Assume I wanted to compute the alpha (in CAPM sense, i.e. $r_i - r_f = \alpha_i + \beta_i(r_m - r_f) + \epsilon_i$) of a stock. So I take, say, monthly returns of a stock $i$ for 1 year. Subtract the riskfree rate from this and call this $Y_i$. Now I take the corresponding benchmarket return and similarly i subtract the risk free rate from this and call it $X_i$. Now, to compute $\alpha_i$ I perform a linear regression for the model: $Y_i=\alpha_i + \beta_i X_i + \epsilon_i$

Now, if I wanted to compute $\alpha$ for an equally weighted portfolio of stocks listed on different exchanges in different countries (assume risk free rate equal). Then we have the portfolio return $$r_p = \frac{1}{n}\sum_{i=1}^{n}r_i = \\\frac{1}{n}\sum_{i=1}^{n}(r_f + \alpha_i + \beta_i(r_m-r_f)) = \\r_f + \frac{1}{n}\sum_{i=1}^{n}\alpha_i + \frac{1}{n}\sum_{i=1}^{n}\beta_i (r_m-r_f)$$ Can I now say that $\alpha_p = \frac{1}{n}\sum_{i=1}^{n}\alpha_i$ and $\beta_p=\frac{1}{n}\sum_{i=1}^{n}\beta_i$?

(Here $r_m$ is the benchmark for each corresponding asset, I didn't use a notation for this though..)

Also, if the portfolio strategy is a buy-and-hold with horizon 1 year before being balanced and the alphas are monthly, is it safe to annualize them in the most naive manner $(1+\alpha)^{12}-1$?

  • $\begingroup$ In your 2 questions, you make the approximations $ \exp{(\alpha)} \approx 1+\alpha$. This is the first order approximation, as all approximations it is imprecise, whether it is suitable or not depends on your needs. $\endgroup$
    – Yulia V
    May 14, 2014 at 16:46
  • $\begingroup$ Well, I was using simple returns, not exponential. $\endgroup$ May 14, 2014 at 17:14

2 Answers 2


Error term

The error term tells the difference between the theoretical and the observed values of the dependent variable. As such it is referred to the single observations. In your equation, as you say, $i$ stands for the $i$-th share, therefore the meaning of $\varepsilon_i$ is unclear (as it can't be the $i$-th observation) and the related equations too. Perhaps you should add to the given equation(s) a second index, say $j$, related to the $j$-th observation.

CAPM and alpha

There is a problem with your question in that the CAPM assumes alphas to be zeros. Your claims is trivially true since the average of many zeros will still be zero, but perhaps this is not what you are looking for.

CAPM test

CAPM assumptions can be violated, therefore one might want to test whether in actual markets the share alphas are zeros or not. When alphas are not zero, you may wonder what is the relation between their value with respect to a portfolio and its single constituents.

Portfolio alpha with a linear model

Consider a generic linear model relating a portfolio and the market return in excess of the risk-free rate: $$ r_{pj} -r_f = \alpha_p + \beta_p (r_{mj} -r_f ) + \varepsilon_{pj} $$ where the equation is related to the $j$-th observation of the portfolio and market excess return.

