pmnorm: probabilites don't sum up to 1

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!

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.mes at cbs.dk  Priv: PDalgd at gmail.com