The CAPM model states that the returns of a stock are-

$r_s=r_f+\beta (r_m-r_f)+\varepsilon_s$

The $\beta$ defined above is then calculated as $\frac{cov(r_s,r_m)}{var(r_m)}$. My question is regarding this formula. This is the regression coefficient if the intercept is NOT set to 0, which is not the case in CAPM since the intercept is set to a constant $r_f$, which is essentially regressing $r_s-r_f$ against $r_m-r_f$ while setting the intercept to 0. That should yield $\beta=\frac{E[(r_s-r_f)(r_m-r_f)]}{E[(r_m-r_f)^{2}]}$, which is not equal to the canonical form.

Please let me know what is the issue here.

  • $\begingroup$ This question seems to have been asked already on the Stats Stack Exchange - stats.stackexchange.com/questions/232839/… The crux of the answers is that CAPM is an economic theory which suggests that the OLS regression between the stock's and the market's excess returns will have a 0 intercept. In practice that translates to performing the regression to estimate the $\beta$, and then ignoring the intercept. $\endgroup$ Apr 2, 2018 at 4:33

3 Answers 3


If you really believed the CAPM's prediction that $\alpha=0$, then imposing $\alpha=0$ in your estimation would indeed lead to your 2nd formula.

The problems?

  • The CAPM doesn't work so imposing a false restriction during estimation is problematic.
  • More generally, taking factor models extremely seriously and imposing $\alpha=0$ in estimation to gain efficiency loses you some robustness because factor models are almost certainly at least somewhat misspecified.

Empirical researchers generally don't restrict a constant to zero during estimation.

Model 1 (without a constant):

Let's assume we have the following regression model (without a constant):

$$ r_{st} - r_{ft} = \beta_1 \left( r_{mt} - r_{ft} \right) + \epsilon_t$$

Assuming the orthogonality condition $\operatorname{E}\left[\epsilon_t \left( r_{mt} - r_{ft}\right)\right] = 0$, then $\beta_1$ would be given by:

$$ \beta_1 = \frac{\operatorname{E}\left[\left( r_{st} - r_{ft} \right)\left(r_{mt} - r_{ft} \right) \right] }{\operatorname{E}\left[\left(r_{mt} - r_{ft}\right)^2\right]}$$

If you really take the CAPM theory seriously, then there is something principled to imposing the restriction $\alpha= 0$ in estimation (which is what we did above). Quoting Cochrane (2004) with regards to more general factor models with normally distributed errors, "The maximum likelihood estimate of $\beta$ is the OLS regression without a constant." As Cochrane describes though, researchers don't generally estimate without a constant because it sacrifices some robustness.

Model 2 (add a constant):

$$ r_{st} - r_{ft} = \alpha_2 + \beta_2 \left( r_{mt} - r_{ft} \right) + \epsilon_t$$

Now with $\alpha_2$ there and assuming the orthogonality conditions $\operatorname{E}[\epsilon_t] = 0$ and $\operatorname{E}\left[\epsilon_t \left( r_{mt} - r_{ft}\right)\right] = 0$, you get:

$$ \beta_2 = \frac{\operatorname{Cov}\left( r_{st} - r_{ft} , r_{mt} - r_{ft} \right) }{\operatorname{Var}\left( r_{mt} - r_{ft} \right)}$$

Model 1 is a special case of Model 2 where $\alpha $ is restricted to 0.

Model 3 (if the risk free rate weren't random):

If the risk free rate isn't random then it drops out:

$$ \beta_3 = \frac{\operatorname{Cov}\left( r_{st}, r_{mt} \right) }{\operatorname{Var}\left( r_{mt} \right)}$$

In periods like the present where the risk free rate is constantly about 0, maybe this bogus assumption is innocuous. I think it's hand-wavy, intro MBA type stuff though.

A comment on the CAPM

Be aware that the CAPM is a zombie theory: long ago shot dead in academia because it doesn't work, the CAPM continues to skulk the earth. Quoting Fama and French (2004), "... the empirical record of the model is poor—poor enough to invalidate the way it is used in applications."


Cochrane, John. 2005. Asset Pricing, p. 273

Fama, Eugene, F., and Kenneth R. French. 2005. "The Capital Asset Pricing Model: Theory and Evidence." Journal of Economic Perspectives, 18 (3): 25-46.

  • 2
    $\begingroup$ What I like about this answer is that explicitly reintroduces the $\epsilon_t$ random terms that were omitted in the original post. What you believe about these terms is key to how you would want to estimate $\beta$. Many people do not believe that these terms average to exactly zero i.i.d in a finite sample and that is why they estimate a separate $\alpha$, which is then thrown away. $\endgroup$
    – Alex C
    Apr 3, 2018 at 17:51
  • $\begingroup$ @AlexC My experience is that sloppily notation is often a symptom or source of confusion. When someone is learning, it's especially important to get all the details right. $\endgroup$ Apr 3, 2018 at 17:59

I am going to start off by saying that your initial equation is not the correct formulation for CAPM. It should be:

$E(r_s) = r_f + \beta (E(r_m) – r_f)$.

In other words, this is a relationship between the expected returns of the security and the expected returns of the market and is a prediction of the CAPM (the Security Market Line). Rewriting the equation like this highlights that there are two separate calculations here and the two should not be mixed.

In the first calculation, $\beta$ is the regression coefficient for a particular security’s returns ($r_s$) on the market returns ($r_m$). That $\beta$ is a function of our assumed distribution of security returns. Typically, it would be calculated using OLS from a time series of returns.

In the second calculation, the relationship between $\beta$ and the expected returns is a prediction of the CAPM. This prediction is only true if the CAPM theory applies and the benchmark is a proper representation of the market. In that case, $E(r_s) = r_f + \beta (E(r_m) – r_f)$. As you can see, this equation deals with the expected returns, whereas the calculation of $\beta$ deals with the full distribution of returns. Typically, you would estimate $E(r_m)-r_f$ by using a universe of securities and regressing their average returns on their $\beta$'s.


If you do a linear regression like $r_s - r_f = \beta (r_m - r_f) $ Then $\beta$ is calculated as $\beta = \frac { Cov(r_m - r_f, r_s-r_f)}{var (r_m-r_f)} $. Using colinearity, and the fact that $r_f$ is not random yields $\beta = \frac{cov(r_s,r_m)}{var(r_m)}$.

  • 1
    $\begingroup$ If the intercept is 0 then $\beta$ is calculated as in thequestion, not as in your answer $\endgroup$
    – zer0hedge
    Apr 1, 2018 at 7:40
  • 1
    $\begingroup$ The OP makes a very solid argument. The best estimate of $\beta$ GIVEN that you know the y intercept seems to be the one involving $r_f$ $\endgroup$
    – dm63
    Apr 1, 2018 at 16:01
  • $\begingroup$ Look up the part where $\beta$ is calculated at en.m.wikipedia.org/wiki/Simple_linear_regression under "Linear regression without the intercept term". $\endgroup$ Apr 1, 2018 at 16:06

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