# Finding UMVUE of $\theta$ when the underlying distribution is exponential distribution

Hi I'm solving some exercise problems in my text : "A Course in Mathematical Statistics".

I'm in the chapter "Point estimation" now, and I want to find a UMVUE of $$\theta$$ where $$X_1 ,...,X_n$$ are i.i.d random variables with the p.d.f $$f(x; \theta)=\theta e^{-\theta x}, x\gt0$$.

I know that $$E(X_i)=1/\theta,$$ for each $$i$$, and also have that $$\bar{X}$$ (or equivalently $$\sum_1^n X_i$$) is a complete sufficient statistic for $$\theta$$. But I cannot go any further here. Somebody can help me?

• – StubbornAtom May 23 '20 at 18:35

You have $\overline{X}$ complete & sufficient and moreover $E[ \overline{X} ] = 1/\theta$; i.e. $\overline{X}$ is the UMVUE for $1/\theta$. It seems reasonable to guess that $1/\overline{X}$ may be the UMVUE for $\theta$. Note that $\sum_{i=1}^n X_i \sim \Gamma(n,\theta)$ since each $X_i$ is exponential rate $\theta$ and they're iid. Let $Z \sim \Gamma(n,\theta)$. \begin{align*} E[1/\overline{X}] = n E[1/Z] &= n \int_0^\infty \dfrac{1}{z} \dfrac{\theta^n}{\Gamma(n)} z^{n-1} e^{- \theta z} \; dz \\ &= n \int_0^\infty \dfrac{\theta^n}{\Gamma(n)} z^{n-2} e^{-\theta z } \; dz \\ &= n \theta \dfrac{\Gamma(n-1)}{\Gamma(n)} \underbrace{\int_0^\infty \dfrac{\theta^{n-1}}{\Gamma(n-1)} z^{n-2} e^{-\theta z } \; dz}_{=1} \\ &= \dfrac{n \theta \Gamma(n-1)}{\Gamma(n)} = \dfrac{n \theta}{n-1} \end{align*}
So $\dfrac{n-1}{n} \cdot \dfrac{1}{\overline{X}} = \dfrac{n-1}{\sum_{i=1}^n X_i}$ is the UMVUE for $\theta$.
• Note that this means that when $n=1$ there is no unbiased estimator of $\theta$. The expression here suggests $0$ which is clearly biased down, while $E[1/X_i] = +\infty$ – Henry Nov 2 '19 at 11:36