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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.

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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.

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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 $\theta\in(0,6).$

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