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 stillnecessary thatP(l|m)+P(s|m)=1 and P(l|m)/P(s|m)=r, which is all thatis requiredfor 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. 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.

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. 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.

`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.`

Jesse