Let $X_1,\cdots,X_n$ be a random sample from $f(x|\theta)=\theta(1-\theta)^x,x=0,1,\cdots; 0 < \theta <1$ is unknown. Find the UMVUE of $\frac{1-\theta}{\theta}$.

My work:

I know that I should apply the Lehmann-Scheffe Theorem here, so I will need an unbiased estimator of $\frac{1-\theta}{\theta}$ and a complete sufficient statistic for $\theta$. I know that, since this distribution is an exponential family, the complete sufficient statistic is $\sum^n_{i=1}x_i$. However, I am having problems finding an unbiased estimator of $\frac{1-\theta}{\theta}$. Is there a systematic approach that I can take to finding an unbiased estimator?

  • 2
    $\begingroup$ What is $\mathbb{E}x$ equal to? And what is an unbiased estimator of the population mean? $\endgroup$
    – jbowman
    Commented Feb 26, 2020 at 3:41
  • $\begingroup$ @jbowman I discovered that this distribution is Negative Binomial with $r=1,p=\theta$, so $E(X)=\frac{1-\theta}{\theta}$, which means any $X_i$ is an unbiased estimator for $\frac{1-\theta}{\theta}$. I just don't see how I can find $E(X|\sum X)$ which is the BUE. $\endgroup$
    – Ron Snow
    Commented Feb 28, 2020 at 5:54
  • 1
    $\begingroup$ Would $X_1$ have a lower variance than ${1 \over n}\sum x_i$ as an estimate of the quantity of interest? You are making this problem harder than it actually is :) $\endgroup$
    – jbowman
    Commented Feb 28, 2020 at 15:25
  • $\begingroup$ @jbowman I definitely made this problem much harder than it is. I posted my answer below. Does it check out? $\endgroup$
    – Ron Snow
    Commented Feb 29, 2020 at 20:46

1 Answer 1


We recognize that $f(x|\theta)$ follows a Negative Binomial distribution with $r=1,p=\theta$.

Following this realization, we notice $E(X)=\frac{1-\theta}{\theta}$, so $W(X)=X$ is unbiased for our quantity, $\tau(\theta)$.

Additionally, we see that $f(x|\theta)$ is in the exponential family, so $T(X)=\sum X_i$ is complete sufficient for $\theta$.

By Lehmann-Scheffe Theorem, $W^*(X)=E[W(X)|T(X)]$ is the UMVUE for $\tau(\theta)$.

Since $E(X)=\tau(\theta), E(\bar{X})=\tau(\theta)$. Since $\bar{X}$ is a function of the complete sufficient statistic $T(X)$, $\bar{X}$ is the UMVUE for $\tau(\theta)$.


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