MLE of the mean of uniforms I'm solving some exercises in Wasserman's book "All of Statistics". Given $X_1,\dots,X_n \sim U(a,b)$ where $a<b$ are unknown parameters I was able to find the method of moments estimators for $a,b$ as well as the MLE of $a,b$. The next question is to find the MLE of $\tau := \int x dF(x)$.
I'm not sure if the exercise is that simple: I have my MLE estimator for $a$ and $b$ and I know given a function $\tau$ and a MLE $\hat{\theta}$ the MLE of $\tau$ is given by $\hat{\tau}=\tau(\hat{\theta})$. So is the MLE of $\tau$ given as $\frac{\hat{b}+\hat{a}}{2}$, where $\hat{a}$ and $\hat{b}$ are denoting the MLE of $a,b$?
 A: Maximum likelihood esitmators have a property of functional invariance (indeed they possess a couple of forms of functional invariance). Quoting part of that Wikipedia link:

The maximum likelihood estimator selects the parameter value which gives the observed data the largest possible probability (or probability density, in the continuous case). If the parameter consists of a number of components, then we define their separate maximum likelihood estimators, as the corresponding component of the MLE of the complete parameter. Consistent with this, if $\widehat{\theta}$ is the MLE for $θ$, and if $g(θ)$ is any transformation of $θ$, then the MLE for $α = g(θ)$ is by definition
$$\widehat{\alpha} = g(\,\widehat{\theta}\,). \, $$

So if $\theta =(\begin{smallmatrix} ^a\\ _b \end{smallmatrix})$, then the MLE of $\theta$ is $(\begin{smallmatrix} \hat{a}\\ \hat{b} \end{smallmatrix})$ and the MLE of $\frac{a+b}{2}$ is $\frac{\hat{a}+\hat{b}}{2}$.
[Based on a quick glance, Wasserman seems to only show it for one-to-one functions. However, I suppose you could always construct a pair $\phi=(\tau,\sigma)^\top$ such that $\tau$ is defined as before and $\sigma=\frac{b-a}{2}$; then $\phi=A\theta$, where $A$ is invertible and so $\hat{\phi}=A\hat{\theta}$.]
