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A random sample $X_1,\ldots,X_n$ is taken from a distribution with probability density function which is $B(\theta,2\theta)$. Show that $\prod X_i(1-X_i)^2$ is a sufficient statistic for $\theta$.

Below is the beta distribution with the parameters referred to:

$$f_X(x;\theta)=\frac{\Gamma(3\theta)}{\Gamma(\theta)(\Gamma(2\theta)}x^{\theta-1}(1-x)^{2\theta-1}$$

To be honest, this question is beyond me. I don't know how to start.

I have been reading up about sufficient statistics, and using factorisation theorem, but I don't know if this is the right approach and I don't really know where to begin with this.

Do I need to work out a joint pdf? Do I need to know the likelihood function? Is factorisation theorem even the right thing to use?

Any help (even if it is just directing me to a source for "sufficient statistics for dummies") is much appreciated!

StubbornAtom
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Maths Barry
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  • You can indeed write down the likelihood function and then find a sufficient statistic using the factorisation theorem – Tim Dikland Apr 14 '18 at 15:12
  • Btw, be careful with your expression for the sufficient statistic. There should be an index, $X_i$ rather than just $X_1$. –  Apr 14 '18 at 16:09
  • Similar: https://math.stackexchange.com/q/321952/321264. – StubbornAtom Feb 03 '21 at 17:51

1 Answers1

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Indeed you can use the factorization theorem and show that $$ \prod_{i=1}^n f(x_i ; \theta)= \frac{1}{B ^n( \theta, 2\theta)} \left( \prod_{i=1}^n x_i ( 1 - x_i) ^ 2 \right) ^ {\theta} \left(\prod_{i=1}^nx_i(1-x_i)\right)^{-1} $$ thus, considering $$T(x) =\prod_{i=1}^n x_i ( 1 - x_i)^2 $$ one sees that $$g_{\theta}(T(x)) = \frac{1}{B ^ n( \theta, 2\theta)} \left( \prod_{i=1}^n x_i ( 1 - x_i) ^ 2 \right) ^ {\theta}$$ so $T(x) $ is the MSS.

EDIT:

You should start with the likelihood function (not the log-likelihood), and then "separate" everything known (the sample or its function) from everything unknown (depends on $\theta$). Then find the function $g$ where $\theta$ depends on some function of the sample - which will be the MSS. Namely, start by reformulating the likelihood itself \begin{align} L(\theta; x_1,...,x_n) &= \prod_{i=1}^n ( B (\theta, 2\theta) ) ^{-1} x_i^{\theta} x_i ^{-1} ( ( 1 - x_i) ^2)^{\theta} (1-x_i) ^{-1} \\ &=\prod_{i=1}^n ( B (\theta, 2\theta) ) ^{-1} (x_i( 1 - x_i ) ^ 2 )^{\theta} ((1-x_i)x_i) ^{-1} \\ & = ( B (\theta, 2\theta) ) ^{-n} \left(\prod_{i=1}^n x_i( 1 - x_i ) ^ 2 \right)^{\theta} \prod_{i=1}^n\left((1-x_i)x_i \right) ^{-1}. \end{align}

Although the MLE is indeed a function of the MSS, however it is "overkill" to derive the MLE in order to find the MSS.

Did
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V. Vancak
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