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In a uniform distribution where $0\leq X \leq \theta$, the pdf is represented as $f(X|\theta) = \frac{1}{\theta}I(0\leq X \leq \theta)$, and the likelihood is $L(\theta) = \prod\frac{1}{\theta}I(0\leq X \leq \theta) = \frac{1}{\theta^n}I(\max \ x_i < \theta, \min \ x_i > 0)$. This makes logical sense to me, as the $x_i$ must be smaller than $\theta$ and larger than 0.

What I don't understand is that I'm supposed to be able to logically conclude from this information that the MLE of $\theta$ is therefore the MAXIMUM of $x_i$. I am reading that the correct answer to the question of "what is the MLE of $\theta$?" for this question is that it is the MAXIMUM of $X$.

That doesn't make sense. Say that $x_1 = 1, x_2 = 2, x_3 = 3$, and these values are plugged in as $\theta$ in the likelihood function to see which of them maximizes our likelihood function. Then clearly 1 is the maximizer here, since for X=1, $\frac{1}{1^n}*I(true)=1$, whereas for X=3, $\frac{1}{3^n}*I(true)=0.333$ at MOST and is even smaller when n > 1, and that is all smaller than what we got for x = 1, so since x = 3 has a smaller likelihood, it wouldn't be the maximum likelihood estimator.

Explain where my logic here is wrong?

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In your example $n=3$, $\min x_i = 1$, & $\max x_i = 3$. When $\theta=1$, $$I(\max x_i \leq \theta, \min x_i \geq 0)=I(3\leq 1,1\geq 0)=I(\mathit{false})=0$$ This factor, & therefore the likelihood, will be zero for any $\theta<\max x_i$.

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