Hi Peter,
Thx for your comment - I guess you're right... I made a drawing to
better understand which quadrants overlap
So I ende up with this Code:
library(mnormt) # can handle multivariate normal distributions
uX = 2
uY = 1
uZ = .5
mean = c(uX, uY, uZ)
LX = matrix(c(1,-1,0,1,0,-1), 2, 3, byrow = TRUE)
LY = matrix(c(-1,1,0,0,1,-1), 2, 3, byrow = TRUE)
LZ = matrix(c(-1,0,1,0,-1,1), 2, 3, byrow = TRUE)
muX = LX %*% mean
muY = LY %*% mean
muZ = LZ %*% mean
Sigma = diag(2)
mean = c(0,0)
pX = pmnorm(LXmean,c(0,0),varcov)/(1-(pmnorm(-LXmean,c(0,0),varcov)))
pY = pmnorm(LYmean,c(0,0),varcov)/(1-(pmnorm(-LYmean,c(0,0),varcov)))
pZ = pmnorm(LZmean,c(0,0),varcov)/(1-(pmnorm(-LZmean,c(0,0),varcov)))
pX + pY + pZ
pX is now expressed as a relation:
uX-uY>0 & uX-uZ> = pmnorm(LXmean,c(0,0),varcov) --> counter
in relation to correspondig surface area =
(1-(pmnorm(-LXmean,c(0,0),varcov))) --> denominator
This now sums up to 1 (or nearly) - I guess this should be the correct
way of doing it, right?
Thanks,
Nicolas
Zitat von peter dalgaard <pda...@gmail.com>:
On Dec 15, 2010, at 11:40 , Nicolas Berkowitsch wrote:
Dear list member,
I struggle with the problem, why the probabilities of choosing one
of three mutually exclusive alternatives don?t sum up to 1!
Let?s assume we have three alternatives X, Y, and Z. Let?s further
assume we know their respective utilities:
uX, uY, uZ. I?m interested in calculating the probability of
choosing X, Y, and Z.
Since I assume that the alternatives are mutually exclusive, the
probabilities p(X), p(Y), and p(Z) have to sum up to one. The
utilities of the 3 alternatives can be expressed in 2 utility
differences and, hence, the multivariate case reduces to a
bivariate normal distribution. If I assume that X, Y, and Z are
independent, their corresponding correlations have to be zero and,
hence, the variance-covariance-matrices are set to be a
diagonal-matrix (i.e., identity-matrix).
To calculate p(X), p(Y), and p(Z) I was using the following R-code:
library(mnormt) # can handle multivariate normal distributions
uX = 2
uY = 1
uZ = .5
mu = c(uX, uY, uZ)
LX = matrix(c(1,-1,0,1,0,-1), 2, 3, byrow = TRUE)
LY = matrix(c(-1,1,0,0,1,-1), 2, 3, byrow = TRUE)
LZ = matrix(c(-1,0,1,0,-1,1), 2, 3, byrow = TRUE)
muX = LX %*% mu
muY = LY %*% mu
muZ = LZ %*% mu
Sigma = diag(2)
mean = c(0,0)
pX = pmnorm(muX, mean, Sigma)
pY = pmnorm(muY, mean, Sigma)
pZ = pmnorm(muZ, mean, Sigma)
pX + pY + pZ
I don?t see why the three probabilities don?t sum up to 1?
I know two ?solutions? to this problem so far. However, neither of
them satisfies me:
1. I can set pZ to 1 ? pX ? pY, but doing so, returns a different
result for pZ, than calculating pZ directly using pmnorm.
2. I could calculate the relationship of pX to the sum of pX + pY +
pZ (? pX/(pX + pY + pZ) )
Can anyone explain to me why the probabilities don?t sum up to 1?
How should I rewrite the R-code to overcome this problem?
Thanks a lot for any advice!
I don't think the pX, pY, pZ are probabilities of choosing X, Y, Z.
If you think that they are, then you need to explain it more
convincingly.
What they are are probabilities of three lower-left quadrants with
different origins (muX, muY, muZ). Such quadrants will in general
overlap, so there is no reason to expect their probabilities to sum
to any particular value. If you were expecting to have a partition
of 2d space into three disjoint regions and calculating the
probability of each region, then pmnorm is not the right tool.
--
Peter Dalgaard
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd....@cbs.dk Priv: pda...@gmail.com
____________
lic. phil. Nicolas A. J. Berkowitsch
Universität Basel
Fakultät für Psychologie
Economic Psychology
Missionsstrasse 62a
CH-4055 Basel
Tel. +41 61 267 05 75
E-Mail nicolas.berkowit...@unibas.ch
Web
http://psycho.unibas.ch/abteilungen/abteilung-details/home/abteilung/economic-psychology/
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