# Show that the variance of the market portfolio is the weighted average of the ovariances between each constituent and the market portfolio itself

Let us assume that the market portfolio consists of n assets. Given that the return of the market portfolio can be written as $$r_m = \sum_{j=1}^{n} w_jr_j$$, we have that $$\sigma^2_m = E(\sum_{j=1}^{n} w_jr_j - E(\sum_{j=1}^{n} w_jr_j))^2$$, but how do I show that $$E(\sum_{j=1}^{n} w_jr_j - E(\sum_{j=1}^{n} w_jr_j))^2 = \sum_{j=1}^{n} w_jCov(r_j,r_m)$$? If I show that the equation above is true, than I can claim that $$E(\sum_{j=1}^{n} w_jr_j - E(\sum_{j=1}^{n} w_jr_j))^2 = \sum_{j=1}^{n} w_jCov(r_j,r_m) = \sum_{j=1}^{n} w_j\beta\sigma^2_m$$

This is how I am trying to prove the result. We know that:

$$\sigma^2_m = E(\sum_{j=1}^{n} w_jr_j - E(\sum_{j=1}^{n} w_jr_j))^2= E[(w_jr_j)^2]-E^2[w_jr_j]$$

Accordingly, we may show that:

$$\sum_{j=1}^{n} w_jCov(r_j,r_m) = E[(w_jr_j)^2]-E^2[w_jr_j]$$

Now: $$\sum_{j=1}^{n} w_jCov(r_j,r_m) = \sum_{j=1}^{n} w_jE[r_jr_m]-\sum_{j=1}^{n} w_jE[r_j]E[r_m]=\sum_{j=1}^{n} w_jE[r_j\sum_{j=1}^{n} w_jr_j]-\sum_{j=1}^{n} w_jE[r_j]E[\sum_{j=1}^{n} w_jr_j]=\sum_{j=1}^{n} w_jE[\sum_{j=1}^{n} w_jr_j^2]-E[\sum_{j=1}^{n} w_jr_j]E[\sum_{j=1}^{n} w_jr_j]$$

It looks like I am not able to prove the result because $$\sum_{j=1}^{n} w_jE[\sum_{j=1}^{n} w_jr_j^2] \neq E[(w_jr_j)^2]$$

Can you help me, please?

## 1 Answer

The variance of the portfolio is $$V_p=\sum_i \sum_j w_i w_j Cov(r_i,r_j)$$

because of the properties of $$Cov(\cdot,\cdot)$$ we can rewrite this as $$V_p=\sum_i w_i Cov(r_i,\underbrace{\sum_j w_j r_j}_{R_P})$$ where $$R_p$$ is the return on the portfolio. So we have

$$V_p=\sum_i w_i Cov(r_i,R_p)$$ QED

(And it is true of every portfolio, not just the market portfolio).

• Thank you. Could I ask you what was wrong with my previous sketch proof, please? – Alchemy Jul 5 at 18:55