I was reading that if we know a portfolios beta we can break the excess return on that portfolio into a market component and a residual component.

 r_p = beta_p * r_m + e_p

 r_p - portfolio excess return
 r_m - market return
 e_p - residual return
 beta_p - portfolio beta

It then goes on to so say, the residual return (e_p) will be uncorrelated with the market return (r_m) and so the variance of the portfolio is

  var_p = beta_p^2 * var_m + var_p_residual

  var_p - variance of portfolio
  beta_p^2 - beta of portfolio squared
  var_m - variance of market
  var_p_residual - variance of portfolio residual

So my question is how can we know the residual return will be uncorrelated with the market return?

I've found this web page which near the top it has a section titled The Key Assumption.

Consider, for example, a case in which the residual return is correlated with factor 1. By adjusting the factor exposure (bi1) appropriately, the correlation of the residual with the factor can be made to equal zero.

I'm not sure if this is linked to my question or not but I still don't follow it either


1 Answer 1


Let us ignore the riskless rate for simplicity of the presentation. If you have (historical or simulated) return series $r_i$ for the portfolio and $r_i^M$ for the market, then the beta is the OLS regression beta: $$ \beta = cov(r_i,r_i^M)/var(r_i^M). $$

Then if you write $r_i = \alpha + \beta r_i^M + \epsilon_i$ on the other hand

$$ \epsilon_i = r_i - ( \alpha + \beta r_i^M). $$ Then the covariance of these erros with the market are given as follows: $$ cov(\epsilon_i, r_i^M) = cov(r_i - ( \alpha + \beta r_i^M),r_i^M) $$ and as $cov(\alpha,r_i^M) = 0$ ($\alpha$ is a constant) we get $$ cov(r_i - ( \alpha + \beta r_i^M),r_i^M) = \\ cov(r_i,r_i^M) - \beta cov(r_i^M,r_i^M) = cov(r_i,r_i^M) - cov(r_i,r_i^M)/var(r_i^M) * var(r_i^M) =0, $$ as $cov(r_i^M,r_i^M) = var(r_i^M)$.

In a geometric sense Beta tells you the projection on the space spanned by the market. The residual is orthogonal by construction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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