Suppose that $Y_1, Y_2, ..., Y_n$ is a random sample from a distribution with density function

$$ f(y) = \begin{cases} \theta y^{\theta - 1}\ \ \ \ 0 < y < 1, \\ 0\ \ \ \ \ \ \ \ \ \ \ elsewhere \end{cases} $$

The parameter $\theta$ is positive. Find the MVUE to estimate the parameter $\theta$ using the factorization criterion and Rao-Blackwell theorem.

I've been trying to solve it but without success.

The likelihood of the sample is

$$ L(y_1, y_2, ..., y_n | \theta) = \theta^n (\prod_{i=1}^n y_i)^{\theta - 1} $$

Since $L(\theta)$ is a function of $\theta$ and $\prod_{i=1}^n y_i$, then

$$ g(\prod_{i=1}^n y_i, \theta) = \theta^n (\prod_{i=1}^n y_i)^{\theta - 1} $$ and $$ h(y_1, y_2, ..., y_n) = 1 $$

By factorization criterion theorem, this implies that the estimator $U = \prod_{i=1}^n y_i$ is sufficient for the parameter $\theta$.

Then, when trying to use Rao-Blackwell theorem, I have that $E[U]=E[\prod_{i=1}^n y_i] = ...$ but I don't know how to solve this in order to get the factor to apply to U and make it unbiased.

Thanks for your time!

EDIT: So far, I have the sufficient estimator $U$. In order to use Rao-Blackwell, I need to solve $E[T|U]$ for some statistic $T$, so that would give the MVUE statistic $T^*$.

What would be the statistic $T$ in this case? And how could I solve the this conditional expectation?


2 Answers 2


Hint: you condition on the SS; you don't take the expectation of it.

The Rao-Blackwell theorem does not say you have to take expectations of the sufficient statistic. Rather, you can take expectations of anything, but they are to be conditional expectations, conditioning on the sufficient statistic. If you want to Rao-Blackwellize some statistic $T$ and you have an SS $U$, then the improved estimator is $T^* = E[T|U]$.


A few years later, now that the question has changed:

  1. $-\log Y_i \sim \text{Gamma}(1,1/\theta)$

  2. $-\sum_{i=1}^n \log Y_i = - \log \prod_{i=1}^n Y_i \sim \text{Gamma}(n, 1/\theta)$.

  3. $\left[-\sum_{i=1}^n \log Y_i\right]^{-1} \sim \text{Inv-Gamma}(n,\theta)$.

  4. $E[T] = E\left( (n-1)\left[-\sum_{i=1}^n \log Y_i\right]^{-1} \right)= \theta$

  5. $E[T \mid \text{sufficient stat}] = T$ because $T$ is a function of the sufficient stat.

Some other interesting things:

-The MLE estimator is $n\left[-\sum_{i=1}^n \log Y_i\right]^{-1}$

-the expectation of $1/(\log Y_i)$ isn't defined, so you have to sum them all together early in the process.

  • $\begingroup$ What would be the statistic $T$ in this case? And how could I solve the this conditional expectation? I'm really lost :/ $\endgroup$ Commented Aug 16, 2017 at 7:45
  • $\begingroup$ I suppose you should start with an unbiased statistic, for example based on method of moments (this ensures the Rao-Blackwellized statistic is UMVUE). Probably it's best to try to guess an unbiased statistic which will have a simple conditional expectation (in that sense calculating the expectation of $U$ might actually be a good idea.) $\endgroup$ Commented Mar 31, 2019 at 22:49
  • 1
    $\begingroup$ @user1587692 the MOM estimator is not a function of the sufficient stat. But if it was, then yes, it is trivial to take the conditional expectation of it $\endgroup$
    – Taylor
    Commented Apr 1, 2019 at 18:16

You want to take the conditional expectation of some unbiased statistic $T$ given $U$. Then $E E[T|U]=ET=\theta$, so indeed it is unbiased, and it will be lower variance by Rao Blackwell. (if $T$ is also complete the resulting conditional expectation will be UMVUE). If you conclude that the product of the observations is sufficient, then you should find a transformation of it that is unbiased for theta. In that way, the conditioning is trivial (this is not always possible, if it is not, the conditional expectation is probably hard to calculate so in many exercises this would work). As you have a product it seems logical to take logs. Then calculating the expectation of $\ln(y_i)$ you easily obtain an unbiased transformation. Edit: @taylor is right in this case it does not work immediately.


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