TITLE: Joint distribution of integer and fractional part of random variable
QUESTION [4 upvotes]: Here's the set up:
Let the random variable $X$ have the following distribution 
$$f(x)=
  \begin{cases}
    e^{-x} & \quad \text{if}\hspace{2mm} 0<x<\infty\\
    0 & \quad \text{elsewhere.}
  \end{cases} $$
Let $Y=[X]$ be the integer part and $Z=X-[X]$ be the fractional part.
Problem:
I'm supposed to find the joint distribution function of $Y$ and $Z$, and the moment generating function of $Y$. I think I have the mgf of $Y$, but don't really know how to approach finding the joint distribution. 
My inner dialogue:
It seems like to find  $\mathbb{P}(Y\leq t)$, we'll be summing $e^{-x}$ in $x$ over the integers up to $\lfloor t \rfloor$. On the other hand, with $\mathbb{P}(Z\leq t)$, will we integrate, just skipping integers?  But $\mathbb{P}(Z\leq t)=\mathbb{P}(Z< t)$ if $Z$ is continuous? Hmm...
I would very much appreciate any suggestions on how to get started here. 
Also, this isn't homework.

REPLY [3 votes]: Let's look at the conditional distribution of the fractional part given that the integer part is $3$.  That's the conditional distribution of $X-3$ given that $3\le X<4$.  The conditional distribution of $X$ itself, given that it's in that interval has density $ce^{-x}$ for $3\le x<4$ and $0$ outside that interval.  Translating that $e$ units leftward, we get $ce^{-(x+3)}$ $=(ce^{-x}\cdot\text{another constant})$.  All probability densities on $[0,1)$ that are proportional to $e^{-x}$ on that interval and $0$ elsewhere are the same!
Therefore, the conditional distribution of the fraction part given that the integer part is $3$ is the same as the conditional distribution of the fraction part given that the integer part is anything else.  In other words: the two are independent.
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