Let $X_1, X_2, \dots, X_n$ be a sample of i.i.d. random variables, with density $$f_\theta=\frac{2}{3\theta}\left(1-\frac{x}{3\theta}\right) $$ for $0 < x < 3\theta$. And $f_\theta=0$ if $ x < 0$ or $ x>3\theta$
Let $\hat{\theta}=\overline{X}$ be an estimate for $\theta$
1.) Determine whether $\theta$ is unbiased?
2.) Determine whether $\theta$ is consistent?
3.) Determine whether $\theta$ is sufficient?
4.) Why doesn't the Cramer-Rao lower bound apply to unbiased estimates of $\theta$ for this distribution?
I tried:
1.) $\theta$ is unbiased because the integral of $$\int_0^{3\theta}{x}\left(\frac{2}{3\theta}\left(1-\frac{x}{3\theta}\right)\right) \, dx = \theta = E[\hat{\theta}]$$Hence the statistic is unbiased.
2.) Yes, because $$\operatorname{VAR}[\hat{\theta}]=\operatorname{VAR}\left[\overline{X}\right]=\frac{\sigma^2}{n}. \lim_{n\to \infty}\left(\frac{\sigma^2}{n}\right)=0$$Hence $\theta$ is consistent.
3.)According to the factorization theorem I have to find a function $g(\hat{\theta},\theta)$ and $h(x_1,x_2,\dots ,x_n)$ so that $gh=f(x_1,x_2,...,x_n; \theta)$
I guess I have to calculate $\prod_{i=1}^{n}{f_\theta} $ and derive some function g. But I don't know how to start. Thank you for your help in advance!
The c.d.f. of $X_1$ is $x\mapsto \dfrac{x(6\theta-x)}{9\theta^2}$ for $0\le x\le 3\theta$ (and is equal to $1$ for $x>3\theta$ and $0$ for $x<0$). It follows that $\dfrac{X(6\theta-X)}{9\theta^2}$ is uniformly distributed on the interval $(0,1)$. It's not hard to show from there that $\dfrac{X(6\theta-X)}{9\theta}$ is uniformly distributed on the interval from $0$ to $\theta$.
For a sample from that distribution, the minimal sufficient statistic for $\theta$ is the sample maximum. Use the fact that this transformation of the random variables is an increasing function to show that the sample maximum from your original untransformed sample is the minimal sufficient statistic for $\theta$.
Since the sample maximum cannot be computed if only $\bar X$ is known, it follows that $\bar X$ cannot be sufficient.