# analytical distribution of the maximum likelihood estimator for a uniform distribution

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?

• You need to review the definition of a probability density function. And $P(X_1\le x)=\frac{x}{\theta}$ for all $0<x<\theta$. – StubbornAtom Mar 31 at 13:56
• I stated that $\theta = 1$, so $P(X_1 \le x) = \frac{x}{\theta} = x$. – lstbl Mar 31 at 13:57
• 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. – StubbornAtom Mar 31 at 14:01
• 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. – lstbl Mar 31 at 14:08