How large of a random sample must be taken in order that Pr(| max{X1, . . . , Xn } − θ| ≤ 0.1θ) ≥ 0.95, for all possible θ?

Question

Problem

Suppose that a random sample $X_1, \cdots, X_n$ is to be taken from the uniform distribution on the interval $[0,θ]$ and that $θ$ is unkown. How large a random sample must be taken in order that

Pr$(\;|\max\{X1,\cdots, X_n \} − θ| ≤ 0.1 \;θ\;\;) ≥ 0.95$,

for all possible $θ$?

Here's what I've tried so far: I wrote down that we're looking for the probability that the max of the sample is between .1 of the parameter.I plotted a number line to see within what range on the line the difference would be within.

I don't understand how to find the n that makes the statement true. I understand that if n were only 1, then the probability that that single observation is within the range would be very low. So in order to make the probability greater than or equal to .95, n would have to be large. But I'm unsure on how to calculate that.

Answer 1

Let $Y_i = \dfrac{X_i}{\theta},$ so that $Y_i \sim \mathsf{Unif}(0,1).$ Furthermore, the probability you are after is $$ \begin{align*} \mathbf{P}_\theta(|\max X_i - \theta| < 0.1 \theta) &= \mathbf{P}_\theta(|\max Y_i - 1| < 0.1) \\ &= \mathbf{P}_\theta(0.9 < \max Y_i < 1.1) \\ &= 1 - \mathbf{P}_\theta(\max Y_i \leq 0.9) = 1 - (0.9)^n. \end{align*} $$ The rest of the exercise should be very easy.

Answer 2

Since $0 \leq X_k \leq \theta$: $$ Pr (|\max(X_1, X_2, ..., X_n)-\theta|\leq 0.1\theta) = Pr (\max(X_1,...,X_n) \geq 0.9\theta) = $$ $$ = 1 - Pr (X_k < 0.9\theta\;\; \forall k) = 1 - 0.9^n \Rightarrow $$ $$ \Rightarrow 0.9^n \leq 0.05 \Leftrightarrow n \geq 29 $$