Suppose we have a linear regression model with stochastic $X$ such that

$$ Y = \beta X + \epsilon,$$

and $X$ and $\epsilon$ follow a bivariate normal distribution with $\mu_x = \mu_{\epsilon} = 0$, unknown variances $\sigma_X^2$ and $\sigma_{\epsilon}^2$, and correlation $\rho = Corr(X, \epsilon)$. Also, let $(X_i, Y_i)$ be iid random samples for all $i \in \mathbb{N}$. How does one find the bias of the least squares estimator $\hat{\beta}$?

It seems to me that we no longer have that $\hat{\beta} = (X^TX)^{-1}X^TY$. First, $\rho = \frac{E[X\epsilon]}{\sigma_x\sigma_{\epsilon}}$, and then the needed assumption doesn't hold

$$E[\epsilon|X] \neq 0 ~~\Rightarrow~~ E[\epsilon|X] = \mu_x + \rho\frac{\sigma_x}{\sigma_{\epsilon}}(Y - \mu_{\epsilon}) = \rho Y\frac{\sigma_{x}}{\sigma_{\epsilon}} = \frac{E[X\epsilon]}{\sigma_{\epsilon}^2}Y.$$

So, I guess I would need to find the new derivation of $\hat{\beta}$ first, but am not sure how to proceed?

  • $\begingroup$ Hi: Is it stated somewhere that $X$ and $\epsilon$ are correlated ? I think you are assuming that but I'm not sure. ? $\endgroup$
    – mlofton
    Commented Mar 10, 2020 at 2:11
  • $\begingroup$ @mlofton Sorry, yes they are correlated: $\rho = Corr(X,\epsilon)$, which reduces a bit given the means are 0. $\endgroup$
    – jj8989
    Commented Mar 10, 2020 at 2:30

1 Answer 1


The least square estimator is always $\hat{\beta} = (X^TX)^{-1}X^TY$. It's just when $X$ and $\epsilon$ are correlated, the estimate is no longer unbiased.

Note that

$\hat{\beta} = (X^TX)^{-1}X^T(X\beta + \epsilon)$

$= \beta + (X^TX)^{-1} X^T\epsilon$

Thus the bias of $\hat{\beta}$ is:

$E[\hat{\beta} - \beta] = E (X^TX)^{-1} X^T\epsilon = E(\sum_i^n X_i^2)^{-1} \sum_i^n X_i \epsilon_i$

While I am not sure how to easily calculate this with finite sample size, at least we can see that when the sample size becomes large, the bias term

$(\sum_i^n X_i^2)^{-1} \sum_i^n X_i \epsilon_i = (\sum_i^n \frac{X_i^2}{n})^{-1} \sum_i^n \frac{X_i \epsilon_i}{n}$ will converge to $\frac{Cov(X, \epsilon)}{\sigma_X^2} = \frac{\rho \sigma_{\epsilon}}{\sigma_X}$

  • 1
    $\begingroup$ Nice answer. I'm pretty sure that the finite sample bias will just be the the empirical counterpart of your expression under : "Thus the bias of $\hat\beta$ is". So, one can calculate the finite sample bias by calculating the $\hat\epsilon_{i}$ and replacing the $\epsilon_{i}$ in your expression with those. Note though that, to calculate the $\hat{\epsilon}_{i}$, one would use the unbiased expression for $\hat{\beta}$, namely $(X^{T}X)^{-1} X^{T} Y$. $\endgroup$
    – mlofton
    Commented Mar 10, 2020 at 19:15
  • $\begingroup$ @mlofton I agree. But that is just an empirical estimate of the bias, and the quality of it heavily depends on the quality of your estimate of $\epsilon$. What I'm saying is that, since we are given the distributions of $X_i$'s and $\epsilon_i$'s, we should be able to work out the bias expression as a function of $\sigma_X, \sigma_{\epsilon}$ and $\rho$, without even seeing any data. $\endgroup$
    – Lii
    Commented Mar 12, 2020 at 3:25
  • $\begingroup$ Thanks Lii: Just to make sure that I understand what you're saying: Do you mean, find the expectation and variance of the expression right before " will converge to". If so, I understand and thanks again. Note that I won't even attempt doing it. I'm happy enough just knowing what the goal would be :). $\endgroup$
    – mlofton
    Commented Mar 13, 2020 at 4:32
  • $\begingroup$ @mlofton Yes, exactly $\endgroup$
    – Lii
    Commented Mar 16, 2020 at 3:04

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