# UMVUE of parameter $\frac{1}{1+\lambda}$

suppose $X_1,X_2,X_3$ are a random sample of exponential distribution with parameter $\lambda$. how can find UMVUE parameter $\frac{1}{1+\lambda}$. note: $(T=\displaystyle\sum_{i=1}^{3}X_i,\ f(x)=\lambda\exp(-\lambda x))$

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I thought at first that this looked like the usual routine exercise in applying the Lehmann–Scheffé theorem. But in this case just finding the initial estimator was not altogether trivial. – Michael Hardy Mar 22 '12 at 18:33
(Actually, it is "trivial", but it is not "trivial".) – Michael Hardy Mar 22 '12 at 18:39

At first I thought of explaining that the Lehmann–Scheffé theorem is what you'll need here, and that $X_1+X_2+X_3$ is the minimal sufficient statistic and is complete. But it seems in this case the "hard" part lies elsewhere. Lehmann–Scheffé says that if the sufficient statistic on which you condition when you apply the Rao–Blackwell theorem is complete, then what you get is the UMVUE. But what is the thing whose conditional expectation you seek? Usually that's the easy part: you pick any crude stupid thing that has the right marginal expected value.

This time I actually needed to look at a table of inverse-Laplace transforms to find the right thing, and by hindsight it's easy to see that it's the right thing.

It is $1-e^{-X_1}$.

The point is that you want $h(X_1)$, where $\displaystyle\int_0^\infty h(x) \lambda e^{-\lambda x}\;dx = \frac{1}{1+\lambda}$, so $h(x)$ is the inverse-Laplace transform of $\dfrac{1}{\lambda(1+\lambda)}$. I'd forgotten all the routine stuff about Laplace transforms, so I looked in the table.

You can check by doing the integral that $\mathbb{E}(1-e^{-X_1}) = \dfrac{1}{1+\lambda}$.

So you need to find $$\mathbb{E}( 1 - e^{-X_1} \mid X_1+X_2+X_3).$$

The conditional distribution of $X_1/(X_1+X_2+X_3)$ given $X_1+X_2+X_3$ is in this case a Beta distribution with parameters $1$ and $2$, which has density $v\mapsto 2(1-v)$ for $0\le v\le 1$. So $$X_1 \mid (X_1+X_2+X_3=x) \sim x\cdot\operatorname{Beta}(1,2).$$

The $x\cdot\operatorname{Beta}(1,2)$ distribution has density $w\mapsto \dfrac{2}{x^2} (x-w)$ for $0\le w \le x$.

So find $$\int_0^x (1-e^{-w}) \frac{2}{x^2} (x-w)\;dw =:g(x).$$ It seems you'll have to integrate by parts to find this function $g$.

Then the UMVUE will be $g(X_1+X_2+X_3)$.

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If this is homework, it would probably be prudent to include proofs of several things that the answer above merely asserts. Or else cite theorems in the book, of which the assertions are instances. – Michael Hardy Mar 22 '12 at 18:42
For guessing the crude estimator one might note that the function we are trying to estimate looks a lot like the MGF of an exponential random variable, evaluated at a suitable point. – guy Mar 22 '12 at 20:01