How are critical values derived for the Kolmogorov-Smirnov Test?
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One appealing feature of the K-S test is that it is distribution-free. So this leads to my question - how are the critical values for the K-S derived, then? Is there a way to express the critical values as an integral, like for percentiles of the standard normal distribution?
Sources that have such information would be very helpful (i.e., a textbook).
See, for example, the table below (from http://people.cs.pitt.edu/~lipschultz/cs1538/prob-table_KS.pdf).
statistics
asked Sep 22 '14 at 2:38
ClarinetistClarinetist
10.9k42778
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add a comment |
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One appealing feature of the K-S test is that it is distribution-free. So this leads to my question - how are the critical values for the K-S derived, then? Is there a way to express the critical values as an integral, like for percentiles of the standard normal distribution?
Sources that have such information would be very helpful (i.e., a textbook).
See, for example, the table below (from http://people.cs.pitt.edu/~lipschultz/cs1538/prob-table_KS.pdf).
statistics
asked Sep 22 '14 at 2:38
ClarinetistClarinetist
10.9k42778
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Your link is dead.
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– Astrid
Jan 6 at 23:18
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Here it is: people.cs.pitt.edu/~lipschultz/cs1538/prob-table_KS.pdf
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– Astrid
Jan 6 at 23:30
add a comment |
$begingroup$
One appealing feature of the K-S test is that it is distribution-free. So this leads to my question - how are the critical values for the K-S derived, then? Is there a way to express the critical values as an integral, like for percentiles of the standard normal distribution?
Sources that have such information would be very helpful (i.e., a textbook).
See, for example, the table below (from http://people.cs.pitt.edu/~lipschultz/cs1538/prob-table_KS.pdf).
statistics
asked Sep 22 '14 at 2:38
ClarinetistClarinetist
10.9k42778
$endgroup$
One appealing feature of the K-S test is that it is distribution-free. So this leads to my question - how are the critical values for the K-S derived, then? Is there a way to express the critical values as an integral, like for percentiles of the standard normal distribution?
Sources that have such information would be very helpful (i.e., a textbook).
See, for example, the table below (from http://people.cs.pitt.edu/~lipschultz/cs1538/prob-table_KS.pdf).
statistics
statistics
asked Sep 22 '14 at 2:38
ClarinetistClarinetist
10.9k42778
asked Sep 22 '14 at 2:38
ClarinetistClarinetist
10.9k42778
edited Jan 7 at 13:19
Clarinetist
asked Sep 22 '14 at 2:38
ClarinetistClarinetist
10.9k42778
asked Sep 22 '14 at 2:38
ClarinetistClarinetist
10.9k42778
asked Sep 22 '14 at 2:38
ClarinetistClarinetist
10.9k42778
10.9k42778
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Your link is dead.
$endgroup$
– Astrid
Jan 6 at 23:18
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Here it is: people.cs.pitt.edu/~lipschultz/cs1538/prob-table_KS.pdf
$endgroup$
– Astrid
Jan 6 at 23:30
add a comment |
$begingroup$
Your link is dead.
$endgroup$
– Astrid
Jan 6 at 23:18
$begingroup$
Here it is: people.cs.pitt.edu/~lipschultz/cs1538/prob-table_KS.pdf
$endgroup$
– Astrid
Jan 6 at 23:30
$begingroup$
Your link is dead.
$endgroup$
– Astrid
Jan 6 at 23:18
$begingroup$
Your link is dead.
$endgroup$
– Astrid
Jan 6 at 23:18
$begingroup$
Here it is: people.cs.pitt.edu/~lipschultz/cs1538/prob-table_KS.pdf
$endgroup$
– Astrid
Jan 6 at 23:30
$begingroup$
Here it is: people.cs.pitt.edu/~lipschultz/cs1538/prob-table_KS.pdf
$endgroup$
– Astrid
Jan 6 at 23:30
add a comment |
1 Answer
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If you are testing fit to a specific distribution, then note that for a sample of data $X$ and hypothesized distribution $F, F^{-1}(X) sim U(0,1)$. Therefore, you need to derive the distribution of the largest vertical difference between your transformed sample and the hypothesized CDF, which will be taken to be the standard uniform. The distribution of this is not mathematically nice. Here is a paper that walks you though why...and why we just use numerical methods.
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1 Answer
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
If you are testing fit to a specific distribution, then note that for a sample of data $X$ and hypothesized distribution $F, F^{-1}(X) sim U(0,1)$. Therefore, you need to derive the distribution of the largest vertical difference between your transformed sample and the hypothesized CDF, which will be taken to be the standard uniform. The distribution of this is not mathematically nice. Here is a paper that walks you though why...and why we just use numerical methods.
$endgroup$
add a comment |
$begingroup$
If you are testing fit to a specific distribution, then note that for a sample of data $X$ and hypothesized distribution $F, F^{-1}(X) sim U(0,1)$. Therefore, you need to derive the distribution of the largest vertical difference between your transformed sample and the hypothesized CDF, which will be taken to be the standard uniform. The distribution of this is not mathematically nice. Here is a paper that walks you though why...and why we just use numerical methods.
$endgroup$
add a comment |
$begingroup$
If you are testing fit to a specific distribution, then note that for a sample of data $X$ and hypothesized distribution $F, F^{-1}(X) sim U(0,1)$. Therefore, you need to derive the distribution of the largest vertical difference between your transformed sample and the hypothesized CDF, which will be taken to be the standard uniform. The distribution of this is not mathematically nice. Here is a paper that walks you though why...and why we just use numerical methods.
$endgroup$
If you are testing fit to a specific distribution, then note that for a sample of data $X$ and hypothesized distribution $F, F^{-1}(X) sim U(0,1)$. Therefore, you need to derive the distribution of the largest vertical difference between your transformed sample and the hypothesized CDF, which will be taken to be the standard uniform. The distribution of this is not mathematically nice. Here is a paper that walks you though why...and why we just use numerical methods.
answered Sep 22 '14 at 18:50
user76844
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Your link is dead.
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– Astrid
Jan 6 at 23:18
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Here it is: people.cs.pitt.edu/~lipschultz/cs1538/prob-table_KS.pdf
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– Astrid
Jan 6 at 23:30