4
$\begingroup$

Suppose I have a family of (continuous) distributions $\mathcal{P}=\{P_\theta(x),\theta\in\mathbb{R}^+\}$. I also have a statistic $T(x)$ that is sufficient for $\theta$.

The value of the parameter $\theta$ is unknown and my interest in it is limited to testing $H_0:\theta=0$ vs. $H_1:\theta>0$.

Clearly, the alternate hypothesis $H_1$ is composite, thus Neyman-Pearson lemma doesn't apply. Also, unfortunately, the "easy" uniformly most powerful (UMP) test formulation doesn't apply as I don't think that the family of densities $p_\theta(x)$ has a monotone likelihood ratio (see Lehmann and Romano p. 65). As the matter of fact, while, for any $\theta <\theta'$, the distributions $P_\theta$ and $P_{\theta'}$ are distinct, the ratio $p_{\theta'}(x)/p_\theta(x)$ is a non-increasing (as opposed to non-decreasing) function of $T(x)$.

I understand that the maximum likelihood estimator (MLE) $\hat{\theta}_{MLE}$ of $\theta$ is a function of $T(x)$. But what is $T(x)$'s relationship to hypothesis testing?

I ask because a trivial test that solves my problem can be constructed as follows: first, pick threshold $\eta$, then compute estimate $\hat{\theta}_{MLE}$, and accept $H_0$ if $\hat{\theta}_{MLE}<\eta$, reject otherwise. The error probabilities then depend on the choice of $\eta$ and the value of $\theta$. However, can anything be said about optimality (in some way, shape or form) of the said test? Are there any conditions on $\mathcal{P}$ other than monotone likelihood function described above that lead to optimality (maybe something weaker than UMP)? The test using $\hat{\theta}_{MLE}$ seems like the best I can do, however, is there a way to prove this statement?

$\endgroup$
4
$\begingroup$

Not sure if this is an answer. But perhaps a few comments. If I am restating what you are probably already aware of, my apologies.

First, based on the Fisher–Neyman Factorization, if $T(\mathbf{x})$ is a sufficient statistic, then the likelihood function factorizes to the product of (1) a function that does not involve $\theta$; times (2) a function that depends on the sample only through the sufficient statistic $T(\mathbf{x})$. So the first function out of the factorization cancels when one looks at the likelihood ratio. In other words, if there is a sufficient statistic $T(\mathbf{x})$, the likelihood ratio's dependency on the sample is only through $T(\mathbf{x})$. Then, assessing the plausibility of a bigger $\theta$ versus a smaller $\theta$ (i.e., whether or not to reject the null hypothesis) based on the sample $\mathbf{x}$ has to be tied to $T(\mathbf{x})$.

Second, if the ratio is "non-increasing (as opposed to non-decreasing)" function of $T(\mathbf{x})$, wouldn't the ratio automatically be a non-decreasing function of $-T(\mathbf{x})$? Doesn't the theorem about UMP test then apply, now using the "different" sufficient statistic $-T(\mathbf{x})$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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