2

Let $X_1,...,X_n$ be a random sample from a uniform distribution $(0,\theta)$.

I have found the maximum likelihood estimator to be $Y_n = \text{max}\{X_1,...,X_n\}$

The solution then claims that $\frac{Y_n}{\theta}$ is a pivot. I don't understand how to show that it is a pivot.

mrnovice
  • 5,773

1 Answers1

3

To show that the function $\frac{Y_n}{\theta}$ is a pivot we show that its distribution does not depend on $\theta$. Let's find the distribution of the cdf of $\frac{Y_n}{\theta}\in(0,1)$ using the cdf method: let $x\in(0,1)$,

\begin{align} P\left(\frac{Y_n}{\theta}\leq x\right) &= P\left((\max\{X_1,\ldots,X_n\})\leq x\theta\right)\\ &= P\left(\left(X_1\leq x\theta\right)\cap\ldots\cap\left(X_n\leq x\theta\right)\right)\\ &= \prod\limits_{i=1}^nP\left(X_i\leq x\theta\right)\quad X_i\mbox{ independent}\\ &= \prod\limits_{i=1}^n \frac{x\theta}{\theta}\\ &= \prod\limits_{i=1}^nx\\ &= x^n \end{align}

Since the cdf doesn't depend on $\theta$, neither will the pdf. Thus the quantity $\frac{Y_n}{\theta}$ is a pivot.

Jürgen Sukumaran
  • 7,645
  • Does $P(X_i\leq x\theta) = \frac{x\theta}{\theta}$ come from the CDF of a uniform distribution? – mrnovice May 20 '18 at 10:20
  • 2
    Yes, since $x\in(0,1)$ we have that $x\theta\in(0,\theta)$, call it $\tilde\theta$ if you like. This gives $P(X_i\leq \tilde\theta)$ which is the cdf of the uniform distribution on $(0,\theta)$ evaluated at $\tilde\theta$. The cdf for such a distribution is $C(\tilde\theta) = \frac{\tilde\theta}{\theta}=\frac{x\theta}{\theta}=x$. – Jürgen Sukumaran May 20 '18 at 10:23