Question: There is a riddle which I heard a while ago and haven't came up with a satisfying approach yet.

Say a book is written by person $A$, two other people, $B$ and $C$ review the text and each of them find 40 and 30 errors respectively and among their founded errors, there are 15 common words. Estimate the true number of errors in the book with this given information.

I know there might be various approaches, so any ideas would be appreciated.

Answer: 
Wilk's theorem assumes that the support of the distribution, i.e., the sample space, does not depend on the unknown parameter. Here, the parameter $b$ determines the width of the sample space so Wilk's theorem cannot be applied.

All the standard asymptotic tests associated with likelihood theory (Wald tests, likelihood ratio tests, score tests) make the same assumption.

To see why this assumption is so important, consider one of the most fundamental theorems used in likelihood theory, the fact that
$$E\left(\frac{\partial}{\partial \theta}\log f(X;\theta)\right)=0$$
where $f(x;\theta)$ is the density of $X$.
This theorem is proved by differentiating the left hand side of
$$
\int \exp(\log f(x;\theta)) dx = 1
$$
with respect to $\theta$ under the integral sign.
But now imagine what happens when the limits of the integral depend on $\theta$:
$$
\int\_{a(\theta)}^{b(\theta)} \exp\left(\log f(x;\theta)\right) dx = 1
$$
Now the derivative of the left-hand-side is far more complex and the theorem is no longer true.
The same sort of problem occurs for all the other identities used in asympototic likelihod theory and for Wilk's theorem in particular.
### Wilks' theorem does not apply here

Before getting to Wilks' theorem, one problem here (as with many questions about uniform distributions on this site) is that you have not taken account of the support of the uniform distribution in your mathematics, and this leads you to an incorrect form for the likelihood-ratio statistic. You didn't specify your alternative hypothesis, but I'm going to assume it is $H\_\text{A}:b>1$ (i.e., a one-sided test). Given data $\mathbf{x} = (x\_1,...,x\_n)$ the likelihood function is:

$$L\_\mathbf{x}(b) = \frac{\mathbb{I}(x\_{(n)} \leqslant b)}{b^n} 
\quad \quad \quad \quad \quad \text{for all } b > 0,$$

so the maximised likelihood over any parameter subspace $\mathscr{B} \subseteq (0, \infty)$ is:

$$\begin{align}
\hat{L}\_\mathbf{x}(\mathscr{B})
\equiv \sup\_{b \in \mathscr{B}} L\_\mathbf{x}(b) 
&= \sup\_{b \in \mathscr{B}} \frac{\mathbb{I}(x\_{(n)} \leqslant b)}{b^n} \\[12pt]
&= \begin{cases}
(\inf \mathscr{B})^{-n} & \text{if } \mathscr{B} \cap [x\_{(n)}, \infty) \neq \varnothing, \\[6pt]
0 & \text{if } \mathscr{B} \cap [x\_{(n)}, \infty) = \varnothing. \\[6pt]
\end{cases} \end{align}$$

(Note that this maximised likelihood is *zero* if there are no values $b \in \mathscr{B}$ with $b \geqslant x\_{(n)}$.) Consequently, using the hypotheses $H\_0:b=1$ versus $H\_\text{A}:b>1$ you get:

$$\begin{align}
\hat{L}\_\mathbf{x}(H\_\text{A})
&= \hat{L}\_\mathbf{x}((1,\infty)) = 1, \\[12pt]
\hat{L}\_\mathbf{x}(H\_0)
&= \hat{L}\_\mathbf{x}([1]) = \mathbb{I}(x\_{(n)} \leqslant 1).
\end{align}$$

This gives the likelihood-ratio statistic:

$$\begin{align}
\Delta\_{LR}
\equiv 2 \log \Bigg( \frac{\hat{L}\_\mathbf{x}(H\_\text{A})}{\hat{L}\_\mathbf{x}(H\_0)} \Bigg) 
&= 2 [ \hat{\ell}\_\mathbf{x}(H\_\text{A}) - \hat{\ell}\_\mathbf{x}(H\_0) ] \\[12pt]
&= 2 [ 0 - \log \mathbb{I}(x\_{(n)} \leqslant 1)] \\[12pt]
&= \begin{cases}
0 & & \text{if } x\_{(n)} \leqslant 1, \\[6pt]
\infty & & \text{if } x\_{(n)} > 1. \\[6pt]
\end{cases}
\end{align}$$

This statistic has a distribution concentrated on two possible values, so it does not obey the approximate chi-squared distribution. The reason that [Wilks' theorem](https://en.wikipedia.org/wiki/Wilks%27_theorem) does not apply to this case is that the maximising parameter values for each maximised likelihood value occur on the boundary of the parameter range rather than the interior of the parameter range.