I have a point $\mathbf{x}$ in 3-dimensional space, which is measured with a degree of uncertainty. The point falls within a unit cube, and the uncertainty is assumed to follow a multivariate normal distribution with mean $\mathbf{\mu}=\mathbf{x}$ and covariance $\mathbf{\Sigma}$, where $$ \mathbf{\Sigma} = \begin{bmatrix} e_x^2 & 0 & 0\\ 0 & e_y^2 & 0\\ 0 & 0 & e_z^2 \end{bmatrix}. $$ To project this point into a given 2-dimensional space I first normalise $\mathbf{x}$ so its values sum to 1, and then use $\mathbf{x}'=\mathbf{V}\mathbf{x}$, where $\mathbf{x}'$ is the projected point and $$ \mathbf{V} = \begin{bmatrix} \sqrt{\dfrac{1}{2}} & 0 & -\sqrt{\dfrac{1}{2}}\\ -\sqrt{\dfrac{2}{3}}\left(\dfrac{1}{2}\right) & \sqrt{\dfrac{2}{3}} & -\sqrt{\dfrac{2}{3}}\left(\dfrac{1}{2}\right) \end{bmatrix}. $$ However I am struggling to also project the covariance matrix. By simulating random 3-dimensional vectors with a given mean and covariance and then projecting these I can visualise the projected covariance: for example, for a covariance matrix with diagonal elements $e_x^2=0.0005$, $e_y^2=0.0003$ and $e_z^2=0.0001$, the figure below shows 10000 simulated points for $\mathbf{x}=[0.6,0.1,0.1]$ (red points), $\mathbf{x}=[0.1,0.6,0.1]$ (green points), and $\mathbf{x}=[0.1,0.1,0.6]$ (blue points):
This can be reproduced using the following R code:
library(mvtnorm) # For rmvnorm
V = rbind(c(sqrt(1/2), 0, -sqrt(1/2)),
c(-sqrt(2/3)*(1/2), sqrt(2/3), -sqrt(2/3)*(1/2)))
x = c(0.6, 0.1, 0.1)
Sigma = diag(c(0.0005, 0.0003, 0.0001))
x.prime = matrix(0, nrow=2, ncol=10000)
for(i in 1:10000) {
r = rmvnorm(n=1, mean=x, sigma=Sigma)
r = r/sum(r) # Normalise
x.prime[,i] = V %*% t(r) # Project
}
plot(x.prime[1,], x.prime[2,], pch=16, col=rgb(0,0,0,0.1), asp=1)
Any help on how to obtain the projected covariance matrix would be very much appreciated.
