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A single observation $x$ is used to test the null hypothesis $\theta=\theta_0$ against alternative hypothesis $\theta=\theta_1<\theta_0$ for a geometric distribution with parameter $\theta$. Use the Neyman-Pearson lemma to find the form most powerful critical region of size $\alpha$.

Initally, I started with $$L_0=\prod_{x=1}^n \theta_0(1-\theta_0)^{x-1} = \theta_0^n(1-\theta_0)^{-n+\sum_{x=1}^n x}$$ and $$L_1=\prod_{x=1}^n \theta_1 (1-\theta_1)^{x-1} = \theta_1^n(1-\theta_1)^{-n+\sum_{x=1}^nx}$$

Then, $\dfrac{L_0}{L_1}=\dfrac{(1-\theta_0)^{\sum_{x=1}^n x}}{(1-\theta_1)^{\sum_{x=1}^n x}}\leq K$ inside the critical region and $\geq K$ outside the critical region. However, this expression is very complex, so I wasn't sure how to simplify this any further.

I then thought since it says a single observation of $x$, can I just let $x=1$? So then I end up with $L_0=\theta_0$ and $L_1=\theta_1$. However, then I get the critical region is $\frac{\theta_0}{\theta_1}\leq k$, which I didn't think made sense becaue it doesn't depend on any variables.

Michael Hardy
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mmm
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  • Your reason for thinking that that result doesn't make sense is certainly correct. $\qquad$ – Michael Hardy Jul 21 '17 at 03:28
  • I cannot answer your question (comment) if you delete it. https://math.stackexchange.com/questions/2372595/find-the-likelihood-ratio-statistic-lambda-for-a-normal-population?noredirect=1#comment4890428_2372595 – callculus42 Jul 26 '17 at 21:31

2 Answers2

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The observation $x$ is a number in $0,1,2,\ldots $. The likelihood function is $$ \theta(1-\theta)^{x-1}$$ so the likelihood ratio is $$ \frac{\theta_0}{\theta_1}\left(\frac{1-\theta_0}{1-\theta_1}\right)^{x-1}$$ and depends on the data $x$. (And you're correct that it would be strange if it didn't.)

spaceisdarkgreen
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  • With the Neyman-Pearson lemma, is it always the case that $\frac{L_0}{L_1}\leq k$ inside the critical region and $\geq$ outside the critical region regardless of if $\theta_0>\theta_1$ or $\theta_1>\theta_0$? – mmm Jul 21 '17 at 03:15
  • @mmm Yes, iirc the lemma says that the critical region for a most powerful test at a given confidence level will have that form (where $k$ is determined by the desired confidence level). That said, often the calculation of the region splits into two cases, depending on what side of $\theta_0$ $\theta_1$ is on. (For instance here it's clear that what you'll get at the end of the day is a region of the form $x<C$ if $\theta_1>\theta_0$ and $x>C'$ in the case where $\theta_1<\theta_0$) – spaceisdarkgreen Jul 21 '17 at 03:20
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My first thought was that where you wrote $\displaystyle\prod_{x=1}^n \theta_0(1-\theta_0)^{x-1}$ and $\displaystyle \sum_{x=1}^n x,$ you need $\displaystyle \prod_{i=1}^n \theta_0(1-\theta_0)^{x_i-1}$ and $\displaystyle \sum_{i=1}^n x_i$ respectively, so multiplication and addition begin at $i=1,$ not at $x=1,$ and you have $x_i$ rather than $x.$

My second thought was that you wrote "a single observation", and that means $n=1,$ so your expressions are certainly more complicated than they need to be.

So you have $$ \frac{L_0}{L_1} = \frac{\theta_0(1-\theta_0)^{x-1}}{\theta_1(1-\theta_1)^{x-1}} = \left(\frac{1-\theta_0}{1-\theta_1}\right)^{x-1}. \tag 1 $$

Since $\theta_1 < \theta_0,$ the expression $(1)$ decreases as $x$ increases. Since small values of the ratio on the left in $(1)$ favor $H_1,$ that means large values of $x$ favor $H_1.$

To figure out how large is large enough, you need to solve $$ \Pr(X>c\mid H_0) \le \alpha \text{ for } c. $$

Michael Hardy
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  • So, when trying to find the rejection region, one must determine whether $L_0/L_1$ increases or decreases with increasing value of x? – mmm Jul 21 '17 at 03:20
  • @mmm : Yes, but in some problems you might want some function of $x$ other than $x$ itself. For example, $x$ might be $(x_1,\ldots,x_n)$ and you use the sum or the product or the sum of squares or the maximum of these. Often $L_0/L_1$ is a complicated function of $x$ and you seek a simple function of $x$ that is a monotone (i.e. always increasing or always decreasing) function of $L_0/L_1.$ That is how you see that the t-test and the F-test are likelihood-ratio tests. $\qquad$ – Michael Hardy Jul 21 '17 at 03:24