The same model with respect to the $i$-th portfolio constituent is: $$ r_{ij} -r_f = \alpha_i + \beta_i (r_{mj} -r_f ) + \varepsilon_{ij} $$ The portfolio is equally weighted, therefore (for each $j$-th observation) \begin{align} r_{pj}=\frac{1}{n}\sum_j^n r_{ij} \tag{*}\label{*} \end{align} By means of some statistical methods one can find the estimated alphas and betas, that is the "best" $\alpha_*, \beta_*$ to minimise the $\varepsilon$-errors between theoretical and observed return values. This is normally found solving: $$\min_{\alpha_i,\,\beta_i} \sum_{j=1}^n \hat{\varepsilon}_{ij}^{\,2} = \min_{\alpha_i,\,\beta_i} \sum_{j=1}^n \left( r_{ij} -r_f - \alpha_i - \beta_i (r_{mj} -r_f ) \right) $$ and $$\min_{\alpha_p,\,\beta_p} \sum_{j=1}^n \hat{\varepsilon}_{pj}^{\,2} = \min_{\alpha_p,\,\beta_p} \sum_{j=1}^n \left( r_{pj} -r_f - \alpha_p - \beta_p (r_{mj} -r_f ) \right) $$ For a general linear model, $ y = \alpha + \beta x$, the solution (estimator) is known to be (see for example here): \begin{align} \tag{**}\label{**} \hat\beta = \frac{ \sum\limits_{j=1}^{N} (x_{j}-\bar{x})(y_{j}-\bar{y}) }{ \sum\limits_{j=1}^{N} (x_{j}-\bar{x})^2 } \end{align} where $\bar{*}$ is the sample mean, e.g. (replacing summation dummy to avoid name clash): $$ \bar{x}=\frac{1}{N}\sum_h^N x_h $$ Substituting to \eqref{**} our excess returns, with respect to the $i$-th share beta, we get: \begin{align} \hat\beta_i &= \frac{ \sum\limits_{j=1}^{N} \left( r_{mj} -r_f - \frac{1}{N}\sum\limits_h^N (r_{mh} -r_f) \right) \left( r_{ij} -r_f - \frac{1}{N}\sum\limits_h^N (r_{ih} -r_f) \right) } { \sum\limits_{j=1}^{N} \left( r_{mj} -r_f -\frac{1}{N}\sum\limits_h^N (r_{mh} -r_f) \right)^2 } \notag\\ &=\frac{ \sum\limits_{j=1}^{N} \left( r_{mj} - \frac{1}{N}\sum\limits_h^N r_{mh} \right) \left( r_{ij} - \frac{1}{N}\sum\limits_h^N r_{ih} \right) } { \sum\limits_{j=1}^{N} \left( r_{mj} -\frac{1}{N}\sum\limits_h^N r_{mh} \right)^2 } \notag \end{align} As for the portfolio beta, we have: \begin{align} \hat\beta_p &=\frac{ \sum\limits_{j=1}^{N} \left( r_{mj} - \frac{1}{N}\sum\limits_h^N r_{mh} \right) \left( r_{pj} - \frac{1}{N}\sum\limits_h^N r_{ph} \right) } { \sum\limits_{j=1}^{N} \left( r_{mj} -\frac{1}{N}\sum\limits_h^N r_{mh} \right)^2 } \notag \end{align} Replacing the portfolio return definition from \eqref{*}, we obtain: \begin{align} \hat\beta_p &=\frac{ \sum\limits_{j=1}^{N} \left( r_{mj} - \frac{1}{N}\sum\limits_h^N r_{mh} \right) \left( \frac{1}{n}\sum\limits_i^n r_{ij} -\frac{1}{N}\sum\limits_h^N \frac{1}{n}\sum\limits_i^n r_{ih} \right) } { \sum\limits_{j=1}^{N} \left( r_{mj} -\frac{1}{N}\sum\limits_h^N r_{mh} \right)^2 } \notag\\ &=\frac{ \sum\limits_{j=1}^{N} \left( r_{mj} - \frac{1}{N}\sum\limits_h^N r_{mh} \right) \frac{1}{n}\sum\limits_i^n \left( r_{ij} -\frac{1}{N}\sum\limits_h^N r_{ih} \right) } { \sum\limits_{j=1}^{N} \left( r_{mj} -\frac{1}{N}\sum\limits_h^N r_{mh} \right)^2 } \notag\\ &= \frac{1}{n}\sum\limits_i^n \frac{ \sum\limits_{j=1}^{N} \left( r_{mj} - \frac{1}{N}\sum\limits_h^N r_{mh} \right) \left( r_{ij} -\frac{1}{N}\sum\limits_h^N r_{ih} \right) } { \sum\limits_{j=1}^{N} \left( r_{mj} -\frac{1}{N}\sum\limits_h^N r_{mh} \right)^2 } = \frac{1}{n}\sum\limits_i^n \hat\beta_i \notag \end{align}

As for alpha the general estimator, this is: $$ \hat\alpha = \bar{y} - \hat\beta\,\bar{x} $$ Therefore: $$ \hat\alpha_i = \frac{1}{N}\sum\limits_h^N (r_{ih} -r_f) - \hat\beta_i\,\frac{1}{N}\sum\limits_h^N (r_{mh} -r_f) $$ and $$ \hat\alpha_p = \frac{1}{N}\sum\limits_h^N (r_{ph} -r_f) - \hat\beta_p\, \frac{1}{N}\sum\limits_h^N (r_{mh} -r_f) $$ Replacing the portfolio return definition from \eqref{*}: \begin{align} \hat\alpha_p &= \frac{1}{N}\sum\limits_h^N \left(\frac{1}{n}\sum\limits_i^n r_{ih} -r_f\right) - \frac{1}{n}\sum\limits_i^n \hat\beta_i \frac{1}{N}\sum\limits_h^N (r_{mh} -r_f) \notag\\ &= \frac{1}{N}\sum\limits_h^N \frac{1}{n}\sum\limits_i^n \left( r_{ih} -r_f\right) - \frac{1}{n}\sum\limits_i^n \hat\beta_i \frac{1}{N}\sum\limits_h^N (r_{mh} -r_f) \notag\\ &= \frac{1}{n}\sum\limits_i^n \left( \frac{1}{N}\sum\limits_h^N ( r_{ih} -r_f ) - \hat\beta_i \frac{1}{N}\sum\limits_h^N (r_{mh} -r_f)\right) = \frac{1}{n}\sum\limits_i^n \hat\alpha_i \notag \end{align}

This is just to give you a general idea of the problem.


Basically you have two equations as follows: -Regression:

$$R_i = α + β \cdot R_m + e_i$$ -CAPM equation:

$$E(R_i) = r_f + β\left[E(R_m) - r_f\right]$$

In CAPM sense, there is no α . It only exists as idiosyncratic return.

You would need lot more than one year of data to estimate the coefficients for regression. You can check p-values to see if the coefficients are significant ie less than or equal to .05. Since you will be doing the multivariate OLS your data requirement will increase even more.

Actually your portfolio consisting of equal weights of securities can mislead thinking. Remember there is diversification effects in a portfolio; portfolio volatility will be smaller, beta of the portfolio will also change. This will lead to a different alpha at the portfolio level. In other words, you cannot just average the alpha of assets to get portfolio alpha.

  • $\begingroup$ Please use math formatting next time.... $\endgroup$
    – SRKX
    May 17, 2014 at 14:52

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.