Suppose you have $Var(Y_i|X_i)=\sigma^2, \forall i$, so that the conditional variance of $Y$ given $X$ is constant.

However, as long as the conditional expectation function is not linear, this does not imply that the residuals of the linear regression are homoskedastic.

  • To see this is sufficient to write (in the case of one regressor plus the constant): $$E[(Y_i-\beta_0-\beta_1X_i)^2|X_i]=E\{[(Y_i-E[Y_i|X])+(E[Y_i|X]-\beta_0-\beta_1X_i)]^2|X_i\}= Var[Y_i|X_i]+(E[Y_i|X_i]-\beta_0-\beta_1X_i)^2$$

which is in general different from $Var[Y_i|X_i]$ as long as $E[Y_i|X_i]$ is not linear.

  • However my concern is the following: whatever is the functional form of the CEF, one can always write $Y_i=\beta_0+\beta_1X_i+u_i$, therefore: $$Var[Y_i|X_i]=Var[(\beta_0+\beta_1X_i+u_i)|X_i]=Var[(\beta_1X_i+u_i)|X_i]\\= Var[\beta_1X_i|X_i]+Var[u_i|X_i]+2\beta_1Cov[X_i,u_i|X_i]$$ since we have (is it true?) $Var[\beta_1X_i|X_i]=0$ and $2\beta_1Cov[X_i,u_i|X_i]=0$, we conclude that $\sigma^2=Var[Y_i|X_i]=Var[u_i|X_i]$ so that errors must be homoskedastic too.
  • $\begingroup$ Since the two conclusion cannot be true at the same time, either I am aplying two different concepts in the two derivations, or I am simply making some confusion somewhere in the computations $\endgroup$
    – GabMac
    Sep 29, 2017 at 16:05
  • $\begingroup$ In the first bullet, you seem to have forgotten that to compute the variance, you must subtract the squared expectation from the expectation of the square. Thus, this question seems predicated on confounding a raw second moment with the variance. $\endgroup$
    – whuber
    Sep 29, 2017 at 16:41

1 Answer 1


When you calculate the mean squared error of the residuals from a biased model, you must remember that MSE = variance + bias^2.

You are right you can express a working "mean model" (possibly misspecified) for $Y$ as $E[Y|X] = \beta_0 + \beta_1 X$. I use the expectation notation here, even when the mean model is possibly wrong, because the parameters $\beta_0$ and $\beta_1$ are well defined, and extend from the usual mathematical expression of expectation.

You can make a somewhat stronger expression than you've stated. Using the working linear model, you can say the following: $Y = \beta_0 + \beta_1 X + e$ with $\mathbb{E}[e] = 0$ and $\mathbb{E}(eX) = 0$. The problem is that these conditions depend on the "design", meaning that we must take expectation over all the values of $X$, which is why I use the empirical DF notation here. For any particular observation, $i$, however it is not necessarily true that $E(e_i) =0$ nor is it true that $\text{Cov}(e_i, X_i)=0$. Suppose the truth is $Y = f(X) + e$, $e \sim_{iid} F$ with $\int_{-\infty}^\infty x F_x = 0$

Note I have used the squared expectation rather than the variance expression because while $X$ is not a random variable, its design impacts the variance.


\begin{eqnarray} E\left|Y- \beta_0 - \beta_1 X\right|^2 &=& E\left|Y- f(X) + f(X) - \beta_0 + \beta_1 X\right|^2 \\ &=& E\left|Y- f(X)\right|^2 + E\left|f(X) - \beta_0 - \beta_1 X\right|^2\\ \end{eqnarray}

The reason why we must use the squared expectation here to correctly account for the addition to the residual error is the following: at each instance of $X_i$, the observed value of $Y_i$ will vary about a constant but that constant is not given by $\beta_0 + \beta_1 X_i$. So each computed residual will be offset by $f(X) - \beta_0 - \beta_1$ and this squared value will be a contribution to the residual. Put in other ways: the mean squared error will be the squared bias $E\left|f(X) - \beta_0 - \beta_1X\right|^2$ plus the variance $E\left|Y - f(X)\right|^2$.


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