RE: Observation selection effects

[Jesse Mazer]

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. 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