# RE: Observation selection effects

```
>-----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:
>>>
>>>>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.```
```
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

```