Cumulative distribution of Gaussian conditional independent random variables

Question

Suppose X, Y, Z are three jointly Gaussian random variables and X and Z are independent given Y. For example, take three r.v. from a OU process. Here is some R code:

nr <- 3
gg <- seq(0, 1, length.out=nr)
C <- exp(-.1*as.matrix(dist(gg)))

The precision matrix has a zero at element 1,3 and 3,1: so the first (X) and third (Z) random variables are conditionally independent given the second (Y):

P <- solve(C) # 1 and 3 are conditionally independent given 2

Now it would seem to me that these equalities hold by the rules of probability and conditioning: $$ \frac{P(X < x , Y < y)}{P(Y < y)} = P(X < x | Y < y) = P(X < x | Y < y, Z < z) = \frac{P(X < x, Y < y, Z < z)}{P(Y < y, Z < z)} $$ because events related to Z should not change our probability assessments of events related to X, due to conditional independence on Y. However, here for example I get drastically different results:

evalx <- c(-1, 1, -1)
# define a cdf for simplicity
Pr <- \(index) mvtnorm::pmvnorm(upper=evalx[ index ], sigma = C[ index, index ])

x <- 1
y <- 2
z <- 3
xy <- 1:2
yz <- 2:3
xyz <- 1:3

Pr(xy) / Pr(y) # P(X < x | Y < y): 0.186655 
Pr(xyz) / Pr(yz) # P(X < x | Y < y, Z < z): 0.3238458

Where am I making a mistake?

Answer 1

That is really perplexing, but i think i have your answer:

The problem is that $Y < y$ does not uniquely identify $Y$ and so the conditional random variable $Y|Y<y$ is not independent from the event $Z < z$, allowing it to still influence $X|Y<y$. So while $ P(X < x|Y = y) = P(X < x|Y = y, Z < z)$:$$ P(X < x|Y < y) \neq P(X < x|Y < y, Z < z) $$

I was not entirely sure, what you wanted to do in your calculation and whether you did it right, so i made a simple simulation which allowed me to plot the very different distributions of $Y|Y<y$ and $Y|Y<y, Z<z$:

R-Code

library(mvtnorm)
nr <- 3
gg <- seq(0, 1, length.out=nr)
C <- exp(-.1*as.matrix(dist(gg)))
solve(C)

n <- 10^5
dat <- data.frame(rmvnorm(n, sigma = C))
colnames(dat) <- c("X", "Y", "Z")
x <- -1
y <- 1
sum((dat$X < x) * (dat$Y < y))/sum(dat$Y < y) # ~0.19
z <- -1
sum((dat$X < x) * (dat$Y < y) * (dat$Z < z))/sum((dat$Y< y) * (dat$Z< z)) # ~0.73, different from yours?

hist(dat$Y[dat$Y< y], breaks = 30, col = rgb(1,0, 0, .5), freq = F, ylim = c(0, 0.85))
hist(dat$Y[as.logical((dat$Y< y) * (dat$Z < z))], breaks = 30, col = rgb(0,0, 1, .5), freq = F, add = T)
legend("topleft", legend = c("Y|Y < 1", "Y|Y < 1, Z < -1"), fill = c( rgb(1,0, 0, .5),  rgb(0,0, 1, .5)))

Answer 2

I figured a way that makes it clear that the CDF just does not factorize like a pmf/pdf and that shows that the intuition in the original question is wrong.

Suppose $X \to Y \to Z$, therefore $X\perp Z \mid Y$. Assume for simplicity and without loss of generality that $Y \in \{ y_1, y_2, y_3 \}$. \begin{align*} &P(X\le x, Y \in \{y_1, y_2\}, Z\le z) \\&= P(Y = y_1) P(X\le x, Z\le z \mid Y = y_1) + P(Y = y_2) P(X\le x, Z\le z \mid Y = y_2)\\ &= P(Y = y_1) P(X\le x \mid Y = y_1)P( Z\le z \mid Y = y_1)\\ & \qquad + P(Y = y_2) P(X\le x \mid Y = y_2)P( Z\le z \mid Y = y_2)\\ &\ne P(Y \in \{y_1, y_2 \}) P(X\le x \mid Y \in \{y_1, y_2\}) P(Z\le z\mid Y \in \{y_1, y_2\}) \end{align*} Where I used the law of total probability in the first equality and then conditional independence in the second equality. The last inequality is what I was really wishing was true.

The intuition is similar to @lukas-lohse's answer: conditioning on the event $Y \in \{ y_1, y_2 \}$ isn't "really conditioning" in the sense that we still don't know what value $Y$ takes and so we need to use total probability and take the average probability for all possible values $Y$ can take.

I think another correct way to state this is that for all $y$, we have that $\sigma(X \mid Y=y) \perp \sigma(Z \mid Y=y)$ which again checks out the sigma algebras argument but relates it to the two random variables $X \mid Y=y$ and $Z \mid Y=y$.

A lot of sources mention how the "probability distribution" factorizes according to the graphical model but I guess that's very imprecise terminology since typically we refer to the cdf as the "probability distribution" and again, the cdf does not factorize. It is the pdf/pmf that factorizes.

Also, I found this https://epub.ub.uni-muenchen.de/1550/1/paper_161.pdf but it seems the stuff at the end of page 4 is wrong for the same reasons