On 6/27/07, John Randall <[EMAIL PROTECTED]> wrote:
It does not matter how accurate your model is except in the degenerate
case of a constant distribution: \bar X is random variable, while \mu is a
number.
Well... I do use cases where X and \bar X have numeric value to test
that I understand the math properly.
It makes no sense to talk about \mu and \bar X being equal.
If they are not comparable I do not see why it makes sense to
talk about their difference.
Put differently, if I accept that it makes no sense to talk about
them being equal, I wind up with concepts like
E((1/n-1)\sum (X_i-\bar X)^2
not depending on \bar X and X_i but instead on mu and E(X).
However, another issue, I think is that E(n) and n should be
different values. With this (and possibly other issues I have
not worked through) I can't reasonably derive sigma or
S from your latex example.
Anyways, here's how I think of a simple "non-degenerate case"
where E(\bar X) and \bar X are different:
data=:1 2 3
E=:] NB. expected data
S=:2&{. NB. sampled data
mean=:+/ % #
eX=: E data NB. E(X)
sX=: S data NB. X
eM=: mean E data NB. \mu = E(\bar X)
sM=: mean S data NB. \bar X
eN=: #eX NB. E(n)
sN=: #sX NB. n
NB. \sum ((X_i-\mu)-(\bar X -\mu))^2
+/(sX-sM)^2
NB. \sum ((X_i-\mu)^2) -n(\bar X-\mu)^2
+/((sX-eM)-(sM-eM))^2
NB. \sum ((X_i-\mu)^2) -n(\bar X-\mu)^2
(+/(sX-eM)^2)-sN*(sM-eM)^2
NB. \sum (X_i-\mu)=(\sum X_i) - n\mu = n(\bar X-\mu)
(+/sX-eM);((+/sX)-sN*eM);sN*sM-eM
NB. E((1/n-1)\sum (X_i-\bar X)^2
(1%eN-1)*+/eX-eM
... anyways, around here I'm not sure if it's worth
proceeding, because I worry that either you have made
assumptions I disagree with or I have made assumptions
you disagree with or both.
--
Raul
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
