Trying to find the MLE of $\tau$ Let $\tau = \int x \,dF(x),$ and I want to find the MLE of $\tau$ given
$X_1,\ldots,X_n \sim \mathrm{Uniform}(a,b).$ I am not entirely sure, but I would imagine that $\tau = \int x \, dF(x) = \int_a^b x \,dx = \left. \frac{1}{2}x^2 \right|_a^b
= \frac{1}{2}(a^2+b^2).$ This, by equivariance of estimators, suggests to me
that
$$\hat{\tau} = \frac{1}{2}(\hat{a}^2 + \hat{b}^2),$$
and thus this problem comes down to finding the MLEs $\hat{a}$ and $\hat{b}.$
I can take the likelihood of seeing that data as
$$L(X^n,a,b) = \prod_{i=1}^n f(x_i \mid  a,b)
= \prod_{i=1}^n \frac{1}{b-a} = \left(\frac{1}{b-a}\right)^n$$
$$\implies \log(L(X^n,a,b)) = -n\log(b-a).$$
However, I am having issues solving for the MLEs of $b$ and $a$ given their
dependence on each other in this problem. Any suggestions on how to fix this?
 A: Perhaps I misunderstand the setup, but isn't $\tau$ the mean of the distribution?
$$\tau = \int x \mathop{dF(x)} =\frac{1}{b-a} \int_a^b x \mathop{dx} = \frac{1}{b-a} \frac{1}{2} [x^2]_{x=a}^b = \frac{1}{2} \frac{b^2-a^2}{b-a} = \frac{a+b}{2}.$$

Your log likelihood computation is almost correct.
\begin{align}
L(X^n,a,b) = (b-a)^{-n} \cdot \mathbf{1}[\max_i X_i \le b] \cdot \mathbf{1}[\min_i X_i \ge a].
\end{align}
This is zero if either indicator is violated, so we would like the indicators to both hold. To maximize the likelihood, we also want $b-a$ to be as small as possible. This suggests $\hat{b} = \max_i X_i$ and $\hat{a} = \min_i X_i$.
A: In the first place, the cumulative distribution function $F$ must be a straight line function of $x$ on the interval $[a,b]$ and go from $0$ to $1$ as $x$ goes from $a$ to $b$; therefore it is
$$
F(x) = \begin{cases} 0 & \text{if } x < a, \\[4pt]
\dfrac{x-a}{b-a} & \text{if } a\le x\le b, \\[4pt]
1 & \text{if } x>b. \end{cases}
$$
So
$$
\tau = \int_{\mathbb R} x \, dF(x) = \int_a^b x \, \frac{dx}{b-a} = \frac 1 2 \cdot \frac{b-a}{b^2 - a^2}  = \frac{a+b} 2. \tag 1
$$
In other words, as the symmetry would suggest, if a random variable is uniformly distributed between $a$ and $b$, then its expected value is half-way between  $a$ and $b$.
In the second place, you are neglecting the piecewise nature of the density function and hence of the likelihood function.  For the joint density, you have
\begin{align}
f_{X_1,\ldots,X_n} (x_1,\ldots,x_n) & = \begin{cases} \dfrac 1 {(b-a)^n} & \text{if all of }x_1,\ldots,x_n \in (a,b), \\[4pt]  0 & \text{otherwise}.  \end{cases} \\[10pt]
& = \begin{cases} \dfrac 1 {(b-a)^n} & \text{if } \max\{x_1,\ldots,x_n\} \le b \text{ and } \min\{x_1,\ldots,x_n\} \ge a, \\[4pt] 0 & \text{otherwise}. \end{cases}
\end{align}
Hence the likelihood function is
\begin{align}
L(a,b) & = \begin{cases} \dfrac 1 {(b-a)^n} & \text{if } b \ge \max \text{ and } a \le \min, \\[4pt] 0 & \text{otherwise}.  \end{cases}
\end{align}
Thus $L(a,b)$ gets bigger as $b$ gets smaller and $a$ gets bigger, until $b$ gets as small as $\max\{x_1,\ldots,x_n\}$ and until $a$ gets as big as $\min\{x_1,\ldots,x_n\}$.
Thus we have
$$
\widehat a = \min\{x_1,\ldots,x_n\} \quad \text{and} \quad \widehat b = \max\{x_1,\ldots,x_n\}. \tag 2
$$
Equivariance conjoined with line $(1)$ above then implies $\widehat{~\tau~} = \dfrac{\widehat a + \widehat b} 2$ and then you can apply $(2)$.
