As part of my excercise i have to show that X and Y are conditionally independent under the condition $(\lfloor X\rfloor,\lfloor Y\rfloor)$ where $X,Y \sim exp(\lambda)$
I previously computed $E[X|\lfloor X\rfloor]$ and I know that I have to show that $f(X,Y|\sim)=f(X|\sim)f(Y|\sim)$ ... but how?
This requires $X$ and $Y$ to be independently distributed.
$\begin{align} \because f_{X,Y}(x,y) & = f_X(x)\;f_Y(y) \\[3ex] f_X(x\mid \lfloor X\rfloor) &= \frac{ f_X(x) \operatorname{\bf 1}_{[\lfloor X\rfloor,\lceil X\rceil)}(x) }{\displaystyle \int_{\lfloor X\rfloor}^{\lceil X\rceil} f_X(s)\operatorname d s } \\[3ex] \therefore f_{X,Y}(x,y\mid \lfloor X\rfloor, \lfloor Y\rfloor) & = \frac{ f_{X,Y}(x,y) \operatorname{\bf 1}_{[\lfloor X\rfloor, \lceil X\rceil)}(x) \operatorname{\bf 1}_{[\lfloor Y\rfloor, \lceil Y\rceil)}(y) }{\displaystyle \int_{\lfloor X\rfloor}^{\lceil X\rceil} \int_{\lfloor Y\rfloor}^{\lceil Y\rceil} f_{X,Y}(s,t) \operatorname d t\operatorname d s } \\[2ex] & = \frac{ f_{X}(x)\;f_{Y}(y) \operatorname{\bf 1}_{[\lfloor X\rfloor, \lceil X\rceil)}(x) \operatorname{\bf 1}_{[\lfloor Y\rfloor, \lceil Y\rceil)}(y) }{\displaystyle \int_{\lfloor X\rfloor}^{\lceil X\rceil} f_{X}(s) \operatorname d s \int_{\lfloor Y\rfloor}^{\lceil Y\rceil} f_{Y}(t) \operatorname d t } \\[2ex] & = f_X(x\mid \lfloor X\rfloor) \;f_Y(y\mid \lfloor Y\rfloor) \\[1ex] & = f_X(x\mid \lfloor X\rfloor, \lfloor Y\rfloor) \;f_Y(y\mid \lfloor X\rfloor, \lfloor Y\rfloor) & \text{(Why?)} \end{align}$
