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Given a sample $X_1,\dots , X_n$ from a population $X\sim \operatorname {Exp} (\lambda )$, I have to calculate Cramer-Rao bounds for the estimation of $\lambda$ and $\frac 1 \lambda$; I also must determine if there are estimators that limit.

Now, we have that $\frac {\partial}{\partial \lambda} \operatorname {ln(\lambda e ^{-\lambda x}) }= \frac 1 \lambda -x$, and so $\mathbb E[ (\frac 1 \lambda -x)^2]=\frac 1 {\lambda^2}$, since it is the variance by definition. So in the case that we are estimating $\lambda$, the Cramer-Rao bound is $\frac {\lambda^2} n$, while in the other case the bound is $\frac {\lambda^2} n \cdot \frac 1 {\lambda^4}= \frac 1 {n\lambda^2}$. It is clear that the sample mean is an estimator for $\frac 1 \lambda$ with variance exactly $\frac 1 {n\lambda^2}$; however I don't know how to proved in the case of estimating $\lambda $. If a estimator $T_n $ for $\lambda $ equals Cramer-Rao bound, the we would have $\sum_i\frac {\partial}{\partial \lambda} \operatorname {ln(\lambda e ^{-\lambda x_i}) }=K (n,\lambda) ( T_n -\lambda )$; so that $\sum (\frac 1 \lambda -x_i)= K (n,\lambda) ( T_n -\lambda )$. Since we want $\lambda $ and not $\frac 1 \lambda$, the only way is to multiply everything for $\lambda^2$; however with this operation we can't obtain a estimator from the $x_i $, because we will still have a dependence from $\lambda$. I'm not sure about th last statement: did I actually prove that there are no estimators that equal Cramer-Rao bound for $\lambda$? Thanks in advance for your help

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Yes you did.

the lower bound for unbiased estimators of $\lambda$ is $V(T)\geq\frac{\lambda^2}{n}$

Using Lehmann-Scheffé Lemma you can find the UMVUE estimator of $\lambda$

$\hat{\lambda}=\frac{n-1}{\sum_i X_i}$

Its Variance is $V(\frac{n-1}{\sum_i X_i})=\frac{\lambda^2}{n-2}$ (for $n>2$) so, as often happens, the optimum estimator does not reach the Cramér Rao lower bound.

If interested I can show all the calculations...but you can find them in many books, e.g. Mood Graybill Boes, Chapter 7

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  • $\begingroup$ Actually we didn't cover that lemme in the lectures, we only spoke about that condition that I used. Anyway, if what I said it's correct I'm fine, thanks! $\endgroup$
    – Dr. Scotti
    Commented May 25, 2020 at 19:42

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