$f_{\theta}(x)=\frac{2}{3\theta}(1-\frac{x}{3\theta}),\ 0<x<3\theta$ Is $\hat{\theta}=\bar{X}$ unbiased/consistent/sufficient estimator for $\theta$?

For 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!

• Related to (i): Look again at the expression for $E(\hat \theta)$: you are not calculating $E(\hat \theta) = E(\bar X)$ but simply $E(X)$. – Alecos Papadopoulos Sep 23 '13 at 23:21

• (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.