- Contemporary messages sorted: [ by date ] [ by thread ] [ by subject ] [ by author ] [ by messages with attachments ]

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

Date: Wed, 06 Oct 2004 17:26:40 -0400

in my last response to Brent Meeker I wrote:

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

*
This archive was generated by hypermail 2.3.0
: Fri Feb 16 2018 - 13:20:10 PST
*