I have the iids $\ X_1,X_2, ... , X_n$ with pmf $\ P(X_i = x_i) = {{m}\choose{x_i}}\theta^{x_i}(1-\theta)^{m-x_i}, 0 \leq x_i \leq m$

I have the unbiased estimator $\ X_1/m$, the sufficient statistic $T = \sum X_i $ and I want to use Rao-Blackwell to find another unbiased estimator for $\theta$.

So I do $\theta' = E(X_1/m \vert {T}) $

$ = 1/m \sum k P(X_1 = k \vert T = t) = 1/m \sum k{{m}\choose{k}}{{mn-m}\choose{t-k}}/{{mn}\choose{t}} $

But then I can't evaluate this sum. Have I gone wrong?

  • 2
    $\begingroup$ Why isn't $X_1/m$ unbiased? $E(X_1/m) = \frac{1}{m}m\theta = \theta$. $\endgroup$
    – knrumsey
    Commented May 7, 2017 at 17:22
  • $\begingroup$ Yeah X1/m is unbiased, but it is not minimum variance $\endgroup$ Commented May 7, 2017 at 17:59
  • 3
    $\begingroup$ Indeed, that's the point of using Rao-Blackwell. I'm fairly sure the OP knows that $T/mn$ is the umvue. I think the question involves the Rao-Blackwellization of the more naive estimator. $\endgroup$
    – knrumsey
    Commented May 7, 2017 at 18:00
  • 2
    $\begingroup$ Hint: compare your expression to the expectation of a hypergeometric distribution. $\endgroup$
    – whuber
    Commented May 7, 2017 at 20:51

1 Answer 1


Update based on whuber's comment.

First some notation. Let $T_{-1} = \sum_{i=2}^nX_i$ and note that $T \sim Binom(nm, \theta)$ and $T_{-1} \sim Binom((n-1)m, \theta)$. Moreover, note that $X_1$ and $T_{-1}$ are independent.

\begin{align*} \phi(T) &= E(X_1/m |T =t) \\ &= \frac{1}{m}E(X_1|T=t) \\ &= \frac{1}{m}\sum_{x=0}^m xP(X_1=x|T=t) \\ &= \frac{1}{m}\sum_{x=0}^m x\frac{P(X_1=x \cap T=t)}{P(T=t)} \\ &= \frac{1}{m}\sum_{x=0}^m x\frac{P(X_1=x \cap T_{-1}=t-x)}{P(T=t)} \\ &= \frac{1}{m}\sum_{x=0}^m x\frac{P(X_1=x)P(T_{-1}=t-x)}{P(T=t)} \\ &= \frac{1}{m}\sum_{x=0}^m x\frac{\binom{m}{x}\theta^x(1-\theta)^{m-x}\binom{(n-1)m}{t-x}\theta^{t-x}(1-\theta)^{(n-1)m-t+x}}{\binom{nm}{t}\theta^t(1-\theta)^{nm-t}} \\ &= \frac{1}{m}\sum_{x=0}^m x\frac{\binom{m}{x}\binom{nm -m}{t-x}}{\binom{nm}{t}} \\ &= \frac{1}{m}\sum_{x=0}^m x f(x;nm, m, t) \quad\text{where $f$ is the pmf of a hypergeometric random variable}\\ &= \frac{1}{m}E(X) \quad \text{where $X$ is a hypergeometric rv} \\ &= \frac{1}{m}\frac{tm}{mn} = \frac{t}{mn} \end{align*}

Recalling that $t$ is the value of $T$, we get $\hat\theta_{UMVUE} = \frac{T}{nm}$ as expected.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.