Few random variables cannot influence $n$ independent others too much? I have $n$ standard normal and independent random variables $X_i$ (In reality I have a large known number of them, but let's just say I have $n$). In my experiment I want to on average get exactly 3 random variables $X_i$ under a threshold $c$. To get that, I can compute $c$ having that property easily, because the average number of $X_i$ that are under a threshold $c$ is $n \Phi(c)$ where $\Phi$ is the cdf of the standard normal distribution.
So I choose $c = \Phi^{-1}(3/n)$ in this case. (Which is a negative number for large $n$.)
But unfortunately I already know the value of two other standard normal random variables $Y$ and $Z$ which may depend on each other and on any number of the other $X_i$.
So my question is: if I know that $Y$ and $Z$ are under the threshold $c=\Phi^{-1}(3/n)$, is it then still true that on average at most a constant number of the other random variables $X_i$ are under the threshold $c$? So by knowing that $Y$ and $Z$ are under the threshold, they can't suddenly make many of the other random variables go under it too.
I am almost certain that they can't, but I don't know how to prove it. Any hints are welcome. Or books where you think this might be in.
 A: In the asymptotic sense seemingly suggested by the phasing of the question, it's not true, but the analysis might be revealing.
We don't even need $Z$.
Let $p$ be the chance of a standard normal variable being $c$ or less; that is, $p = \Phi(c)$.  Then the chance that at least $k$ or more of the $X_i$ are less than or equal to $c$ is given by a Binomial distribution
$$\sum_{i=0}^{n-k} \binom{n}{i} p^{n-i}(1-p)^i\text{.}$$
Because this sum runs from $p^n \lt p$ to $1 \gt p$, there exists a $k$ between $1$ and $n-1$ where the sum is as large as possible but still less than $p$.
For future reference, note that as $n$ grows large, $k$ is approximately equal to $p n$.  This is a consequence of the Central Limit Theorem (for Binomial variates), because the sum is approximately equal to $\Phi((n-k - (1-p) n) / \sqrt{n p (1-p)})$.  If eventually $k$ were less than $p n$, say $k \lt (p - \epsilon)n$ for $\epsilon \gt 0$, then the sum would approximate $\Phi(\epsilon \sqrt{n} / \sqrt{p(1-p)})$, which approaches $1$ as $n$ increases, but the sum is constructed to stay below $p$.  Similarly, if $k \gt (p + \epsilon)n$, the sum would go to zero, again contradicting the construction of $k$ (to be as large as possible).
Define $q$ (which depends implicitly on $c$ and $n$) to be the value of the sum for such a $k$.  Let $b = \Phi^{-1}(q) \le c$.
Conditional on at least $k$ of the $X_i$ not exceeding $c$, let $Y$ have a truncated standard normal distribution ranging from $-\infty$ to $b$.  This happens with probability $q$.  Otherwise, let $Y$ have a truncated standard normal distribution ranging from $b$ to $+\infty$.  This gives $Y$ a standard normal distribution but it depends on the $X_i$.
If $Y \le c$, the chance that $Y \le b$ equals $q/p$.  With sufficiently large $n$, an easy estimate shows this value is close to $1$.  Given that $Y \le b$, we know at least $k$ of the $X_i$ are below $c$, by construction of $Y$.  Therefore the expected number of such $X_i$ is at least $p n$ (asymptotically in $n$).  This quantity grows without bound, it is not limited by a constant (independent of $n$).
You can work this analysis in reverse: if the expected number of $X_i$ below the threshold $c$, conditional on $Y \le c$, is much larger than $p n$, then the probability that $Y \le c$ would have to be greater than $\Phi(c)$, implying $Y$ does not have a standard Normal distribution.  In this sense the preceding construction is a worst case: asymptotically, it achieves the largest possible expected number of $X_i$ below $c$ consistent with the assumptions on $X_i$ and $Y$.
