Covariance Between Two Frontier Portfolios

Question

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.

Answer 1

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

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