# Error term, R-square and perfectly holding CAPM?

It is known that if the CAPM holds then the E(R) = CAPM predicted return and all securities lie on the SML. However, in each period, there is an error term that is non-zero in every single observation but that is 0 on average (as in any regression) In my view, if the CAPM perfectly holds and we run a regression between the market portfolio and a security, the R square should be 1, however, we would still see deviations from the line of best fit (because of the idiosyncratic risk represented by the error term). However, examples show that if the R square is 100%, then all points are on the line of best fit: .

Is my logic flawed? How would it look graphically if security returns are regressed on the mean-variance efficient market portfolio (theoretically, not empirically)? Wouldn't we see points deviating from the line of best fit (with Beta as slope)?

• What do you think about my answer? If it is helpful and clear, you may accept it by clicking on the tick mark to the left. Otherwise, you may ask for further clarification. (A helpful answer can also be upvoted by clicking on the upward-pointing arrow.) This is how Quantiative Finance SE works. Apr 17 at 10:42

The CAPM states that $$\mathrm{E}(R_i)-r_f=\beta_i[\mathrm{E}(R_m)-r_f].$$ None of these quantities are observable, excerpt for $$r_f$$. The data that is typically used in relation to the CAPM are realized returns $$r_{i,t}$$ and $$r_{m,t}$$ for $$t=1,\dots,T$$ from which we can estimate the three unknown quantities. We do not expect the estimated version of the CAPM, $$\hat{\mathrm{E}}(R_i)-r_f=\hat\beta_i[\hat{\mathrm{E}}(R_m)-r_f],$$ to hold precisely because of the estimation imprecision. Therefore, we do not expect the estimates $$\hat{\mathrm{E}}(R_i)$$ and $$\hat\beta_i$$ to lie precisely on a straight line, regardless of whether $$\mathrm{E}(R_i)$$ and $$\beta_i$$ do (i.e. regardless of whether the CAPM actually holds or not).
This is even further from the truth, because the CAPM does not say $$r_{i,t}-r_f=\beta_i[r_{m,t}-r_f]$$ for some or all $$t=1,\dots,T$$. Even if the hypothesized relationship between the expected values and the beta, $$\mathrm{E}(R_i)-r_f=\beta_i[\mathrm{E}(R_m)-r_f]$$, holds, this does not imply a perfect linear relationship for the realizations $$r_{i,t}$$ and $$r_{m,t}$$ of the random variables $$R_i$$ and $$R_m$$.