Consider two populations, $A$ and $B$. $A + B = C$. Meaning $A$ and $B$ are mutually exclusive and collectively exhaustive of $C$.
Given the above if I calculate the 90th percentile of some metric for all populations is it possible that
$P90_C \ge P90_A$ and
$P90_C \ge P90_B$
I'm not good at combining P90s in my head and can't mathematically prove it.
it's possible
set.seed(0)
N = 1000
i = 0
max_iter = 100
while (i < max_iter) {
i = i + 1
p = runif(1)
g = runif(N) < p
x = rnorm(N)
q = quantile(x, 0.90) # group C
q1 = quantile(x[g], 0.90) # group A
q2 = quantile(x[!g], 0.90) # group B
if ((q > q1) && (q > q2)) {
print(sprintf("i:%i, q:%.4f, q1:%.4f, q2:%.4f", i, q, q1, q2))
break
}
}
result is
i:11, q:1.2368, q1:1.2351, q2:1.2360
This is possible when we consider computations of quantiles that are not exact.
Example
A = c(1,2,3,4,5)
B = c(1,2,3,4,5)
C = c(x,y)
quantile(A,0.9) # this computes as 4.6
quantile(B,0.9) # this computes as 4.6
quantile(C,0.9) # this computes as 5
It if we have instead the condition that the populations A and B are with $10k_A +1$ and $10k_B +1$ members* then the 90-th percentile should be equal to or in-between the 90-th percentiles of A and B.
*in this case the 90-th percentiles are exactly the (9k+1)-th order statistics.
It is certainly possible.
As an example, consider finite populations or two finite samples. Let $x^A_{(0.9n)}$ be the 90th percentile of sample $A$ that has $n$ members, and $x^B_{(0.9m)}$ be the 90th percentile of sample $B$ that has $m$ members.
Suppose now that it holds $${\rm condition\;1:}\quad x^A_{(0.9n)} < x^B_{(0.9m)}.$$
Suppose also that the following holds
$${\rm condition\; 2:}\quad \forall\, x^A: x^A > x^A_{(0.9n)} \implies x^A > x^B_{(0.9m)}.$$
Namely, it so happens that all members of $A$ with value higher than its 90th percentile, are also higher in value than the 90th percentile of sample $B$.
If we pool the two samples creating $C$, there will be $0.1\times (n+m)$ elements higher than $x^B_{(0.9m)}$ in $C$.
Hence we will get $$x^A_{(0.9n)} < x^B_{(0.9m)} = x^C_{(0.9(n+m))}.$$
