I have a few questions with regards to a solution to the problem below:

Question and Answer:

  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!!!

  • $\begingroup$ 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. $\endgroup$ – Glen_b -Reinstate Monica Feb 2 '14 at 11:29
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    $\begingroup$ 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. $\endgroup$ – Alecos Papadopoulos Feb 2 '14 at 11:30
  • $\begingroup$ U can't say that x = (0,0,1,1...) because as u mentioned pdf which means continuous distribution => prob at a point = 0. $\endgroup$ – Hoan Tran Aug 16 '19 at 22:32
  1. 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.

  2. 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.

  3. 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.

  • $\begingroup$ It's worth noting that the pdf is a uniform distribution on the open interval between θ and θ + 1. Both you and the solution seem to have a slight error here. $\endgroup$ – Roland Feb 2 '14 at 11:00
  • $\begingroup$ @Roland I think it’s an unfortunate (though valid) definition in the textbook. Whether you define the support of X as an open, half-open or closed interval, the distribution of X is the same; that is, the cumulative distribution functions are identical. $\endgroup$ – Karl Ove Hufthammer Feb 2 '14 at 11:10
  • $\begingroup$ I see that the cdf are identical, and this makes sense; I haven't thought that far. However, it seems to make a difference for the likelihood function, as only an open interval would allow strict inequalities - cf. my answer to question 2. $\endgroup$ – Roland Feb 2 '14 at 11:16
  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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