Order statistics and conditionnal probability

Question

The problem is as follows: let $S_X = (X_1, ... ,X_n)$ and $S_Y = (Y_1, ... ,Y_n)$ two iid samples whose probability distributions are compactly supported on $(0,1)$ with a.c. densities $(f_X,f_Y)$ w.r.t the Lebesgue measure.

We now order the global sample $S = (X_1, ... ,X_n, Y_1...,Y_n)$ by increasing values. We define $S_k$ as the number of samples coming from the first sample $S_X$ among the $k$ lower values of $S$.

Is it true that for any integer $k$ lower than $n, l \to P(S_{k+1}=l+1 / S_k=l)$ is decreasing?

The result is intuitive, but I cannot develop any rigorous proof of such a result...

Answer 1

No, if you define the function $$ f_k(l) = P(S_{k+1}=l+1 \vert S_k = l ), $$ I wonder wheter $f_k(l) \geq f_k(l+1)$ ...

I have run a lot of simulations which seems to validate this conjecture, and the extremal choice when $X$ and $Y$ shares the same distribution is $$ f_k(l)=\frac{n-l}{n-k}, $$ which is of course decreasing in $l$.

Answer 2

I have found an example where the last property fails (the density assumption is not fullfilled but it is possible to extend the picture to the initial setting).

Let $X \sim \delta_{1/2}$ and $Y=\sim \frac{1}{2} [ \delta_{0}+\delta_{1}]$.

Consider now the case $n=2$. It is immediate to check that when $S_1=1$, then necessarily $S_2=2$ since $Z$ conditionnally to $S_1$, we have $Z=(1/2,1/2,1,1)$: $$ \mathbb{P}[S_2=2 \vert S_1=1] = 1. $$

Of course, it is easy to check that when $S_1=0$: $$ \mathbb{P}[S_2=1 \vert S_1=0] = 1/2. $$

Now, one can regularize the previous distribution with a small convolution and the property remains false.