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I know that , For Uniformly Distributed random variables $X_1,X_2,\dots,X_n$ $\in \mathcal{R}$, the p.d.f is given by:

$f(x_i) = 1/θ$ ; if $0≤x_i≤θ$

$f(x) = 0$ ; otherwise

If the uniformly distributed random variables are arranged in the following order

$0≤X_1≤X_2≤X_3\dots ≤X_n≤θ$,

I understand that The likelihood function is given by:

$L(θ)=\prod_{i=1}^{n}f(x_i)=θ^{−n}$

The log-likelihood is:

$lnL(θ)=−nln(θ)$

Setting its derivative with respect to parameter $\theta$ to zero, we get:

$\frac{d}{d\theta}lnL(\theta)=-n\theta$

which is $< 0$ for $θ > 0$

Hence, $L(θ)$ is a decreasing function and it is maximized at $θ = X_{(n)}$

The maximum likelihood estimate is thus,

$\hat{θ} = X_{(n)}$

What my question is what if I use Supremum to solve this?

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    $\begingroup$ You may want to have a look at formatting your questions in LaTeX, which is possible quite easily here on CV. $\endgroup$ – Christoph Hanck Jun 25 '20 at 11:18
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    $\begingroup$ What is your question? $\endgroup$ – whuber Jun 25 '20 at 13:34
  • $\begingroup$ Please use MathJax for typesetting. $\endgroup$ – StubbornAtom Jun 25 '20 at 14:45
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    $\begingroup$ The likelihood function is given by: $$L(θ)=\prod_{i=1}^{n}f(x_i)=θ^{−n}$$ this is incorrect and should be $$L(θ)= \begin{cases} \prod_{i=1}^{n}f(x_i)=θ^{−n} & \text{if} &\forall i : 0<x_i<\theta \\ 0 & \text{else}\\ \end{cases} $$ $\endgroup$ – Sextus Empiricus Jun 26 '20 at 10:43
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    $\begingroup$ "and it is maximized at $\theta = X_{(n)}$" this has not to do with the derivative of the likelihood, but with the likelihood being zero if $\theta < x_{(n)}$. $\endgroup$ – Sextus Empiricus Jun 26 '20 at 10:52
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The result is correct, but the reasoning is somewhat inaccurate. You need to keep track of the property that the density is zero outside $[0,\theta]$. This implies that the likelihood is zero to the left of the sample maximum, and jumps to $\theta^n$ in the maximum. It indeed decreases afterwards, so that the maximum is the MLE.

This also entails that the likelihood is not differentiable in this point, so that finding the MLE via the "canonical" route of the score function is not the way to go here.

A more detailed formal derivation is, e.g., given here

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  • $\begingroup$ But shouldn't MLE be X(n)? Ain't this theory wrong? How about using supremum to find it from the interval? $\endgroup$ – Muskaan Madan Jun 25 '20 at 11:19
  • $\begingroup$ It would be useful to define your notation, both $Xn$, $X(n)$ show up. The MLE is the sample maximum $\endgroup$ – Christoph Hanck Jun 25 '20 at 11:21
  • $\begingroup$ I got it. Thanks~ $\endgroup$ – Muskaan Madan Jun 25 '20 at 11:22
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Assume $x\ge 0$ so that

$f(x; \theta) = \frac{1}{\theta}I(x \le \theta)$ and

$L(x; \theta) = \prod_{j=1}^J \theta^{-1}I(x_j\le \theta) = \theta^{-J}I(\max_j x_j \le \theta)$

This function is clearly decreasing in $\theta$ so the smallest possible value for $\theta$ maximizes the likelihood. The largest observation is not allowed to exceed $\theta$ because then the LL is zero; this gives $\hat{\theta} = \max_j x_j$.

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