0

Suppose $(X_1,...,X_n)$ is a sample from a $Exp(\lambda)$ population. Try to find an UMVUE of $\lambda$. Remember $S = \sum_{i=1}^n X_i$ is both complete and sufficient for the Exponential Family.

I'm trying to use the Lehmann-Scheffé theorem, but I can't manage to get it done easily. I also tried to use Laplace-inverse transforms but all I get is a Dirac Delta function from WolframMathematica.

StubbornAtom
  • 17,932
José Jiménez
  • 111
  • Since you don't specify whether $\lambda$ is the mean or rate, see https://math.stackexchange.com/q/2034206/321264 or https://math.stackexchange.com/q/2819978/321264. – StubbornAtom Jan 28 '20 at 15:08

2 Answers2

1

Note that $\sum x_i$ follows a Gamma distribution with $(n,\lambda)$. Thus, $\frac{1}{\sum x_i}$ follows an Inverse-Gamma distribution with the same $(n,\lambda)$ parameters. However, the mean of this distribution is $\frac{\lambda}{n-1}$. Taken together we can get an unbiased estimator $\frac{n-1}{\sum x_i}$. Now, we know that $\sum x_i$ is a complete sufficient statistic, which means that we can use the Lehmann-Scheffe theorem to show that our suggested estimator is an UMVUE.

davidama
  • 11
1

Well, if you find a statistic $T=t(S)$ that is a function of $S=\sum_{i=1}^{n} X_i$ and which is an unbiased estimator of $\lambda$, you've got your UMVUE by Lehmann-Scheffe.

For $T$ to be an unbiased estimator we want: $E[(t(S))]=\lambda$, of course we know that for a $X\tilde{} exp(\lambda)$ we already have $E[X]=\lambda$ and in general $E[\bar X]=E[X]$.

So, we see that $\bar X$ is an unbiased estimator of $\lambda$ and because $\bar X$ is an one-to-one function of $S$ (by $\bar X=S/n$), we're done.

Miqail
  • 13