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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?

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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).

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

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