Neyman-Pearson lemma with pdf of $X$ vs Neyman-Pearson Lemma with likelihood functions
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As written in the subject line, I’m confused about the statement made in the Neyman-Pearson Lemma, largely because one of my two textbooks states the Lemma in terms of the pdf of $X$ while the other textbook states the Lemma in terms of the likelihood function of $X$. I don’t believe that the two statements are logically equivalent. Any thoughts on this discrepancy?
statistics statistical-inference hypothesis-testing
asked Nov 28 '18 at 16:52
DavidDavid
337210
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add a comment |
$begingroup$
As written in the subject line, I’m confused about the statement made in the Neyman-Pearson Lemma, largely because one of my two textbooks states the Lemma in terms of the pdf of $X$ while the other textbook states the Lemma in terms of the likelihood function of $X$. I don’t believe that the two statements are logically equivalent. Any thoughts on this discrepancy?
statistics statistical-inference hypothesis-testing
asked Nov 28 '18 at 16:52
DavidDavid
337210
$endgroup$
$begingroup$
If you have a sample $X=(X_1,ldots,X_n)$, then it is the joint density of $(X_1,ldots,X_n)$ (likelihood function) and if you have a single observation $X$ (i.e. for $n=1$), then it is the pdf of $X$. What's the difference?
$endgroup$
– StubbornAtom
Nov 29 '18 at 15:46
add a comment |
$begingroup$
As written in the subject line, I’m confused about the statement made in the Neyman-Pearson Lemma, largely because one of my two textbooks states the Lemma in terms of the pdf of $X$ while the other textbook states the Lemma in terms of the likelihood function of $X$. I don’t believe that the two statements are logically equivalent. Any thoughts on this discrepancy?
statistics statistical-inference hypothesis-testing
asked Nov 28 '18 at 16:52
DavidDavid
337210
$endgroup$
As written in the subject line, I’m confused about the statement made in the Neyman-Pearson Lemma, largely because one of my two textbooks states the Lemma in terms of the pdf of $X$ while the other textbook states the Lemma in terms of the likelihood function of $X$. I don’t believe that the two statements are logically equivalent. Any thoughts on this discrepancy?
statistics statistical-inference hypothesis-testing
statistics statistical-inference hypothesis-testing
asked Nov 28 '18 at 16:52
DavidDavid
337210
asked Nov 28 '18 at 16:52
DavidDavid
337210
asked Nov 28 '18 at 16:52
DavidDavid
337210
asked Nov 28 '18 at 16:52
DavidDavid
337210
asked Nov 28 '18 at 16:52
DavidDavid
337210
337210
$begingroup$
If you have a sample $X=(X_1,ldots,X_n)$, then it is the joint density of $(X_1,ldots,X_n)$ (likelihood function) and if you have a single observation $X$ (i.e. for $n=1$), then it is the pdf of $X$. What's the difference?
$endgroup$
– StubbornAtom
Nov 29 '18 at 15:46
add a comment |
$begingroup$
If you have a sample $X=(X_1,ldots,X_n)$, then it is the joint density of $(X_1,ldots,X_n)$ (likelihood function) and if you have a single observation $X$ (i.e. for $n=1$), then it is the pdf of $X$. What's the difference?
$endgroup$
– StubbornAtom
Nov 29 '18 at 15:46
$begingroup$
If you have a sample $X=(X_1,ldots,X_n)$, then it is the joint density of $(X_1,ldots,X_n)$ (likelihood function) and if you have a single observation $X$ (i.e. for $n=1$), then it is the pdf of $X$. What's the difference?
$endgroup$
– StubbornAtom
Nov 29 '18 at 15:46
$begingroup$
If you have a sample $X=(X_1,ldots,X_n)$, then it is the joint density of $(X_1,ldots,X_n)$ (likelihood function) and if you have a single observation $X$ (i.e. for $n=1$), then it is the pdf of $X$. What's the difference?
$endgroup$
– StubbornAtom
Nov 29 '18 at 15:46
add a comment |
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$begingroup$
If you have a sample $X=(X_1,ldots,X_n)$, then it is the joint density of $(X_1,ldots,X_n)$ (likelihood function) and if you have a single observation $X$ (i.e. for $n=1$), then it is the pdf of $X$. What's the difference?
$endgroup$
– StubbornAtom
Nov 29 '18 at 15:46