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