View Full Version : inverse transform method
July 24th 2006, 07:30 PM
I see from several released Exam M's a simulation technique called "inverse transform method", but I have no idea what that is and have never read it on the suggested texts...Would anyone tell me what that is, or if we're supposed to have read it somewhere?
July 24th 2006, 08:38 PM
Also known as inversion method. No longer on Exam M; moved to Exam C.
July 24th 2006, 08:40 PM
(Goes to get Mahler's notes on this that I almost threw out this weekend but didn't because I had eleventy other things going on)
As mentioned, it's on Exam C and not on Exam M - but since I went to the trouble of digging the material out ... :D
Inverse Transform Method - Continuous Distributions
The general idea is that you're given a distribution function F(x), a random number on (0,1), and are told whether large random numbers correspond to large or small losses. (This is important; if no distinction is given, assume large numbers correspond to large losses - but this should be specified on the exam.) The random number determines the size of the loss.
The steps (if large numbers correspond to large losses):
1. Let u = random number on (0,1).
2. Set u = F(x)
3. Solve for x.
Example: Let F(x) be an exponential distribution with mean 1000. You are given .988 as a random draw on (0,1). What is the size of the loss?
Answer: .988 = 1 - e^(-x/1000). Solving for x, we get -x/1000 = - ln(0.12) = -4.42285
==> x = 4422.85
If large numbers correspond to small losses, then set u = S(x). You can also set this up as 1-u = F(x); either way, then solve for x.
Inverse Transform Method - Discrete Distributions
Same idea here; generally you're solving for the number of events (claims) that occur. Again, you get a distribution function, a random number on (0,1) and are told if large random numbers correspond to large (or small) values.
Given the random number u,
-- If large random numbers correspond to large values: find the smallest x such that F(x) > u
-- If large random numbers correspond to small values: find the smallest x such that F(x) > 1-u, or if you prefer S(x) > u.
Example: Given a Poisson with mean 3 and the random number .673 with large numbers corresponding to large values, find the number of claims simulated.
Answer: (using the POISSON function in Excel)
F(0) = .049787 < .673 - Keep going.
F(1) = .199148 < .673 - Keep going.
F(2) = .423190 < .673 - Keep going.
F(3) = .647232 < .673 - Keep going.
F(4) = .815263 > .673 - Stop.
==> x = 4.
July 25th 2006, 01:44 AM
Thank you, Abraham Weishaus and Irish Blues :]
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