1 回答
1.
In the case of a Gaussian distribution,
sampling from (X1,...,Xn) under the condition that X1+...+Xn=s
is just sampling from a
conditional Gaussian distribution.
The vector (X1,X2,...,Xn,X1+...+Xn) has a Gaussian distribution, with zero mean, and variance matrix
1 0 0 ... 0 1
0 1 0 ... 0 1
0 0 1 ... 0 1
...
0 0 0 ... 1 1
1 1 1 ... 1 n.
We can therefore sample from it as follows.
s <- 1 # Desired sum
n <- 10
mu1 <- rep(0,n)
mu2 <- 0
V11 <- diag(n)
V12 <- as.matrix(rep(1,n))
V21 <- t(V12)
V22 <- as.matrix(n)
mu <- mu1 + V12 %*% solve(V22, s - mu2)
V <- V11 - V12 %*% solve(V22,V21)
library(mvtnorm)
# Random vectors (in each row)
x <- rmvnorm( 100, mu, V )
# Check the sum and the distribution
apply(x, 1, sum)
hist(x[,1])
qqnorm(x[,1])
For an arbitrary distribution, this approach would require you to compute the conditional distribution, which may not be easy.
2. There is another easy special case: a uniform distribution.
To uniformly sample n (positive) numbers that sum up to 1, you can take n-1 numbers, uniformly in [0,1], and sort them: they define n intervals, whose lengths turn sum up to 1, and happen to be uniformly distributed.
Since those points form a Poisson process, you can also generate them with an exponential distribution.
x <- rexp(n)
x <- x / sum(x) # Sums to 1, and each coordinate is uniform in [0,1]
This idea is explained (with a lot of pictures) in the following article: Portfolio Optimization for VaR, CVaR, Omega and Utility with General Return Distributions, (W.T. Shaw, 2011), pages 6 to 8.
3.
(EDIT) I had initially misread the question, which was asking about sum(pnorm(x)), not sum(x). This turns out to be easier.
If X has a Gaussian distribution, then pnorm(X) has a uniform distribution:
the problem is then to sample from a uniform distribution, with a prescribed sum.
n <- 10
s <- 1 # Desired sum
p <- rexp(n)
p <- p / sum(p) * s # Uniform, sums to s
x <- qnorm(p) # Gaussian, the p-values sum to s