# Calculating the MLE for mu(x) in a regression model

Say we have the following regression model: $$Y_i = \alpha + \beta(x_i - \mathrm{mean}(x)) + R_i$$

where $R_1,\ldots,R_{20} \sim G(0, \sigma)$

If we have $\mu(x) = \alpha + \beta(x - \mathrm{mean}(x))$, how do I go about finding the MLE of $\mu(5)$?

I have a given data set with some calculations done for me, but not sure how to approach this?

-
What distribution are you calling "$G$"? – Michael Hardy Jun 28 '12 at 22:29
Whatever might be the answer to the question I posted above, if you find the MLEs for $\alpha$ and $\beta$ and plug those in to $\alpha+\beta(x-\mathrm{mean}(x))$, that should be the MLE for $\mu(x)$. MLEs have that kind of invariance (or more precisely, maybe I should call it "equivariance"). – Michael Hardy Jun 28 '12 at 22:31
How do you know this? Also G is the Gaussian distribution. – DillPixel Jun 28 '12 at 22:33
Usually $N$ or $\mathcal{N}$ is used for the normal or "Gaussian" distribution. – Michael Hardy Jun 28 '12 at 22:35

https://instruct1.cit.cornell.edu/courses/econ620/reviewm5.pdf

Look at the document above and search for "functional invariance". If the MLE for $\alpha$ is $\hat\alpha$ then the MLE for $\cos\alpha$ is $\cos\hat\alpha$, and so on. So if $\hat\alpha$ and $\hat\beta$ are the respective MLEs of $\alpha$ and $\beta$, then $8\hat\alpha+6\hat\beta$ is the MLE for $8\alpha+6\beta$, etc. That's the sort of function you have here.

This property of MLEs is quite easy to prove. You don't need calculus; you just need to know definitions of things like "increasing function" and "maximum".

-

Michael Hardy is correct. Let l(θ) be the log of the likelihood function given the observed x and the parameter θ. Then you maximized the likelihood by finding θ such that ∂/∂θ l(θ) =0. Now suppose we take the function g(θ) (in your case θ =(α, β) and g(α, β)=α+β(x−mean(x)). Consider l(g(θ)). This is maximized by set partial derivatives with respect to α and β to 0. For simplicity lets look at the one parameter case. ∂/∂θ l(g(θ))= ∂/∂θ l(g) ∂/∂θ g(θ) by the chain rule. The loglikelihood is has partial derivative 0 for g(θ) at the same theta that the loglikeihood for theta has partial derivative =0. Hence the maximum likelihood estimator for a differentiable function of theta is maximized at the function evaluated at the maximum likelihood estimator for theta.

-
But as I noted, no calculus is needed to prove this property of MLEs. And in fact it still works in cases where differentiability does not exist---e.g. suppose the parameter space is discrete. – Michael Hardy Jun 28 '12 at 22:59
I was just trying to give a simple proof to convey the idea. Did not intend to say that it was in full generality. – Michael Chernick Jun 28 '12 at 23:33