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Let's say in the exponential distribution, the cdf if $$f(x|\lambda)=\frac{1}{\lambda}\exp\left\{-\frac{x}{\lambda}\right\}$$ If we have $n$ observations: $X_1, X_2, \cdots, X_n$. Then we know since it belongs to the exponential family we have the UMVUE for $\lambda$ is $$T({\bf X})=\sum^n_{i=1}X_i$$ And $$E(T)=E\left(\sum^n_{i=1}X_i\right)=\sum^n_{i=1}E(X_i)=n\lambda$$ Then $$\hat{\lambda}_{UMVUE}=\frac{T}{n}=\bar{X}$$ Then, what if we want to find the UMVUE of $\frac{1}{\lambda}$?

Matata
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  • About notation: As defined in this Question, $\lambda$ is the mean; often denoted $\mu.$ Then the rate is $1/\mu,$ often denoted as $\lambda.$ UMVUE for $\mu$ is $\bar X.$ UMVUE for the rate is as shown in @Momo's Answer (+1). – BruceET Jun 28 '18 at 04:19
  • Use the fact that $T$ is a Gamma variable to calculate $E(1/T)$. Then adjust the lhs to get exactly $1/\lambda$ in the rhs. – StubbornAtom Jun 28 '18 at 04:49

1 Answers1

1

Hint:

First prove that: $$E\left[\frac{1}{X_1+X_2}\right]=\frac{1}{\lambda}$$

Then user Rao-Blackwellization to prove that $\frac{n-1}{\sum X_i}$ is the UMVUE

... or you can directly calculate the expectation of the UMVUE and use that is a function of $T$

Momo
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