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?
 A: 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 $(\beta, \sigma)$?
A: 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.
