Multivariate Normal Distribution: Relationship between two conditional probabilities. 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!
 A: 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.
A: Here's an incomplete answer. In some special circumstances the guess is true, but in some others it is not.
Assuming the components $Z_j$, $j=1,\ldots,n$ are uncorrelated (and thus independent, since they're jointly normal) this reduces to a problem on Bernoulli random variables: Let
$$
Y_j = \begin{cases} 1 &\text{if }Z_j>x_j, \\
0 & \text{otherwise}. \end{cases}
$$
Then $Y_j\sim\mathrm{Bernoulli}(p_j)$ for $j=1,\ldots,n$, where $p_j = \Pr(Z_j>x_j)$, and $Y_j$, $j=1,\ldots,n$ are independent.  Then the question is whether $\Pr(Y_1=1\mid Y_1+\cdots+Y_n=y)$ is an increasing function of $y$.
The answer to that is "yes", and I am just realizing I don't know the most elegant way to prove that, so maybe I'll be back.  I've commented out some preliminary scratchwork below and I may return to finish it later.

Here's one narrow special case in which is it easy to show the answer is "yes": suppose $Z_1,\ldots,Z_n$ are independent (which is not generally true of components of a multivariate normal random variable) and have expected value $0$ (also not generally true) and variance $1$ (also not generally true) and $x_1=\cdots=x_n$.  In that case we can let
$$
Y_j = \begin{cases} 1 & \text{if }Z_j>x_j, \\  0 & \text{otherwise}. \end{cases}
$$
Then
\begin{align}
& \Pr(Y_1=1 \mid Y_1+\cdots+Y_n=y) = \frac{\Pr(Y_1=1)\Pr(Y_2+\cdots+Y_n=y-1)}{\Pr(Y_1+\cdots+Y_n=y)} \\[10pt]
= {} & \frac{p \cdot \dbinom{n-1}{y-1} p^{y-1}(1-p)^{n-y} }{\dbinom n y p^y (1-p)^{n-y}} = \frac y n,
\end{align}
and that certainly increases as $y$ increases.
However, one must consider negative correlations.  Suppose, for example, that $W_1,\ldots,W_n\sim\mathrm{i.i.d.}\,N(0,1)$ and $\bar W=(W_1+\cdots + W_n)/n$ and $Z_j= W_j-\bar W$ for $j=1,\ldots,n$. Then the vector $(Z_1,\ldots,Z_n)$ satisfies the constraint that $Z_1+\cdots+Z_n=0$ and has a multivariate normal distribution whose variance is a singular matrix in which all off-diagonal measures are negative.  So the $Z$s are negatively correlated with each other.  Now suppose $x_1=\cdots=x_n=0$.  Then $\Pr(Z_n>x_n) = 1/2$ but $\Pr(Z_n>x_n \mid \forall i\le n-1\  Z_i>x_i) =0$.
So in some cases the answer is "no".
