1

The uniform (0,3θ) distribution has pdf

$f\left(x\right)=\frac{1}{3θ}$ when $0\le x\le 3θ$ and $0$ otherwise

where $θ>0$

let $x_1,....x_n$ be a random sample from this population distribution where $x_{(n)}$ is the maximum

let $\hat{\theta}$ = $\frac{x_{(n)}}{3}$ be an estimator for θ

Show that this is biased

I know how to do this usually but the $x_{(n)}$ is confusing me.

user520403
  • 459
  • $\theta$ almost surely cannot be less than a third of the maximum value observed since otherwise that observation would not satisfy $0 \le x \le 3\theta$, and $\theta$ is probably more than this, so the expected value of $\frac{x_{(n)}}{3}$ will be strictly less than $\theta$. Try to calculate $\mathbb E\left[\frac{x_{(n)}}{3} \mid \theta\right]$ – Henry Mar 13 '19 at 17:10
  • yes Im just not sure how to compute $E(x_{(n)})$ – user520403 Mar 13 '19 at 17:24
  • Could just look up expected value of maximums of uniform variables. https://math.stackexchange.com/questions/150586/expected-value-of-max-of-iid-variables?noredirect=1&lq=1 $3\theta*n/(n+1)$ – Matthew Liu Mar 13 '19 at 17:49

1 Answers1

1

$$ F_{X_{(n)}}(x) = [F_X(x)]^n = \left( \frac{x}{3\theta} \right)^n, \quad $$ for $ 0 \le x < 3 \theta$, hence $$ \mathbb{E}[X_{(n)}/3]=\frac{1}{3(3\theta)^n}\int_{[0, 3\theta]} x nx^{n-1}dx= \frac{3^{n+1}\theta^{n+1}n}{(n+1)3^{n+1}\theta^n} = \theta\frac{n}{n+1} < \theta. $$

V. Vancak
  • 16,444