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).

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