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I have a random sample $(X_1, X_2,...,X_n)$ and I have an estimator $\bar{X_n}=\sum_{i=1}^{n} X_i$

I need to compute the Fisher information of $\bar{X_n}$. The Fisher information is defined as $-E\left(\frac{d^2}{d\theta^2}logL\right)$, where $L$ is the likelihood function.

My question is: to compute the Fisher information of the estimator (NOT the random sample, but instead a function of the random sample), should we take the likelihood function of the random sample or the likelihood function of the distribution?

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Could you tell us what you mean by a "likelihood function of [a] distribution"? –  whuber Jan 16 '14 at 22:57
Well. The likelihood function of the random sample would be the joint distribution of the n random variables. The likelihood function of the estimator would be the probability of observing a specific value of the estimator... –  user114618 Jan 16 '14 at 23:34
OK, thanks for the clarification. Consider what the difference in $\frac{d^2}{d\theta^2}\log L$ would be between the two likelihoods. –  whuber Jan 16 '14 at 23:40
They are not always the same! Consider $f_{x_i}(x_i,\theta)=exp[-(x_i-\theta)]\cdot I_{(\theta,+\infty)} (x_i)$. The maximum likelihood estimator would be $min(X_i)$ and the two likelihood functions would be $L(X_1,...,X_2)=exp(n\theta-\sum x_i)\cdot \prod I_{(\theta,+\infty)} (x_i)$ and $L(min(X_i))=n \cdot exp(-n\cdot min(X_i)+n \theta)\cdot I_{(0,+\infty)} (min(X_i)-\theta)$ –  user114618 Jan 17 '14 at 9:51
Excellent example! That helps me understand what you are asking. –  whuber Jan 17 '14 at 15:26

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