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!


2 Answers 2


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.

  • $\begingroup$ I should have clarified in my post that I was assuming the covariance matrix was positive definite. If we add that assumption does that allow me to make my conclusion or do we need the independence? I don't think so but maybe? $\endgroup$ Sep 28, 2015 at 6:08
  • 3
    $\begingroup$ Positive-definiteness is not enough. Just add small independent noises to all variables in my example to go from semi-definite to positive definite covariance. Small noise will affect probabilities in a small way, so inequality will stay not in the direction you desire. If you require non-negative pairwise correlations, you might be correct. $\endgroup$
    – A.S.
    Sep 28, 2015 at 7:58
  • $\begingroup$ Any idea on how I could prove it for the non-negative pairwise correlations? $\endgroup$ Oct 7, 2015 at 2:33
  • 1
    $\begingroup$ I was able to show this in case of exchangeable $Z$ - which allows for small negative correlations between $Z_i$'s. In Michael Hardy's notation, you need to show that $$\frac {P(Y_1=1 \wedge Y_2+\cdots+Y_n=k-1)}{P(Y_1=0 \wedge Y_2+\cdots+Y_n=k)}$$ is an increasing function of $k$ and in exchangeable case it's easy to show just by counting it's equal to $\frac{k}{n-k}$ - similarly to computation above. $\endgroup$
    – A.S.
    Oct 7, 2015 at 14:00

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".


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