# 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 $a,b$?

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 $$\hat{\phi}=A\hat{\theta}$$.]