\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
From the definitions, \(X \sim \Exponential (\lambda )\), and \(Y = \floor *{X}\).
Hence, for \(n \geq 0\), \begin {align*} \Prob (Y = n) & = \Prob (\floor *{X} = n) \\ & = \Prob (n \leq X < n + 1) \\ & = \int _{n}^{n + 1} f(x) \Diff x \\ & = \int _{n}^{n + 1} \lambda \cdot e^{-\lambda x} \Diff x \\ & = \left [- e^{-\lambda x}\right ]_{n}^{n + 1} \\ & = - e^{-\lambda (n + 1)} + e^{-\lambda n} \\ & = e^{-n\lambda } \left (1 - e^{-\lambda } \right ), \end {align*}
as desired.
Since \(Z = X - Y\), we know that \(Z = \{X\}\) where \(\{x\}\) stands for the fractional part of \(x\).
Hence, for \(0 \leq z \leq 1\), we have \begin {align*} \Prob (Z < z) & = \Prob (\{X\} < z) \\ & = \Prob (X - Y < z) \\ & = \sum _{n = 0}^{\infty } \Prob (X < Y + z, Y = n) \\ & = \sum _{n = 0}^{\infty } \Prob (n \leq X < n + z) \\ & = \sum _{n = 0}^{\infty } \int _{n}^{n + z} \lambda \cdot e^{-\lambda x} \Diff x \\ & = \sum _{n = 0}^{\infty } \left [- e^{-\lambda x}\right ]_{n}^{n + z} \\ & = \sum _{n = 0}^{\infty } \left [- e^{-\lambda (n + z)} + e^{-\lambda n}\right ] \\ & = \sum _{n = 0}^{\infty } e^{-n\lambda } \left (1 - e^{-\lambda z} \right ) \\ & = \left (1 - e^{-\lambda z} \right ) \sum _{n = 0}^{\infty } e^{-n\lambda } \\ & = \left (1 - e^{-\lambda z} \right ) \cdot \frac {1}{1 - e^{-\lambda }} \\ & = \frac {1 - e^{-\lambda z}}{1 - e^{-\lambda }}, \end {align*}
as desired.
It must be the case that \(0 \leq Z < 1\), and the cumulative distribution function of \(Z\) is given by, for \(0 \leq z \leq 1\), \[ F_{Z}(z) = \frac {1 - e^{-\lambda z}}{1 - e^{-\lambda }}. \]
By differentiating with respect to \(z\), we get the probability density function of \(Z\) is given by, for \(0 \leq z \leq 1\), \begin {align*} f_{Z}(z) & = F'_{Z}(z) \\ & = \DiffOp {z} \frac {1 - e^{-\lambda z}}{1 - e^{-\lambda }} \\ & = \frac {1}{1 - e^{-\lambda }} \cdot \left (\lambda \cdot e^{-\lambda z}\right ) \\ & = \frac {\lambda e^{-\lambda z}}{1 - e^{-\lambda }}, \end {align*}
and zero everywhere else.
Hence, the expectation is given by \begin {align*} \Expt (Z) & = \int _{0}^{1} z f_{Z} (z) \Diff z \\ & = \int _{0}^{1} \frac {\lambda z e^{-\lambda z}}{1 - e^{-\lambda }} \Diff z \\ & = \frac {\lambda }{1 - e^{-\lambda }} \int _{0}^{1} z e^{-\lambda z} \Diff z \\ & = - \frac {1}{1 - e^{-\lambda }} \int _{0}^{1} z \Diff e^{-\lambda z} \\ & = - \frac {1}{1 - e^{-\lambda }} \left [\left (z e^{-\lambda z}\right )_{0}^{1} - \int _{0}^{1} e^{-\lambda z} \Diff z\right ] \\ & = - \frac {1}{1 - e^{-\lambda }} \left [z e^{-\lambda z} + \frac {e^{-\lambda z}}{\lambda }\right ]_{0}^{1} \\ & = - \frac {1}{1 - e^{-\lambda }} \left [\left (e^{-\lambda } + \frac {e^{-\lambda }}{\lambda }\right ) - \left (0 + \frac {1}{\lambda }\right )\right ] \\ & = \frac {\frac {1}{\lambda } - \frac {e^{-\lambda }}{\lambda } - e^{-\lambda }}{1 - e^{-\lambda }} \\ & = \frac {1 - e^{-\lambda } - \lambda e^{-\lambda }}{\lambda \left (1 - e^{-\lambda }\right )}. \end {align*}
Since \(0 \leq z_1 < z_2 \leq 1\), we have \(n \leq n + z_1 < n + z_2 \leq n + 1\), and hence \begin {align*} \Prob (Y = n, z_1 < Z < z_2) & = \Prob (Y = n, z_1 < X - Y < z_2) \\ & = \Prob (n + z_1 < X < n + z_2) \\ & = \int _{n + z_1}^{n + z_2} \lambda \cdot e^{-\lambda x} \\ & = \left [- e^{-\lambda x}\right ]_{n + z_1}^{n + z_2} \\ & = e^{-\lambda (n + z_1)} - e^{-\lambda (n + z_2)} \\ & = e^{-\lambda n} \left [e^{-\lambda z_1} - e^{-\lambda z_2}\right ]. \end {align*}
On the other hand, notice \begin {align*} \Prob (Y = n) \Prob (z_1 < Z < z_2) & = \Prob (Y = n) \left (\Prob (Z < z_2) - \Prob (Z - z_1)\right ) \\ & = (1 - e^{-\lambda }) e^{-n\lambda } \cdot \left [\frac {1 - e^{-\lambda z_2}}{1 - e^{-\lambda }} - \frac {1 - e^{-\lambda z_1}}{1 - e^{-\lambda }}\right ] \\ & = e^{- n \lambda } \left [\left (1 - e^{-\lambda z_2}\right ) - \left (1 - e^{-\lambda z_1}\right )\right ] \\ & = e^{-n \lambda } \left [e^{-\lambda z_1} - e^{-\lambda z_2}\right ]. \end {align*}
Hence, we have \[ \Prob (Y = n, z_1 < Z < z_2) = \Prob (Y = n) \Prob (z_1 < Z < z_2), \] and we can conclude that \(Y\) and \(Z\) are independent.