Composition of percentiles

Question

Assume we have two vectors $v_1$ and $v_2$, such that the $i$-th percentile of both is $p$. Is it true that the vector $v$ obtained by concatenating $v_1$ and $v_2$ has the $i$-th percentile equal to $p$?

Answer 1

Yes, that's correct. Here are two ways to see this.

1. Using a counting argument. Let's say $v_j$ has $n_j$ entries. Then there are exactly $n_j \cdot i/100$ entries in $v_j$ larger than $p$ in $v_j$. The concatenated vector has $n_1 + n_2$ entries of which exactly $i \cdot n_1/100 + i \cdot n_2/100$ entries are larger than $p$. Hence the $i$-th percentile of the concatenated vector is also $p$.

2. Using conditional probabilities. Let $P_j$ be the uniform probability measure on $v_j$ and $P_{12}$ be the uniform probability measure on the concatenated vector. Write $\alpha = i/100$ for simplicity. The assumption is that $$ P_j(x > p) = \alpha $$ for $j = 1, 2$, where $x$ is drawn from the entries of $v_j$. Now draw from the concatenated vector. Let $A_j$ be the event that an entry is drawn that camr from $v_j$. Then since $P(A_1) + P(A_2) = 1$, $$ P_{12}(x > p) = P_1(x > p)P(A_1) + P_2(x > p)P(A_2) = \alpha(P(A_1) + 1 - P(A_1)) = \alpha . $$