# 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:

1. 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?

2. 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 $10<1$?

3. 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!!!

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Did you notice that $\text{max}(x_i) < \text{min}(x_i)+1$? What happens when you subtract 1 from both sides? This relates to both (1) and (2). It looks like you need to pay careful attention to how the density has been defined. Pick $\theta=1.5$, say, and draw the density. Now try $\theta=2.7$. If you do both of those, the answers to several of your questions should be more obvious. In particular, you should see why the question just before (3) makes no sense. –  Glen_b Feb 2 at 11:29
If you actually have a sample where its min value is 1 and its max value is 10, then simply, this sample cannot be a realization from the assumed distribution, and you have a case of misspecification. –  Alecos Papadopoulos Feb 2 at 11:30

1. 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 $\max x_i \leq \theta + 1 \leq \min x_i +1$.
2. 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.

3. 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.

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