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We have a random sample $X_1,X_2,\ldots,X_n$ from a probabilitiy distribution with density $f(x;\theta) = \theta x^{-\theta-1} $ given that $x > 1$, and $0$ else. where $\theta >1 $ is an unknown estimator

  1. derive the CRLB for unbiased estimators of $\theta$.

  2. Is there an unbiased estimator of $\theta$ that reaches the CRLB?

  3. show that $S = \sum^n_{i=1} \ln(X_i)$ is a complete sufficient statistic and use this result to derive thje UMVUE for $\frac{1}{\theta}$

for 1. i did the following: $$ \begin{align} f_\bar{x}(x_1,...,x_n) &= \prod_{i=1}^n \theta x_i^{-(\theta+1)} \\ l(\theta) &= n\ln(\theta) - (\theta+1) \sum_{i=1}^n \ln(x_i) \\ \frac{\delta}{\delta \theta} l(\theta) &= -\frac{\theta+1}{\sum_{i=1}^n x_i} \end{align} $$ however setting this to 0 really doesnt help me. I have been told that looking at the $l(\theta)$ function and seeing how it behaves should help, but I really am not getting any wiser.

for 2. I would just fill in the CRLB equation, should i use $\tau(\theta) = \hat{\theta}$ ?

for 3. what is the difference between a complete sufficient statistic and a sufficient statistic?

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what is Cramer-Rao Lower bound? –  Lost1 May 28 '13 at 15:50
    
$$ Var(T) \geq \frac{[\tau'(\theta)]^2}{nE[\frac{d}{d\theta} ln(f(x;\theta)]^2} $$ The 'best' estimator, i guess? But the useage of T and $\tau$ confuses me. –  WiseStrawberry May 28 '13 at 15:52
    
I dont understand your T and $\tau$ either. en.wikipedia.org/wiki/Cram%C3%A9r%E2%80%93Rao_bound –  Lost1 May 28 '13 at 16:01
    
Isn't $\tau =\theta$, so $\tau'(\theta) = 1$? –  Lost1 May 28 '13 at 16:02
    
@Lost1, yes. that is true –  WiseStrawberry Jun 13 '13 at 8:46

1 Answer 1

up vote 3 down vote accepted

This is not the complete solution, but maybe helpful on your way.
What you have is the Pareto distribution with the scale parameter $x_m=1$.
First of all you have an error while computing the derivative of the log-likelihood function. $$l(\theta) = n\ln(\theta) - (\theta+1) \sum_{i=1}^n \ln(x_i)$$ Taking the derivative with the respect to $\theta$ one can get: $$\frac{\partial l(\theta) }{\partial \theta}=\frac{n}{\theta}-\sum_{i=1}^n \ln(x_i)$$ Second, you do not need to set it to zero (unless you are after the ML estimator). Moreover to compute the Cramer-Rao bound you need to take the second derivative: $$\frac{\partial^2 l(\theta) }{\partial \theta^2}=-\frac{n}{\theta^2}$$ Then taking the expectation: $$\mathrm{E}\left\{ \left(\frac{\partial l(\theta) }{\partial \theta}\right)^2\right\}=-\mathrm{E}\left\{ \frac{\partial^2 l(\theta) }{\partial \theta^2}\right\}=\frac{n}{\theta^2}\int_1^\infty \theta x^{-\theta-1} \mathrm dx=\frac{n}{\theta^2}$$ Here as Lost1 indicated we are asked to find the bound for the parameters' estimate not its' function, so: $$\mathrm var(\theta)\geq\frac{\theta^2}{n}$$



By the way if you need the ML estimator then you can set the first derivative to zero and get: $$\hat{\theta}_{ML}:\left.\frac{\partial l(\theta) }{\partial \theta}\right\vert_{\theta=\hat{\theta}_{ML}}=0 \quad \Rightarrow \quad\hat{\theta}_{ML}=\frac{n}{\sum_{i=1}^n \ln(x_i)}$$

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