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Find the Rao-Cramer lower bound if the random sample $X_1,X_2,...,X_n$ is taken from the distribution with the p.d.f. $$f(x;\theta)=\frac{1}{\theta}x^{\frac{1-\theta}{\theta}}$$ where $0<x<1$ and $0<\theta<\infty$.

I know that I have to compute $$\frac{1}{-n\mathbb{E}[\frac{d^2}{d\theta^2}\ln(f(x;\theta))]} ~~\text{or}~~ \frac{1}{-n\mathbb{E}[\frac{d}{d\theta}\ln(f(x;\theta))]^2}$$ I tried this: \begin{align*} \ln f(x;\theta)&=-\ln(\theta)+\left(\frac{1-\theta}{\theta}\right)\ln(x)\\ \frac{d}{d\theta}\ln f(x;\theta)&=-\frac{1}{\theta}-\left(\frac{1}{\theta^2}\right)\ln(x)\\ \frac{d^2}{d\theta^2}\ln f(x;\theta)&=\frac{1}{\theta^2}+\left(\frac{2}{\theta^3}\right)\ln(x)\\ \end{align*} In both cases I have to compute $\mathbb{E}(\ln(x))$. What is it?

Mike90
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1 Answers1

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The likelihood should be of the form:

$$\displaystyle \prod_{i=1}^n \frac{1}{\theta}x_i^{\frac{1-\theta}{\theta}}=\exp\{\theta^{-1}\sum\log x_i-\sum\log x_i-n\log\theta\}$$

and you find the second derivative of the log likelihood to be: $$\frac{2}{\theta^3}\sum_{i=1}^n\log x_i+\frac{n}{\theta^2}$$

As you said, we need the pdf of $\log x_i$, so we use the transformation of variables, let $Y=\log X$ which becomes $X=e^Y$.

$$f_Y(y)=f_X(e^y)\left| \frac{dx}{dy}\right| =\frac{1}{\theta}(e^y)^\frac{1-\theta}{\theta}\left| e^y\right|$$ and simplify.

Then you can find the expectation of $Y$.

lightfish
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  • I do not understand the solution. Why do you compute the likelihood function? In the definition of the Rao-Cramer bound is just the function not the likelihood. If I compute the Rao-Cramer bound for $f_Y(y)$ is it the same as for $f_X(x)$? If yes, why? – Mike90 Mar 07 '14 at 01:46
  • you need the log likelihood to find the Fisher information, which is needed to find the Cramer Rao lower bound: http://en.wikipedia.org/wiki/Cram%C3%A9r%E2%80%93Rao_bound – lightfish Mar 07 '14 at 01:56