Raul Miller wrote: > On 6/27/07, John Randall <[EMAIL PROTECTED]> wrote:
> 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. This is trivial. Suppose X1,...,Xn are independent random variables, and a1,...,an are numbers. Then E(a1 X1+...+an Xn)=a1 E(X1)+...+an E(Xn), that is, E is linear (indeed affine) on independent variables. This just follows from the linearity of summation or integration. > > 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). > Are the components of your value vector independent? If so, you can go component by component. > 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. > OK, now I understand what you are getting at. If X is a random variable, X=5 (interpreted as a 0/1 result) is also a random variable. If X is a vector, and the components of X are independent, so are the components of X=5. Best wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
