Showing that the maximum likelihood estimator (MLE) exists but is not unique I have a few questions with regards to a solution to the problem below:



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*How is it possible to have $max_{1\leq{i}\leq{n}}x_i-1<\theta<min_{1\leq{i}\leq{n}}x_i$?
How can a value of $\theta$ be both greater than a larger value and less than a smaller value? Isn't that contradictory?

*How is it that $max_{1\leq{i}\leq{n}}x_i-1 < min_{1\leq{i}\leq{n}}x_i $?
For example, what if my $max_{1\leq{i}\leq{n}}x_i = 10$ and $min_{1\leq{i}\leq{n}}x_i = 1$? Would this then mean that $10<1$?

*Finally, if $x_1,...,x_n= 1,0,1,0,1,0,...,1,0$, where the maximum is $1$ and the minimum is $0$, wouldn't it mean that $0<\theta<0$ and so implies that $\theta = 0$ and thus is unique?
Thanks everyone!!!
 A: *

*The value of θ is not greater than a larger value and less than a smaller value. Note that you subtract 1 from the maximum value, and the difference between the maximum and minimum can never be greater than 1.

*Your example is not consistent with the probability distribution. The probability distribution says that all possible values of X are between θ and θ + 1 (for some fixed number θ). Note that you actually have a uniform distribution on this interval.

*Yes. In this case θ itself (not just its estimate) must be 0. The same thing will happen whenever the largest value in the samples is 1 greater than the smallest value. This has of course probability zero of occurring, but is theoretically possible.
A: *

*You might be missing the $-1$ after the max: $\theta$ lies above the maximum value of the $x_i$s $-1$ and the minimum value of the $x_i$s. Another way is two write $\max x_i \leq \theta + 1 \leq \min x_i +1$.

*The likelihood function of $\theta$ given $n$ independent observations $x_1, \dots, x_n$ is given by the product of their probabilities - in this case $\Pi_{i=1}^n f(x_i\vert \theta)$, which is the the product of the indicator functions of the interval $(\theta, \theta+1)$. This means the following: If one of the $x_i$s doesn't lie in this interval, the likelihood is zero. The case where the likelihood is not equal to zero can be translated to: The largest value of the $x_i$s is less than  $\theta +1$, and the smallest is more than $\theta$, i.e. $$\max_{i=1, \dots n}x_i< \theta, \quad \min_{i=1, \dots n}x_i> \theta + 1.$$ Subtracting $1$ from both sides gives us the desired formula.

*Taking the inquality from the solution for the case where the likelihood is equal to $1$ is, as you correctly observed, $0<\theta <0$. Since no number is smaller than itself, this case simply can't happpen. Thus, for your example, all $\theta$s have the same likelihood, namely $0$, and all values of $\theta$ maximize the likelihood, albeit in a rather unsatisfactory way. The argument from the solution is concerning the case where there is some $\theta$ such that the likelihood is equal to one.
