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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!

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You ask whether $\theta$ is unbiased, etc. You must have meant $\hat\theta$. – Michael Hardy Sep 27 '12 at 22:03
I'm inclined to doubt that $\hat\theta$ is sufficient, since $3\theta$ is one of the boundaries of the support of the distribution. – Michael Hardy Sep 27 '12 at 22:11
up vote 0 down vote accepted

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.

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How can you say that the cdf is $\frac{x(6\theta-x)}{9\theta^2}$? – Applied mathematician Sep 29 '12 at 11:44
Can anybody please look at the way i would like to solve? Which is just try to look for a function g and f so that the factorization theorem does his job, and if it's not working, show that it cannot be factored into two functions according to the theorem, concluding that $\hat{\theta}$ is not sufficient – Applied mathematician Sep 29 '12 at 11:47
You get the c.d.f. (capital) $F$ by integrating the density function (lower-case) $f$, thus: $F(x)=\displaystyle\int_{-\infty}^x f(u)\,du$. In this case, the density is $0$ on $(-\infty,0)$, so we can write $F(x)=\displaystyle\int_0^x f(u)\,du$. – Michael Hardy Sep 29 '12 at 15:40
Thanks. Can you maybe give some reasoning with factorization theorem in mind? – Applied mathematician Oct 1 '12 at 13:34
Hello. I'm going to think about this further. I'm not altogether happy with this way of looking at it. – Michael Hardy Oct 1 '12 at 16:31

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