TITLE: Multivariate Normal Distribution: Relationship between two conditional probabilities.
QUESTION [8 upvotes]: Suppose I have a multivariate normal random variable $Z$ which has $n$ dimensions.  Suppose I have a vector $x$. Set $i$ as a number between $1$ and $n$ and $k$ as a number between $1$ and $n-1$.  Can I say the following about the relationship between the following two probabilities?
\begin{align}
& P(Z_i>x_i \mid Z_j>x_j \text{ for exactly $k$ components $j$}) \\[10pt]
< {} & P(Z_i>x_i \mid Z_j>x_j \text{ for exactly $k+1$ components $j$})
\end{align}
This seems logical to me. Essentially we are saying that the fact $k+1$ components cross the threshold as opposed to just $k$ components increases the probability that a specific one crossed its threshold.  However, I have no idea why this has to be true just why I think it should be true.  I would really appreciate a formal explanation or counterproof.Thanks!

REPLY [5 votes]: Your conjecture is false. Consider $Z=(X_1,-X_1,-X_1)$ where $X_1\sim N(0,1)$ and $x=(0,0,0)$. Then because $Z_2$ and $Z_3$ are of the same sign $$P(Z_1>0 \mid Z_j>0 \text{ for exactly 1 component $j$})=P(Z_1>0\mid Z_1>0, Z_2\leq 0, Z_3\leq 0)=1$$ 
while
$$P(Z_1>0 \mid Z_j>0 \text{ for exactly 2 components $j$})=P(Z_1>0\mid Z_1\leq 0, Z_2>0, Z_3>0)=0$$
Your reasoning is false because components of $Z$ can be anti-correlated - so the more you push one of them up, the more another one has to come down. If components are independent, you are probably correct.