Suppose I have a sequence of iid random variables $X_1, \ldots, X_n$ following the pdf:

$$ f_\theta (x) = \theta x^{\theta-1} $$

for $\theta >0$ and $0 <x<1$.

I would like to find a sufficient statistic $T(X)$, such that the family $f_\theta (x)$ has a monotone likelihood ratio (MLR) in $T(X)$.

I do this by having:

$$ \frac{f(x|\theta_1)}{f(x|\theta_2)} = \frac{\prod_{i=1}^{n}\theta_1x_i^{\theta_1-1}}{\prod_{i=1}^{n}\theta_2x_i^{\theta_2-1}} = \left(\frac{\theta_1}{\theta_2}\right)^n \prod_{i=1}^n\left(x_i\right)^{\theta_1-\theta_2} = \left(\frac{\theta_1}{\theta_2}\right)^n \left(\prod_{i=1}^nx_i\right)^{\theta_1-\theta_2} $$

At this point, is the statistic corresponding to the MLR $\prod_{i=1}^nx_i$? If this is the case, what is the distribution of $\prod_{i=1}^n$ which is HARD to find?

Or would it be:

$$ \frac{f(x|\theta_1)}{f(x|\theta_2)} = \left(\frac{\theta_1}{\theta_2}\right)^n \left(\prod_{i=1}^nx_i\right)^{\theta_1-\theta_2} = \left(\frac{\theta_1}{\theta_2}\right)^n \left(e^{\sum_{i=1}^n \log(x_i)}\right)^{\theta_1-\theta_2} $$

and hence the sufficient statistic is $\sum_{i=1}^n \log(x_i)$ which has Gamma distribution?

  • $\begingroup$ One problem you don't explain where y comes from. $\endgroup$ Commented Dec 25, 2016 at 22:20
  • $\begingroup$ I don't think the function is a density for all theta >0 without further constraint on the range of x. Maybe that is where y comes in. $\endgroup$ Commented Dec 25, 2016 at 22:23
  • $\begingroup$ @MichaelChernick Sorry, $y$ should be $x$ here, I fixed it above $\endgroup$
    – user321627
    Commented Dec 25, 2016 at 22:26
  • $\begingroup$ I think that will help a lot. $\endgroup$ Commented Dec 25, 2016 at 22:34

1 Answer 1


You are correct that$$\prod_{i=1}^n x_i$$is sufficient. But since any bijective transform of a sufficient statistic is also sufficient$$\sum_{i=1}^n \log(x_i)=\log\left(\prod_{i=1}^n x_i\right)$$is also sufficient.

You are also correct that$$y=-\log(x)\sim\mathcal{G}a(1,\theta)$$and thus that the sufficient statistic$$-\sum_{i=1}^n \log(x_i)=\sum_{i=1}^n y_i\sim\mathcal{G}a(n,\theta)$$

  • $\begingroup$ What is confusing me is that if we instead used the first statistic, $\prod_{i=1}^n x_i$, its distribution will change and no longer be Gamma. How can a monotone likelihood property admit multiple sufficient statistics up to a bijection but also yield different distributions of that sufficient statistic? In other words, why isn't there a unique distribution when asked to find the sufficient statistic for a monotone likelihood property? Thanks!! $\endgroup$
    – user321627
    Commented Dec 27, 2016 at 16:40
  • 1
    $\begingroup$ This is clearly stated in my answer: any bijective transform like $\log$ is also sufficient, hence there is no unique distribution for a sufficient statistic. $\endgroup$
    – Xi'an
    Commented Dec 27, 2016 at 21:33

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