# Covariance Between Two Frontier Portfolios

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.

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}