Linear regression: any non-normal distribution giving identity of OLS and MLE?

Question

This question is inspired from the long discussion in comments here: How does linear regression use the normal distribution?

In the usual linear regression model, for simplicity here written with only one predictor: $$ Y_i = \beta_0 + \beta_1 x_i + \epsilon_i $$ where the $x_i$ are known constants and $\epsilon_i$ are zero-mean independent error terms. If we in addition assume normal distributions for the errors, then the usual least squares estimators and the maximum likelihood estimators of $\beta_0, \beta_1$ are identical.

So my easy question: do there exist any other distribution for the error terms such that the mle are identical with the ordinary least squaeres estimator? The one implication is easy to show, the other one not so.

Answer 1

In maximum likelihood estimation, we calculate

$$\hat \beta_{ML}: \sum \frac {\partial \ln f(\epsilon_i)}{\partial \beta} = \mathbf 0 \implies \sum \frac {f'(\epsilon_i)}{f(\epsilon_i)}\mathbf x_i = \mathbf 0$$

the last relation taking into account the linearity structure of the regression equation.

In comparison , the OLS estimator satisfies

$$\sum \epsilon_i\mathbf x_i = \mathbf 0$$

In order to obtain identical algebraic expressions for the slope coefficients we need to have a density for the error term such that

$$\frac {f'(\epsilon_i)}{f(\epsilon_i)} = \pm \;c\epsilon_i \implies f'(\epsilon_i)= \pm \;c\epsilon_if(\epsilon_i)$$

These are differential equations of the form $y' = \pm\; xy$ that have solutions

$$\int \frac 1 {y}dy = \pm \int x dx\implies \ln y = \pm\;\frac 12 x^2$$

$$ \implies y = f(\epsilon) = \exp\left \{\pm\;\frac 12 c\epsilon^2\right\}$$

Any function that has this kernel and integrates to unity over an appropriate domain, will make the MLE and OLS for the slope coefficients identical. Namely we are looking for

$$g(x)= A\exp\left \{\pm\;\frac 12 cx^2\right\} : \int_a^b g(x)dx =1$$

Is there such a $g$ that is not the normal density (or the half-normal or the derivative of the error function)?

Certainly. But one more thing one has to consider is the following: if one uses the plus sign in the exponent, and a symmetric support around zero for example, one will get a density that has a unique minimum in the middle, and two local maxima at the boundaries of the support.

Answer 2

If we define the OLS as the solution to $$\arg_{\beta_0,\beta_1}\min\sum_{i=1}^n (y_i-\beta_0-\beta_1x_i)^2$$ any density $f(y|x,\beta_0,\beta_1)$ such that $$\arg_{\beta_0,\beta_1}\min\sum_{i=1}^n \log\{f(y_i|x_i,\beta_0,\beta_1)\}=\arg_{\beta_0,\beta_1}\min\sum_{i=1}^n (y_i-\beta_0-\beta_1x_i)^2$$is acceptable. This means for instance that densities of the form $$f(y|x,\beta_0,\beta_1)=f_0(y|x)\exp\{-\omega(y_i-\beta_0-\beta_1x_i)^2\}$$ are acceptable since the factor $f_0(y|x)$ does not depend on the parameter $(\beta_0,\beta_1)$. There is therefore an infinity of such distributions.

Another setting where both estimators coincide is when the data comes from a spherically symmetric distribution, namely when the (vector) data $\mathbf{y}$ has conditional density$$h(||\mathbf{y}-\mathbf{X}\beta||)$$ with $h(\cdot)$ a decreasing function. (In this case the OLS is still available although the assumption of the independence of the $\epsilon_i$'s only holds in the Normal case.)

Answer 3

I didn't know about this question until @Xi'an just updated with an answer. There is a more generic solution. Exponential family distributions with some parameters fixed yield to Bregman divergences. For such distributions mean is the minimizer. OLS minimizer is also the mean. Therefore for all such distributions they should coincide when the linear functional is linked to the mean parameter.

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.75.6958&rep=rep1&type=pdf