# Mathematical proof that the covariance between two portfolios is $w_A^\top\Sigma w_B$

How to prove in a line-by-line derivation that the covariance between two mean-variance efficient portfolios is equal to

$$w_A^\top\Sigma w_B$$

where $$w_i$$ is a unique portfolio weight vector, and $$\Sigma$$ is the covariance matrix of asset returns.

Is there a source that goes over this in detail for portfolios in general, not just the minimum-variance and max Sharpe portfolios?

• What have you tried yourself so far? You can easily show this by either starting from the definition of covariance for two portfolios whose return is defined as$\sum_i w_ir_i$ or you start from your equation and simply write it out. Nov 16, 2020 at 7:06
• Is the "covariance for two portfolios" the same thing as "the covariance matrix of asset returns"? Nov 16, 2020 at 14:36
• Hint: How would you compute the return covariance of portfolios $A,b$, i.e. $Cov(r_A,r_B)$? The covariance matrix belongs to the asset returns. A portfolio's variance is then driven by its weights in the single assets. Nov 16, 2020 at 15:00
• I have the feeling we are in different languages. A good course of action could be for you to present what you have done or tried so far. Nov 16, 2020 at 16:08
• @develarist: If you take steveo america's nice answer and replace $w$ with $v$ ( IMHO, it's clearer if steveo america used variance below instead of covariance ), you get the variance of a portfolio with weights $v$.. The only difference between the two cases you referred to is the first one uses the same return vector when calculating the covariance matrix and the second one uses two different return vectors. In the end, you get a scalar in both cases. Rudd and Clasing might be a good reference for this but it's been too long to be sure. I bet steveoamerica could give us a good one ? Nov 19, 2020 at 11:16

When $$x_i$$ is the return of the $$i$$th asset, the returns of portfolio $$\vec{w}$$ are $$\sum_i w_i x_i$$. The covariance of the returns of two portfolios, $$\vec{w}$$ and $$\vec{v}$$ are then $$\sum_i \sum_j w_i v_j \operatorname{cov}\left(x_i, x_j\right).$$ Now note that $$\Sigma_{i,j} = \operatorname{cov}\left(x_i,x_j\right)$$. The rest is confirming that this expression is the bilinear form $$\vec{v}^{\top}\Sigma \vec{w}$$.