On 6/27/07, John Randall <[EMAIL PROTECTED]> wrote:
Raul Miller wrote:
> I wrote:
>> 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.
>
> I should add that I'm not trying to drop the subject -- but the
> above J expression would always be zero, regardless of the
> underlying data (and regardless of its cardinality), and regardless
> of the sampling technique used, which seems to make its use
> a problem, in a sequence of equivalences intended to show something
> about that magnitude of the cardinality of the sampled data in some
> context.
If you ignore the leading E, I don't see why \sum (x_i-\bar x)^2 is zero
for particular sample values. For example, take n=2, x1=0, x2=2, \bar
x=1. Then \sum (x_i-\bar x)^2 =2.
Oops, because I forgot to square them. Thanks for catching that.
That said, note that I believe when n is inside the E it should represent
the cardinality of the complete set of all potential values rather than the
cardinality of the sampled variables.
In other words, I am extremely uncomfortable with a proof about
the magnitude of n which moves n from inside the E() to outside
the E() with no comment whatsoever as to why this is a valid
operation.
On 6/27/07, John Randall <[EMAIL PROTECTED]> wrote:
Raul Miller wrote:
> Well... I do use cases where X and \bar X have numeric value to test
> that I understand the math properly.
This does not work. I have tried to emphasize that these are random
variables, not numbers.
I have been modelling random variables as a vector of potential
values or as functions applied to these vectors. I interpret "these
are random variables, not numbers" not to mean that I cannot use
arithmetic notation in this context, but instead as a caution that
the cardinality of these vectors is independent of the cardinality
of any sample vectors.
>> 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.
Sure it does. Let X be the random variable counting the number of heads
when 10 coins are tossed. Then Y=X-5 is a perfectly good random variable.
Granted. But that's a different issue, from my point of view.
Here, the cardinality of Y (and of X) matches the cardinality of my
potential value vector. However, both \mu and \bar X are
singletons (since the corresponding potential value vector has
been summed).
That said, I think I should also be able to do something like X=5, and
have a vector (of the same cardinality as X) of true/false values.
--
Raul
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