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?


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).

| improve this answer | |
  • $\begingroup$ Thank you. Could I ask you what was wrong with my previous sketch proof, please? $\endgroup$ – Alchemy Jul 5 '19 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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