Maybe if you can't find a closed form inversion of a mixed distribution, you
would want to select the component distribution based on the first random
draw?
If you can find a closed form inversion, I can't imagine why you would want
to "waste" a random number, since these appear to be finite from the
simulation algorithms.
Mahler's example doesn't follow that mold, because the distributions can be
combined and inverted without trouble.
Anyone else have an idea?
-----Original Message-----
From: Sipes, Summer [SMTP:SSIPE@Allstate.COM]
Sent: Monday, October 18, 1999 6:28 PM
To: studygroup4b@lists.casact.org
Subject: Spring 97 question 26
It's also the CSM question A11 in Herzog S.
I understand how they calculate the answer, but according to
Mahler's notes
in Section 8 of Simulation under Mixed Distributions:
"In order to simulate a random draw from a mixed
distribution, first one would simulate a random number from 0 to 1
and see
whether it is less than or equal to p, the weight applied to the
first
distribution. If so then one simulates a random draw from the first
distribution."
Mahler then goes on to give an example of this. You cannot do this
method,
though. You get the wrong answer. (And they don't give you a second
random
number.) Which tipped me off not to do it the way I wanted to.
When would you use the method he describes? It uses inversion as
well, so I
assume they'd be the same.
How am I looking at this wrong?