Let $X_1, X_2 : \Omega \rightarrow \mathbb{Z}_{+}$
How to write $\mathbb{P}(X_1 = s \mid X_2 = s )$ using c.d.f.s of $X_1$ and $X_2$?
I can get
$\mathbb{P}(X_1 = s \mid X_2 = s)$ $= \frac{\mathbb{P}(X_1 = s \cap X_2 = s)}{\mathbb{P}(X_2 = s)}$
But is here anything that I can express as c.d.f.s?
If $X_1, X_2$ are discrete random variables then
$\begin{align}\Bbb P(X_1{=}s\mid X_2{=}s) & =\dfrac{\Bbb P(X_1{=}s, X_2{=}s)}{\Bbb P(X_2{=}s)} \\[1ex] & = \dfrac{\Bbb P(X_1{=}s, X_2{\leq}s)-\Bbb P(X_1{=}s, X_2{<}s)}{\Bbb P(X_2{\leq}s)-\Bbb P(X_2{<}s)} \\[1ex]&=\dfrac{\Bbb P(X_1{\leq}s{,}X_2{\leq}s){-}\Bbb P(X_1{<} s{,}X_2{\leq} s){-}\Bbb P(X_1{\leq}s{,}X_2{<}s){+}\Bbb P(X_1{<}s{,}X_2{<}s)}{\Bbb P(X_2{\leq}s){-}\Bbb P(X_2{<}s)} \end{align}$
If they are integer-valued random variables, where $F_{1,2}(,)$ is the joint CDF of $X_1,X_2$, and $F_2()$ is the CDF of $X_2$, then :$$\begin{align}\Bbb P(X_1{=}s\mid X_2{=}s) &=\dfrac{F_{1,2}(s,s)-F_{1,2}(s-1,s)-F_{1,2}(s,s-1)+F_{1,2}(s-1,s-1)}{F_2(s)-F_2(s-1)}\end{align}$$
Two questions:
Are X1 and X2 discrete?
Are X1 and X2 independent?
If the answer to Q1 is no, then the answer is zero/not defined. If the answer to Q2 is no, then the joint probability cannot be separated
The answer
$= \frac{(F_{X_1}(s)-F_{X_1}(s-1))(F_{X_2}(s)-F_{X_2}(s-1))}{F_{X_2}(s)-F_{X_2}(s-1)}$
