MLE of a function of a parameter I am working on a problem where we are interested in finding the MLE for a function of two parameters.
I am having problems with going about finding this. Intuitively, the idea makes sense. I am just wondering about the definition of the MLE of a function of two parameters (Google isn't turning up much). The question is as follows:
Question: Suppose that $X_1,\ldots,\,X_n$ are iid $N(\mu, \sigma^2)$ with unknown $\mu,\sigma^2$. Find the MLE for $\frac{\mu}{\sigma}$. 
Note that this is not just a homework problem, but part of a take home final. I really am not looking for much of an answer, but more or less the idea for such problems. 
Edit
Apparently MLE's are invariant under function. TY
 A: The question was answered with a link in a comment, so let me just give here the argument from the link, for future completeness.  
We assume a statistical model for data $X$ is parameterized by a parameter $\theta$ (which can be scalar or vector, or even more general).  Let the likelihood function be $L(\theta)$ and the value of $\theta$ maximizing that be the maximum likelihood estimator $\hat{\theta}$ (mle). We do assume that estimator exists and is unique. Wanted is the mle of $g(\theta)$, a function of $\theta$. First we assume that $g$ is one-to-one. Then we can write
$$
    L(\theta) = L(g^{-1}(g(\theta))
$$
and both functions are clearly maximized by $\hat{\theta}$, so
$$
   \hat{\theta} = g^{-1}(\hat{g(\theta)})
$$  or
$$
  g(\hat{\theta}) = \hat{g(\theta)}  
$$
If $g$ is many-to-one, then $\hat{\theta}$ which maximizes $L(\theta)$ still corresponds to $g(\hat{\theta})$, so $g(\hat{\theta})$ still corresponds to the maximum of $L(\theta)$. (Argument paraphrased from the link in the comment above).
A: I just came up with a solution that I personally like more (for wide range of reasonable estimators):
Assume that $f$ is convex (likelihood function, the solution is unique) and $A$ is monotonous (function of parameter, $\exists A^{-1}$), then $f(A(x))$ is also convex.
What we want to prove is that 
$$ \arg\max_{x\in X} f(A(x)) = A^{-1}(\arg\max_{z \in A(X)} f(z))$$
Now assume that there exists such a pair $(\tilde x, \tilde z)$ s.t. they are optimal for corresponding problems but $A(\tilde x) \neq \tilde z$. 
From optimality:
$f(\tilde z) > f(z') \quad \forall z'$ and $f(A(\tilde x)) > f(A(x')) \quad \forall x'$
including $z' = A(\tilde x)$ and $x' = A^{-1}(\tilde z)$, then
$$ f(\tilde z) > f(A(\tilde x)) > f(A(A^{-1}(\tilde z)) = f(\tilde z)$$
- contradiction!
