# RE: Observation selection effects

From: Jesse Mazer <lasermazer.domain.name.hidden>
Date: Tue, 05 Oct 2004 16:45:00 -0400

Brent Meeker wrote:

>>-----Original Message-----
>>From: Jesse Mazer [mailto:lasermazer.domain.name.hidden]
>>Sent: Tuesday, October 05, 2004 6:33 PM
>>To: meekerdb.domain.name.hidden
>>Cc: everything-list.domain.name.hidden
>>Subject: RE: Observation selection effects
>>
>>Brent Meeker wrote:
>>
>>>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.
>>
>>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
Received on Tue Oct 05 2004 - 16:53:20 PDT

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