I wish to prove the following theorem:
If $\phi = g(\theta)$ is an injective function then the mle of $\phi$ is $\hat{\phi}=g(\hat{\theta})$.
This is my proof so far:
$\hat{\phi}$ = arg max {$L(\phi;x_{i})$} = arg max {$L(g(\theta);x_{i})$}.
Thus, we proceed to calculate arg max {$L(g(\theta);x_{i})$}.
\begin{align} L(g(\theta)) &=\prod_{i=1}^{n}f_{g(\theta)}(x_{i})\\ &= \prod_{i=1}^{n} f_{\theta}(g^{-1}(x_{i}))\cdot\left|\frac{d\theta}{dg(\theta)}\right| \tag{1} \end{align}
Thus,
\begin{align} \textit{l}(g(\theta))&= \sum_{i=1}^{n}\ln\left|\frac{d\theta}{dg(\theta)}\right| + \sum_{i=1}^{n} \ln(f_{\theta}(g^{-1}(x_{i}))) \\ &= \sum_{i=1}^{n}\ln\left|\frac{d\theta}{dg(\theta)}\right| + \textit{l}(\theta) \end{align}
Where (1) occurs due to injective density transformation and $\left|\frac{d\theta}{dg(\theta)}\right|$ is evaluated at $\theta=g^{-1}(x_{i})$ for $i=1,2,...,n$
It is at this point that I am unsure what the next step in the proof is.