# Joint distribution of integer and fractional part of random variable

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.

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If $t$ is an integer, $\mathbb{P}(Y \leq t) = \mathbb{P}( \lfloor X \rfloor \leq t) = \mathbb{P} ( X < t+1) = \int_0^{t+1} e^{-x}\, dx$, not summing at the integer points. – Calvin Lin Jan 16 '13 at 1:09
The integer and fractional parts are independent, and that's about 98% of the answer. See my answer below. – Michael Hardy Jan 16 '13 at 1:21
Yeah, I definitely misinterpreted the set up. – eeeeeeeeee Jan 16 '13 at 1:50

This is not a complete answer, but just to state what $Y, Z$ should be.

If $t$ is an integer, $\mathbb{P}(Y \leq t) = \mathbb{P}( \lfloor X \rfloor \leq t) = \mathbb{P} ( X < t+1) = \int_0^{t+1} e^{-x}\, dx$, not summing at the integer points.

If $0 \leq r \leq 1$, then $\mathbb{P}(Z \leq r) = \mathbb{P}( X - \lfloor X \rfloor < r) = \sum_{n=0}^\infty \mathbb{P} (n < X < n+r)$.

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Oh, okay. Thanks for the clarification. – eeeeeeeeee Jan 16 '13 at 1:22

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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Sorry, I don't completely follow. Why does the conditional distribution of $X$, given that it's in the interval have $ce^{-x}$ as its density? – eeeeeeeeee Jan 16 '13 at 1:33
Oh, nevermind. I think I got it. I like this approach. Thanks! – eeeeeeeeee Jan 16 '13 at 1:54
Actually, I recently answered that same question about conditional densities here: math.stackexchange.com/questions/278762/… – Michael Hardy Jan 16 '13 at 2:01