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Let ${X_1},...,{X_n}$ be a random sample from uniform distribution $U\left[ {0,\theta } \right]$. Find the MLE of $\theta $.

I wrote the likelihood function of this distribution, that being a uniform one has $f\left( x \right) = {1 \over {b - a}} = {1 \over \theta }$ as $$l\left( \theta \right) = \prod\limits_{i = 1}^n {{1 \over \theta }} $$

which has an equivalent loglikelihood function $$\log \left( l \right) = \sum\limits_{i = 1}^n {\left[ {\log \left( {{\theta ^{ - 1}}} \right)} \right]} = - n\log \left( \theta \right)$$

Now if I want to find the MLE, I'll set its first derivative to zero: $$\eqalign{ & {\partial \over {\partial \theta }}\left( { - n\log \theta } \right) = 0 \cr & {{ - n} \over \theta } = 0 \cr} $$

And that's where things stopped making sense for me, for I can't solve for $\theta $ like I normally can (for other distributions). Checking a bunch of different threads in this community, apparently writing $\theta = {X_{(n)}}$, that is, the highest observation from the sample, is enough as an answer.

However, I can't stop thinking about the fact that no ${X_{(n)}}$ on the planet would make the first derivative be equal to zero, so how can that be a valid estimator?

Pedro Alonso
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  • You have $f(x) = \theta^{-1} I_{0 \leq x \leq \theta}$. The indicator is important in this situation to derive the MLE (which is indeed $X_{(n)}$). – periwinkle Jan 25 '20 at 12:14
  • Do search the site. Here is one of several posts of this question: https://math.stackexchange.com/q/649678/321264. – StubbornAtom Jan 25 '20 at 14:36
  • After the Indicator use has been pointed out, I've taken at look at some answers again, but I can't quite understand the indicator use yet. – Pedro Alonso Jan 25 '20 at 14:39

1 Answers1

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As already noted in a comment, you’re missing an indicator function, since the factor corresponding to $X_i$ is zero for $\theta\lt X_i$. If you include this, you get a discontinuous likelihood function, which can take its maximum either at a point with zero derivative or at a discontinuity. You’ve excluded the first possibility, so all that’s left is to check that the only discontinuity occurs at $X_{(n)}$ and that the likelihood is indeed maximal at that point. To show this, you can make use of the fact that you’ve already shown that the derivative of the log likelihood is everywhere negative.

joriki
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  • I'm not familiar with the Indicator notion. From what I could tell, it's either 1 or 0, right? How can it sometimes be 0 without messing the the whole product? – Pedro Alonso Jan 25 '20 at 14:35