$f_{\theta}(x)=\frac{2}{3\theta}(1-\frac{x}{3\theta}),\ 0For my statistics homework:
Let $X_1 , \dots X_n$ be a sample of independent, identically distributed random variables, with density:
$$ f_{\theta}(x)=  \left\{ \begin{array}{l} \frac{2}{3\theta}\left(  1- \frac{x}{3\theta} \right)\ \ \ \ \ \  \text{       if  }0 < x<3\theta  \\ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   \text{       else  } \end{array}  \right. $$
Let $\hat{\theta}=\bar{X}$ be an estimate of $\theta$.
Question: (i) is $\hat{\theta}$ unbiased?
(ii) is $\hat{\theta}$ consistent?
(iii) is $\hat{\theta}$ sufficient?
(iv) why doesn't the Cramer-Rao lower bound apply?
Answer: 
(i) To determine if $\hat{\theta}$ is unbiased we check that $E(\hat{\theta}) = \theta$ for all $\theta$:
$$E(\hat{\theta})   =   \int\limits_{0}^{3\theta} x \frac{2}{3\theta}\left(  1- \frac{x}{3\theta} \right)dx  =    \int\limits_{0}^{3\theta}  \left(\frac{2x}{3\theta} - \frac{2x^2}{9\theta ^2} \right)dx  =  \left[   \frac{x ^2}{3\theta} - \frac{2x^3}{27 \theta^2}  \right]_{x=0}^{x=3\theta}    =  \left(     \frac{9\theta^2}{3\theta} - \frac{54\theta^3}{27\theta^2}   \right)=\theta  $$
Thus $\hat{\theta}$ is an unbiased estimator of $\theta$.
(iv) The support (domain for which $f_{\theta}>0$ is dependent on $\theta$ thus the regularity conditions are not met and the Cramer-Rao lower bound on does not necessarily apply.
For (ii) and (iii) I can't really get started. The definition of consistent and sufficient are not really clear enough to me so I get confused. if anyone could help me with a tip or a start in the right direction that would be great! Thanks!
 A: Answering for practice :)


*

*(i) It is indeed unbiased:
$$
\mathbb{E}\left[\hat{\theta}\right]
=
\mathbb{E}\left[ \frac{1}{n}\sum_i x_i \right]
=
\frac{1}{n}\sum_i \mathbb{E}\left[ x_i \right]
=
\frac{1}{n}\sum_i \int_0^{3\theta} x_i\frac{2}{3\theta}
\left(1−\frac{x_i}{3\theta}\right) \,dx_i
=
\frac{1}{n}\sum_i \theta
=
\theta
$$

*(ii) For an unbiased estimator, the estimator is consistent if the variance of the estimator vanishes as $n\rightarrow\infty$. But:
\begin{align*}
\lim_{n\rightarrow\infty}\mathbb{V}\left[\hat{\theta}\right] 
&= 
\lim_{n\rightarrow\infty}\mathbb{E}\left[\hat{\theta}^2\right] - \mathbb{E}\left[\hat{\theta}\right]^2 \\
&= 
\lim_{n\rightarrow\infty}\mathbb{E}\left[ \frac{1}{n^2}\left(\sum_i x_i\right)^2 \right] - \theta^2 \\
&= 
\lim_{n\rightarrow\infty}\frac{1}{n^2}\mathbb{E}\left[ \sum_i\sum_j x_ix_j \right] - \theta^2 \\
&= 
\lim_{n\rightarrow\infty}\frac{1}{n^2}\sum_i\sum_j\mathbb{E}\left[  x_ix_j \right] - \theta^2 \\
&= 
\lim_{n\rightarrow\infty}\frac{1}{n^2}
\left(
\sum_i\sum_j \mathbb{E}\left[  x_i \right] \mathbb{E}\left[x_j \right] (1-\delta_{ij})
+
\sum_i\mathbb{E}\left[  x_i^2 \right]\delta_{ij} 
\right)
- \theta^2 \\
&=
\lim_{n\rightarrow\infty}\frac{1}{n^2}\left[ (n^2-n)\theta^2 + n\frac{3}{2}\theta^2 \right] - \theta^2 \\
&=
\theta^2-\theta^2\\
&= 0
\end{align*}
where $\delta_{ij}$ is the Kronecker delta and using the second moment
$$
\mathbb{E}\left[  x^2 \right] 
= 
\int_{0}^{3\theta}x^2\frac{2}{3\theta}
  \left( 1 - \frac{x}{3\theta} \right) dx = \frac{3}{2}\theta^2
$$
So it is consistent.

*(iii) Normally, one would use the Fisher–Neyman factorization theorem to determine whether the statistic is sufficient.
However, in this case, I would say that one can say it is simply enough to see that the statistic $\bar{X}$ cannot capture the information in $T=\max_i x_i$, especially with respect to the estimation of $\theta$, because $3\theta$ bounds the support
(e.g. see here).
For instance, $\vec{x}=[1,2,3]$ vs $\vec{x}=[1.9,2.1,2]$ give the same mean, but clearly the max is more useful here. 
The question itself answers part (iv); hopefully someone will tell me if I've made a calculation error.
