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The Capital Asset Pricing model proposes that, $$ R_i=R_f+\beta(R_m-R_f) $$

where $R_i$ is the return of the i-th asset, $R_f$ is the risk-free rate and $R_m$ is the Market returns. $\beta$ is generally estimated using simple linear regression, but the covariates are random variables which violates the assumptions of regression.

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First, the CAPM does not state that. It states that (using your notation) $$ \mathbb{E}(R_i)=R_f+\beta(\mathbb{E}(R_m)-R_f). $$ Second, $$ \beta:=\frac{ \text{Cov}(R_i,R_m) }{ \text{Var}(R_m) } $$ is well defined when $R_i$ and $R_m$ are random variables, as long as $\text{Var}(R_m)$ exists. Also, having a random covariate is not a violation of regression assumptions in general. There is no single set of regression assumptions. There are different sets for obtaining different properties of different estimators. This is a related thread that goes into more detail.

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  • $\begingroup$ Thank you for your answer. This estimator of $\beta$ arises as the least squares solution of $R_{i,t}=\alpha_i+\beta(R_{m,t})+\epsilon_t$. One can indeed forego the assumption of covariates being fixed, but then one would have to pose it as a measurement error model. The second answer in the link you shared mentions that. $\endgroup$
    – Who cares
    Commented Apr 3, 2023 at 9:27
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    $\begingroup$ @Whocares, might you be confusing (however slightly) the CAPM with the definition of $\beta$? For estimation of $\beta$, the regression equation in your comment is spot on. It allows estimating $\beta$ and is not problematic from a statistical perspective. For CAPM, it replaces $\mathbb{E}(R_m)$ with $R_{m,t}$, and that indeed introduces measurement error. So use it for estimating $\beta$ directly, but do not use it for implementing the CAPM without taking care of the measurement error. $\endgroup$
    – Richard Hardy
    Commented Apr 3, 2023 at 9:36
  • $\begingroup$ @Whocares, we do treat $R_{m,t}$ as a random variable, but that does not introduce measurement error for $\beta$. Why do you think it should contain measurement error? $\beta$ is defined based on $R_i$ and $R_m$, and we measure both of these precisely (no error). $\endgroup$
    – Richard Hardy
    Commented Apr 3, 2023 at 9:56
  • $\begingroup$ Since $R_{m,t}$ is a random variable, we can break it down as $R_{m,t}=u+\gamma_t$, where $u$ is a constant and $\gamma_t$ is a random variable with mean 0. So $\epsilon_t$ and $R_{m,t}$ may be correlated. It is in this sense that it carries a measurement error, bias and maybe even consistency of $\hat{\beta}$ is lost. $\endgroup$
    – Who cares
    Commented Apr 3, 2023 at 10:12
  • $\begingroup$ @Whocares, no, in the regression equation for $\beta$ the error term is defined to be uncorrelated with $R_{m,t}$. Otherwise the regression coefficient would not be $\beta$, it would be something else. $\endgroup$
    – Richard Hardy
    Commented Apr 3, 2023 at 10:20

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