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Obviously the MLE of $\theta$ for a distribution $X_1, X_2, \dots, X_n \sim Uniform(0,\theta)$ is $\hat{\theta} = max(X_1, X_2,\dots,X_n)$

Now, assume $\theta = 1$. If you take repeated samples with $n=50$. What would the distribution of $\hat{\theta}$ be?

I assume it would be: $$f(x) = P(\hat{\theta} = x) = P(X_1 \le x) P(X_2 \le x) \dots P(X_{50} \le x) = x^{50}$$

given that $x≤1$ always since

However, if you integrate this distribution, it does not sum to 1: $$\int_0^1x^{50}dx = \frac{1}{51}$$

so, would the "true" pdf be $$f(x) = 51x^{50}$$ or would it be a different function altogether?

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    $\begingroup$ You need to review the definition of a probability density function. And $P(X_1\le x)=\frac{x}{\theta}$ for all $0<x<\theta$. $\endgroup$ – StubbornAtom Mar 31 at 13:56
  • $\begingroup$ I stated that $\theta = 1$, so $P(X_1 \le x) = \frac{x}{\theta} = x$. $\endgroup$ – lstbl Mar 31 at 13:57
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    $\begingroup$ Okay. But the blunder is '$f(x)=P(\hat\theta=x)=\cdots$' assuming $f$ is the pdf of $\hat\theta$. It is actually the cdf $F(x)=P(\hat\theta\le x)=(P(X_1\le x))^n$. Now find pdf from cdf. $\endgroup$ – StubbornAtom Mar 31 at 14:01
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    $\begingroup$ AHHHHHH! Totally makes sense. I guess I was thinking about this a little backwards. What is the probability that $\hat{\theta} = x$... well I guess that probability is 0 for any continuous distribution function. However you CAN make a statement about $P(X≤x)$ for a continuous function. $\endgroup$ – lstbl Mar 31 at 14:08

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