Show that $Y$ and $Z$ are independent and find their distributions Suppose that $X\sim\exp(\lambda=1)$. Let $Y$ be the integer part and $Z$ be the fractional part. Show that $Y$ and $Z$ are independent and find their distributions. 
This one is kinda confusing. Any help?
 A: For $n\in\mathbb{N},n\geq 0$, we have
$$
\Pr(Y\leq n)=\Pr(X< n+1)=1-e^{-n-1}.
$$
For $z\in[0,1)$, we have
\begin{align*}
\Pr(Z\leq z)&=\sum_{m=0}^\infty\Pr(X\in[m,m+z])\\
&=\sum_{m=0}^\infty(\exp(-m)-\exp(-m-z))\\
&=(1-e^{-z})\sum_{m=0}^\infty(1/e)^m\\
&=\frac{e-e^{1-z}}{e-1}
\end{align*}
Finally,
\begin{align*}
\Pr(Y\leq n,Z\leq z)&=\sum_{m=0}^n\Pr(m\leq X\leq m+z)\\
&=\sum_{m=0}^n(\exp(-m)-\exp(-m-z))\\
&=(1-e^{-z})\sum_{m=0}^n(1/e)^m
\end{align*}
which you can verify to be equal to $\Pr(Y\leq n)\Pr(Z\leq z)$.
A: Because $Y$ is the integer part of random variable $X$, the probability mass distribution for $Y\in\{0,...\}$ is therefore:
$$\begin{align}
p_Y(y) & = \mathsf P(y\leq X < y+1)
\\[1ex] & = \int_0^1 f_X(y+t)\operatorname d t
\\[1ex] & = \mathsf e^{-\lambda y}\cdot\int_0^1 \mathsf e^{-\lambda t}\operatorname d t
\\[1ex] & = \underline{\qquad}
\end{align}$$

Because $Z$ is the fractional part of the random variable $X$, the probability density distribution for $Z\in[0;1)$ is
$$\begin{align}
f_Z(z) & = \sum_{t\in\Bbb Z} f_X(s+z)
\\[1ex] & = \sum_{t=0}^1 \lambda\mathsf e^{-\lambda(t+z)}
\\[1ex] & = \lambda\, \mathsf e^{-\lambda z} \sum_{t=0}^\infty {(e^{-\lambda})}^t
\\[1ex] & = \underline{\qquad}
\end{align}$$

The joint mass-density is: 
$$p_{Y,Z}(y,z) = f_{X}(y+z) = \lambda\,\mathsf e^{-\lambda(y+z)}$$
Because the joint event of $Y=y$ and $Z=z$ uniquely describes $X=y+z$, since any real is the sum of an unique pair of an integer and a unit-interval real.    $\forall x\in \Bbb R\; \exists! y\in \Bbb Z\; \exists! z\in [0;1]\;\big(x=y+z\big)$ 

So clearly: $p_{Y,Z}(y,z) = p_Y(y)\;f_Z(z)$ and thus the random variables $Y$ and $Z$ are independent.
