This question already has an answer here:

  1. (20%) Let $X_1, \dotsc, X_n$ be an iid sample from a distribution with pdf $f(x;\theta) = \theta x^{\theta-1}, 0< x< 1$, zero elsewhere, where $\theta >0$. Find a sufficient statistic for $\theta$ and show that a UMP test of $H_0: \theta = 6$ against $H_1:\theta <6$ is based on this statistic.

I used the exponential family form to get that the summation of $\ln(x)$ is a sufficient statistic for $\theta$, but I do not know how to find a UMP Test based on this statistic. I believe it has something to do with likelihood ratios. Thanks in advance for your help.

My question is different from that question because I do not already have a MP test, these are different numbers, the pdf is different, the significance level is not given, and the alternative hypothesis is an inequality.


marked as duplicate by Seyhmus Güngören, NCh, dantopa, Claude Leibovici, user223391 May 19 '17 at 17:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Fisher's factorization theorem says $T(X_1,\ldots,X_n)$ is sufficient for a family of distributions if the joint density can be written as $$ g_\theta(T(x_1,\ldots,x_n)) h(x_1,\ldots,x_n) $$ and only the first factor depends on $\theta$. In this case the joint density is $\theta^n (x_1\cdots x_n)^{\theta-1},$ so you can take $T(x_1,\ldots,x_n)$ to be the product $x_1\cdots x_n$ and $h(x_1,\ldots,x_n)$ to be $1$. (For most of the families of distributions considered in introductory courses you get $h$ constant; an exception is the Poisson distribution.)

Thus in this case the product is sufficient; or equivalently the sum of the logarithms is sufficient.

For any particular positive number $\theta < 6$ you have a likelihood ratio $$ \frac{L_0}{L_1} = \frac{6^n (x_1\cdots x_n)^{6-1}}{\theta^n(x_1\cdots x_n)^{\theta-1}} = \left( \frac 6 \theta \right)^n (x_1\cdots x_n)^{6-\theta}. $$ The likelihood ratio test rejects the null hypothesis if this ratio is too small. Since $6-\theta>0,$ this ratio gets smaller as the product $x_1\cdots x_n$ gets smaller, or equivalently, if the sum of the logarithms gets smaller. The Neyman–Pearson lemma says that test is more powerful than any other test regardless of the value of $\theta\in(0,6).$


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