>-----Original Message----- >From: Jesse Mazer [mailto:[EMAIL PROTECTED] >Sent: Tuesday, October 05, 2004 8:45 PM >To: [EMAIL PROTECTED]; [EMAIL PROTECTED] >Subject: RE: Observation selection effects > > >Brent Meeker wrote: > >>>-----Original Message----- >>>From: Jesse Mazer [mailto:[EMAIL PROTECTED] >>>Sent: Tuesday, October 05, 2004 6:33 PM >>>To: [EMAIL PROTECTED] >>>Cc: [EMAIL PROTECTED] >>>Subject: RE: Observation selection effects >>> >>>Brent Meeker wrote: >>> >>>>On reviewing my analysis (I hadn't looked at for about four >>>>years), I think it works without the restrictive >assumption that >>>>the range of distirbutions not overlap. It's still >>>necessary that >>>>P(l|m)+P(s|m)=1 and P(l|m)/P(s|m)=r, which is all that >>>is required >>>>for the proof to go thru. >>> >>>Hmm, I think I misread your analysis, I had somehow >>>gotten the idea that you >>>were assuming a uniform probability of the >>>envelope-stuffer picking any >>>number between x1 and x2 for the first envelope, but I >>>see this isn't >>>necessary, so your proof is a lot more general than I >>>thought. Still not >>>completely general though, because the envelope-stuffer >>>can also use a >>>distribution which has no upper bound x2 on possible >>>amounts to put in the >>>first envelope, like the one I mentioned in my last post: >>> >>>>For example, he could use a >>>>distribution that >>>>gives him a 1/2 probability of putting between 0 and 1 >>>>dollars in one >>>>envelope (assume the dollar amounts can take any >>>>positive real value, and he >>>>uses a flat probability distribution to pick a number >>>>between 0 and 1), a >>>>1/4 probability of putting in between 1 and 2 dollars, a >>>>1/8 probability of >>>>putting in between 2 and 3 dollars, and in general a >>>>1/2^n probability of >>>>putting in between n-1 and n dollars. This would insure >>>>there was some >>>>finite probability that *any* positive real number could >>>>be found in either >>>>envelope. >>> >>>Likewise, he could also use the continuous probability >>>distribution P(x) = >>>1/e^x (whose integral from 0 to infinity is 1). And if >>>you want to restrict >>>the amounts in the envelope to positive integers, he >could use a >>>distribution which gives a 1/2^n probability of putting >>>exactly n dollars in >>>the first envelope. >>> >>>Jesse >> >>That doesn't matter if I can do without the no-overlap >>assumption - which I think I can. Do you see a flaw? >> >>When I first did it I was drawing pictures of >distributions and I >>thought I needed non-overlap to assert that P(l|m)/P(s|m)=r and >>P(l|m)+P(s|m)=1. But now it seems that the last >follows just from >>the fact that the amount was either from the larger or the >>smaller. The ratio doesn't depend on the ranges not >overlapping, >>it just depends on the fact that the larger amount's >distribution >>must be a copy of the smaller amount's distribution >stretched by a >>factor of r. > >But in order for the range of the larger amount to be >double that of the >smaller amount, you need to assume the range of the >smaller amount is >finite.
I only had to assume it was finite to avoid overlap in the ranges. Dropping that assumption the range of the lower amount can be infinite. >If the range of the smaller amount is infinite, >as in my P(x)=1/e^x >example, then it would no longer make sense to say that >the range of the >larger amount is r times larger. Sure it does; r*inf=inf. P(s)=exp(-x) -> P(l)=exp(-x/r) > >Also, what if the envelope-stuffer is only picking from >a finite set of >numbers rather than a continuous range? For example, he >might have a 1/3 >chance of putting 100, 125 or 150 dollars in the first >envelope, and then he >would double that amount for the second envelope. In >this case, your >assumption "no matter what m is, it is r-times more >likely to fall in a >large-amount interval than in a small-amount interval" >wouldn't seem to be >valid, since there are only six possible values of m >here (100, 125, 150, >200, 250, or 300) and three of them are in the smaller range. Yes, that's true. The proof depends on smooth, integrable distributions. Brent