conditional density of an exponential variable and its integer part

Question

Assume Y is the integer part of an exponential variable X~Exp($\lambda$), if y is given, how are we able to find the conditional density of this?

I know y < x < y+1 in this case, so I tried to use $P(X \leq y+1 | y)$ = $\frac{P(y \leq X \leq y+1)}{P(y \leq X)}$ = $1 - e^{-\lambda}$ but this is clearly not the correct conditional distribution as the density will become zero.

Thanks in advance.

Answer 1

$P(X\leq x|Y=y)=P(X\leq x |y \leq X <y+1)=\frac {P(y<X <\min \{y+1,x\})} {P(y<X<y+1)}$. For $x \notin (y,y+1)$ the numerator is independent of $x$ so the pdf is $0$. For $y<x<y+1$ we get $P(X\leq x|Y=y)=\frac {\int_y^{x} \lambda e^{-\lambda t}dt} {{\int_y^{y+1} \lambda e^{-\lambda t}dt}}$ so the pdf is $\frac {\lambda e^{-\lambda x}} {{\int_y^{y+1} \lambda e^{-\lambda t}dt}}$.