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In my book: $\mathbf{X}=(X_1,\ldots,X_n)$ $f(\mathbf{x})$ is the joint density, where $f$ is either $f_0 \text{ or } f_1$.

Suppose we want to test $H_0: f=f_0$ or $H_1: f=f_1$. The test, whose test function is

$$\phi(\mathbf{X})=1\text{ if }\frac{f_1}{f_0}\geq k;$$

$$\phi(\mathbf{X})=0 \text{ otherwise,}$$

(for some $0<k<\infty$) is a most powerful test of $H_0$ versus $H_1$ at level $E_0(\phi(\mathbf{X}))$.

My question is how is $k$ defined? Can I interpret the lemma as if $\forall k \in (0,\infty)$ there will be a test function $\phi(\mathbf{X})$ such that it will determine the size for which the test with test function $\phi(\mathbf{X})$ is most powerful?

I'm just trying to understand which kind of relationship $k$ and the size of the test have between each other.

EDIT:

Here is a citation from the book I'm using: «the Neyman-Pearson lemma as stated here does not guarantee the existence of an MP $\alpha$ level test but merely states that the test that rejects $H_0$ for $T(X)\geq k$ will be an MP for some level $\alpha$» This makes me want to interpret as $\forall k \exists \alpha$. May I?

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  • $\begingroup$ Please don't cross-post your questions here and on stats.stackexchange. $\endgroup$
    – Stefan Hansen
    Commented Jul 30, 2014 at 15:01

2 Answers 2

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I don't think your interpretation is correct. What the lemma is saying is that if you use the likelihood ratio as your test function (or something that is equivalent to it), then it is the most powerful test of the null vs. alt. hypothesis for a particular type I error rate. In particular, it has the highest power of all tests that have a significance level of $E_0(\phi(\mathbf{X}))$

The value $k$ defines the "rejection region" of the likelihood ratio test, and it determines the expected value of $\phi(\mathbf{X}): \phi(\mathbf{X})=1\text{ if }\frac{f_1}{f_0}\geq k; \;\;0 \;\;\text{otherwise}$

This is spelled out by the expression $E_0(\phi(\mathbf{X}))$ in the text; which says that the expected value of the test function un*der the null distribution* (i.e., that's why the subscript is $0$) is the level of the test. We can make this clearer by re-writting this as $E_0(\phi(\mathbf{X})|k)$ to make it clear that it depends on $k$. Now, $E_0(\phi(\mathbf{X})|k)=\alpha=$Type I error rate (as per Michael Hardy) in typical "hypothesis test" terminology.

The larger you make $k$, the smaller you make $E_0(\phi(\mathbf{X})|k)$ and hence you reduce your chances of rejecting the null when it is in fact true.

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  • $\begingroup$ Thanks for your answer. If the size isn't given, can I choose any $k \in (0,\infty)$? $\endgroup$
    – An old man in the sea.
    Commented Jul 30, 2014 at 15:07
  • $\begingroup$ @Anoldmaninthesea. For a given $f_0,f_1$, each value of $k$ determines a specific test "size" or type I error rate. if you don't care about the type I error rate, then why do any hypothesis test at all? Usually, you want to find $k:E_0(\phi(\mathbf{X})|k)\leq\alpha$ where $\alpha$ is your desired test size. $\endgroup$
    – user76844
    Commented Jul 30, 2014 at 15:14
  • $\begingroup$ Here is a citation from the book «the Neyman-Pearson lemma as stated here does not guarantee the existence of an MP$\alpha$ level test but merely states that the test that rejects $H_0$ for $T(X)\geq k$ will be an MP for some level $\alpha$» This makes me want to interpret as $\forall k \exists \alpha$. May I? $\endgroup$
    – An old man in the sea.
    Commented Jul 30, 2014 at 15:23
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    $\begingroup$ @Anoldmaninthesea. Your latter formulation is correct. Each $k$ can be associated with a most powerful test of level $\alpha$ for some $\alpha$ $\endgroup$
    – user76844
    Commented Jul 30, 2014 at 16:16
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One chooses $k$ usually by deciding how much probability of Type I error one will tolerate (i.e. falsely rejecting $H_0$), balanced against the power of the test, which is the probability of rejecting $H_0$ when $H_0$ is false. And that is a subjecctive economic decision.

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  • $\begingroup$ ,I thank you for your answer. Your answer seems to be from a more applied perspective. I was looking for an answer more based on the the text of the lemma itself. Could you help me in that way? $\endgroup$
    – An old man in the sea.
    Commented Jul 30, 2014 at 14:12

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