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If I rewrite my original questions 2 and 3 to be consistent with me
re-writing of question 1...
I can calculate N sample standard deviations from N sets of samples,
where each element of each sample is independently taken from the same
known but arbitrary probability distribution. Does a general method
exist for calculating the expected distribution of the N sample standard
deviations
2A. In the case where each of the N sets contains exactly M
elements?
2B. In the case where each of the N sets contains a different
number of elements?
Also, if I replace "standard deviation" in the above with any higher
moment about the mean...
3. Does such a method exist?
...can I assume that the answers stay the same?
-- Andrew
Herman Rubin wrote:
>Andrew Morse <[EMAIL PROTECTED]> wrote:
>
>
>
>> 2. Does a general method exist for calculating the expected
>>distribution of standard deviations of repeated trials of N samples
>>taken from an arbitrary, known probability distribution?
>>
>>
>
>This can only be done in closed form for a few
>distributions. This even applies to the simpler problem of
>the second moment about 0. Using complex variable methods,
>it may well be possible to do it numerically in a reasonable
>amount of time.
>
>The general method for this, if it can be done, is to compute
>the characteristic function of the moment, raise it to the N-th
>power, and invert.
>
>
>
>
>> 3. Does a general method exist for calculating the expected
>>distribution of any of the moments of repeated trials of N samples taken
>>
>>
>>from an arbitrary, known probability distribution?
>
>The same remarks as above hold.
>
>If the 2k-th moment exists, the Central Limit Theorem gives
>the asymptotic normal distribution of the k-th moment about 0.
>Similar results hold for ths standard deviation of th fourth
>moment exists. These asymptotic results are very old.
>
>
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If I rewrite my original questions 2 and 3 to be consistent with me
re-writing of question 1... <br>
<br>
I can calculate N sample standard deviations from N sets of
samples, where each element of each sample is independently taken from
the same known but arbitrary probability distribution. Does a general
method exist for calculating the expected distribution of the N sample
standard deviations<br>
2A. In the case where each of the N sets
contains exactly M
elements?<br>
2B. In the case where each of the N sets
contains a different
number of elements? <br>
<br>
Also, if I replace "standard deviation" in the above with any
higher moment about the mean...<br>
3. Does such a method exist?<br>
<br>
...can I assume that the answers stay the same?<br>
<br>
-- Andrew<br>
<br>
<br>
<br>
Herman Rubin wrote:<br>
<blockquote type="cite" cite="[EMAIL PROTECTED]">
<pre wrap="">
Andrew Morse <a class="moz-txt-link-rfc2396E" href="mailto:[EMAIL
PROTECTED]"><[EMAIL PROTECTED]></a> wrote:
</pre>
<blockquote type="cite">
<pre wrap=""> 2. Does a general method exist for calculating the expected
distribution of standard deviations of repeated trials of N samples
taken from an arbitrary, known probability distribution?
</pre>
</blockquote>
<pre wrap=""><!---->
This can only be done in closed form for a few
distributions. This even applies to the simpler problem of
the second moment about 0. Using complex variable methods,
it may well be possible to do it numerically in a reasonable
amount of time.
The general method for this, if it can be done, is to compute
the characteristic function of the moment, raise it to the N-th
power, and invert.</pre>
</blockquote>
<blockquote type="cite" cite="[EMAIL PROTECTED]">
<pre wrap="">
</pre>
<blockquote type="cite">
<pre wrap=""> 3. Does a general method exist for calculating the expected
distribution of any of the moments of repeated trials of N samples taken
</pre>
</blockquote>
<pre wrap=""><!---->>from an arbitrary, known probability distribution?
The same remarks as above hold.
If the 2k-th moment exists, the Central Limit Theorem gives
the asymptotic normal distribution of the k-th moment about 0.
Similar results hold for ths standard deviation of th fourth
moment exists. These asymptotic results are very old.
</pre>
</blockquote>
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