T\Sigma^{-1}\mathbb{\mu}\\ d&=ac-b^2 \end{align} $$

Canonically, the minimum variance portfolio's optimal weights are

$$ w_0=\frac{\Sigma^{-1}\mathbb{1}}{\mathbb{1}^T\Sigma^{-1}\mathbb{1}}=\frac{\Sigma^{-1}\mathbb{1}}{a} $$

Its expected return is $\mathrm{E}(R_0)=w_0^T\mu=b/a$, its variance is $\mathrm{Var}(R_0)\equiv \sigma_0^2=w_0^T\Sigma w_0=1/a$.

The weights of the tangency portfolio are

$$ w_m=\frac{\Sigma^{-1}\mathbb{\mu}}{\mathbb{1}^T\Sigma^{-1}\mathbb{\mu}}=\frac{\Sigma^{-1}\mathbb{\mu}}{b} $$

Its expected return is $\mathrm{E}(R_m)=w_m^T\mu=c/b$, its variance is $\mathrm{Var}(R_m)\equiv \sigma_m^2=w_m^T\Sigma w_m=c/b^2$. The covariance between the two is $\sigma_{m,0}=1/a=\sigma_0^2$

For any portfolio on the efficient frontier, $R_i$, its expected return is a combination of these two portfolios (or any other two portfolios):

$$ \begin{align} \mathrm{E}(R_i)&=\alpha_i\mathrm{E}(R_m)+(1-\alpha_i)\mathrm{E}(R_0)\\ \Rightarrow\qquad \alpha_i&=\frac{\mathrm{E}(R_i)-\mu_0}{\mu_m-\mu_0}\\ &=\frac{\mathrm{E}(R_i)-b/a}{c/b-b/a}\\ &=\frac{ab}{d} \left(\mathrm{E}(R_i)-b/a\right) \end{align} $$

We can now calculate the covariance as

$$ \begin{align} \mathrm{Cov}(R_i,R_j)&=\left[\alpha_i w_m+\left(1-\alpha_i\right)w_0\right]\Sigma\left[\alpha_j w_m+\left(1-\alpha_j\right)w_0\right]\\ &=\alpha_i\alpha_j w_m^T\Sigma w_m\\ &+\alpha_i(1-\alpha_j)w_m^T\Sigma w_0\\ &+(1-\alpha_i)\alpha_j w_0^T\Sigma w_m\\ &+(1-\alpha_i)(1-\alpha_j)w_0^T\Sigma w_0\ \end{align} $$

From here on, we can plug in the values for $\alpha_i,\alpha_j$, substitute the (co-)variances and arrive at

$$ \begin{align} \mathrm{Cov}(R_i,R_j)&=\sigma_0^2+\alpha_i\alpha_j (\sigma_m^2-\sigma_0^2)\\ &=\frac{1}{a}+\frac{a^2b^2}{d^2}\left(\frac{c}{b^2}-\frac{1}{a}\right)\left(\mathrm{E}(R_i)-b/a\right)\left(\mathrm{E}(R_j)-b/a\right)\\ &=\frac{1}{a}+\frac{a}{d}\left(\mathrm{E}(R_i)-\frac{b}{a}\right)\left(\mathrm{E}(R_j)-\frac{b}{a}\right) \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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