# MLE estimate of $\beta/\sigma$ - Linear regression

I have a question regarding Maximum Likelihood Estimate in linear regression model without intercept. I have a model: $$Y_i=\beta X_i +\epsilon_i, \ \ i=1,...,n$$ where $\epsilon_i$ are i.i.d. $N(0, \sigma^2)$

I concluded that the log-likelihood function looks like this: $$l(\beta,\sigma)=\sum_{i=1}^n \left( \ln(\frac{1}{\sqrt{2\pi}}) -\ln(\sigma) -\frac{(y_i-(\beta x_i))^2}{2\sigma^2} \right)$$ Easy part of this question is MLE of $\sigma$ and MLE of $\beta$. But what I really don't know how to evaluate is MLE of $\frac{\beta}{\sigma}$. What crossed my mind is to only set MLE of $\frac{\beta}{\sigma}=\frac{\beta_{MLE}}{\sigma_{MLE}}$. But I believe that this is not the right solution.

Any hints?

• I have provided a hint as an answer. Also, this looks like a homework problem, so if it is, please add the [self-study] tag to ensure appropriate discussion of the solution. Commented Apr 11, 2016 at 19:25

A maximum likelihood estimator has the nice property that it is invariant under transformations. This means that if $\theta_{MLE}$ is the MLE for $\theta$, then for a function $g(\theta)$, $g(\theta_{MLE})$ is the MLE for $g(\theta)$.

This can be directly applied to your problem. Hint: what is the MLE for $(\beta, \sigma)$?

• Then, if I understand correctly, MLE of that ratio really is just the ratio of MLE estimates. Then my idea was not that far away from truth.. Thanks! And of course, next time I will include self-study tag :) Commented Apr 11, 2016 at 20:13
• Juraj, it may help future visitors if you were to mark this answer as accepted: that influences site searches, for instance. You also get a little bump in reputation for the effort :-).
– whuber
Commented Apr 13, 2016 at 20:56

Greenparker has already answered the substance of your question. I will just add that it is also a good idea to simplify your log-likelihood as much as possible before proceeding with maximisation. Removing the additive constant and simplifying the remaining sums gives:

$$l(\beta,\sigma)= -n \ln (\sigma) - \frac{1}{2 \sigma^2} \sum_{i=1}^n (y_i-\beta x_i)^2 = -n \ln (\sigma) - \frac{1}{2 \sigma^2} || \boldsymbol{y} - \beta \boldsymbol{x} ||^2 .$$

Conditional on $\hat{\sigma}$ you need only minimise the quantity $|| \boldsymbol{y} - \beta \boldsymbol{x} ||^2$. You are correct that once you have found the MLEs for the individual parameters, the MLE of their ratio is the ratio of the MLEs.