MLE of $\frac{1}{\theta}$ Uniform distribution. Let $X_1,X_2...X_n$ be a random sample from $U(\theta-5,\theta+5)$ where $\theta \in(0,\infty) $ is unknown. Let $T=\max(X_1,X_2...,X_n)$ and $U=\min(X_1,X_2...,X_n)$. Then which of the following statements is TRUE?
$(A) \dfrac{2}{T+U}$ is M LE of $\dfrac{1}{\theta}$
$(B) MLE$ of $\dfrac{1}{\theta}$ doesnt exist
$(C)U+8$ is MLE of $\theta$
$(D)\dfrac{T+U}{2}$ is unique MLE of $\theta$
$\theta-5<x_{(1)} < x_{(2)}....x_{(n)}< \theta+5$
$\theta<x_{(1)}+5$ and $\theta>x_{(n)}-5$
$x_{(n)}-5<\theta<x_{(1)}+5$
Any statistic satisfying above condition is MLE of $\theta$ so there is no unique MLE here option $(D)$ rules out
$x_{(1)}+8<x_{(1)}+5$
Option $(C)$
$8<5$ which is false so option $C$ is false.
I am not sure how to prove $(A)$ and $(B)$ but in the exam, I have ticked $(B)$ so am I right?   If I am wrong can anyone prove how?
 A: Because of the equivariance of MLEs, the MLE of $1/\theta$ is the reciprocal of the MLE of $\theta.$
The joint density is
$$
\begin{cases} 1/10^n & \text{as long as $\theta$ is so located that} \\
& \text{the interval $(\theta\pm5)$ includes} \\
& \text{both $U$ and $T$,}  \\[6pt] 0 & \text{otherwise.} \end{cases}
$$
So $\theta$ must be in the interval $(T-5,U+5)$ in order that the likelihood be positive; otherwise the likelihood is $0.$ Except that we have this complication: If it is known that $\theta>0,$ then one must use instead $(\min\{T-5,0\},\, U+5).$
Thus one can say there is a non-unique MLE: Any number in that interval will serve.
A: If $\theta\in \mathbb R$, any MLE of $\theta$ can be expressed as a convex combination $X_{(n)}-5$ and $X_{(1)}+5$:
$$\hat\theta_{\alpha}=\alpha(X_{(n)}-5)+(1-\alpha)(X_{(1)}+5)\,,$$
where $\alpha\in (0,1)$ is free of $\theta$.
Under the restriction $\theta>0$, MLE is $\hat\theta_{\alpha}$ provided $\hat\theta_{\alpha}>0$. If $\hat\theta_{\alpha}<0$, then technically MLE of $\theta$ and hence MLE of $1/\theta$ does not exist.
So for $\alpha=\frac12$, one MLE of $\theta$ is  $\hat\theta=\frac{X_{(n)}+X_{(1)}}{2}$ if $X_{(n)}+X_{(1)}>0$. Then $1/\hat\theta$ is guaranteed to be one MLE of $1/\theta$ provided $X_{(n)}+X_{(1)}>0$.
So I think the correct option among (A) and (B) depends on the sample observed.
A: This question really boils down to whether you're allowed to divide by 0 or not. Suppose that you have a sample $X_1,...,X_n$ with $U=\min\{X_1,...,X_n\}=-1$ and $T=\max\{X_1,...,X_n\}=1$. Then we can agree that any $\theta\in[-4,4]$ is an MLE of $\theta$ because the indicator variables in the likelihood is 1 (as opposed to 0). That is, the inequality
$$T-5\le\theta\le U+5$$
holds. Then by the invariance property $\left(\frac 1 \theta\right)_{MLE}=\frac 1 {\theta_{MLE}}$. That is, $\left(\frac 1 \theta\right)_{MLE}\in(-\infty,-\frac 1 4]\cup[\frac 1 4,\infty)$. In particular, $\frac{T+U}{2}=0$ is an MLE of $\theta$, but $\frac{2}{T+U}$ is not an MLE of $\frac 1 \theta$ because it is undefined. However $\left(\frac 1 \theta\right)_{MLE}$ do exist, as previously mentioned being in the interval of anything smaller than -1/4 or greater than 1/4.
It is helpful to determine the difference between maximum likelihood estimators and maximum likelihood estimates. Note that $\frac{T+U}{2}$ is a MLE of $\theta$ that works for all samples - in fact it is the only estimator that works for all samples. By the invariance property of maximum likelihood estimators, we can conclude that $\frac{2}{T+U}$ is a maximum likelihood estimator of $\frac 1 \theta$, even though as in the example in the previous paragraph, its maximum likelihood estimates that it produces may be undefined when U=-T, whilst other MLEstimates exist for 1/theta.
Conclusion
$\frac 2{T+U}$ is a Maximum Likelihood Estimator of $\frac 1 \theta$. Since it does not exist when the denominator is 0, it is also true that it is a MLEstimator of $\frac 1 \theta$ when this happens.
Eliminate:
(B) It is too specific. A MLE of 1/theta only doesn't exist when U=-5 and T=5. 2/(U+T) doesn't work for when U=-T and $U\ne-5$, but MLEs of 1/theta always exist in these cases.
