RE: Observation selection effects

From: Jesse Mazer <lasermazer.domain.name.hidden>
Date: Wed, 06 Oct 2004 17:26:40 -0400

in my last response to Brent Meeker I wrote:

>As for your statement that "P(s)=exp(-x) -> P(l)=exp(-x/r)", that can't be
>true. It doesn't make sense that the value of the second probability
>distribution at x would be exp(-x/r), since the range of possible values
>for the amount in that envelope is 0 to infinity, but the integral of
>exp(-x/r) from 0 to infinity is not equal to 1, so that's not a valid
>probability distribution.

thinking a little more about this I realized Brent Meeker was almost right
about the second probability distribution in this case--it would actually be
(1/r)*e^(-x/r), so he was just off by a constant factor. In general, if the
probability distribution for the envelope with the smaller amount is f(x),
then the probability distribution for the envelope with r times that amount
should be (1/r)*f(x/r)...this insures that if you integrate f(x) over an
interval (a,b), giving the probability the smaller envelope contains an
amount between a and b, then this will be equal to the integral of
(1/r)*f(x/r) over the interval (r*a, r*b).

Jesse
Received on Wed Oct 06 2004 - 17:38:16 PDT