\(% 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}}\)
By the definition within the question, we have that \(X, Y \sim \Geometric (p)\), and for \(t \geq 1\), \begin {align*} \Prob (X = t) = \Prob (Y = t) = q^{t - 1} p. \end {align*}
For \(S = X + Y\), we have for \(s \geq 2\), \begin {align*} \Prob (S = s) & = \Prob (X + Y = s) \\ & = \sum _{t = 1}^{s - 1} \Prob (X = t, Y = s - t) \\ & = \sum _{t = 1}^{s - 1} \Prob (X = t) \Prob (Y = s - t) \\ & = \sum _{t = 1}^{s - 1} q^{t - 1} p q^{s - t - 1} p \\ & = \sum _{t = 1}^{s - 1} q^{s - 2} p^2 \\ & = (s - 1) q^{s - 2} p^2. \end {align*}
For \(T = \max \{X, Y\}\), we have for \(t \geq 1\), \begin {align*} \Prob (T = t) & = \Prob (X = Y = t) + \Prob (X = t, Y < X) + \Prob (Y = t, X < Y) \\ & = \Prob (X = t, Y = t) + 2 \Prob (X = t, Y < X) \\ & = \Prob (X = t) \Prob (Y = t) + 2 \sum _{r = 1}^{t - 1} \Prob (X = t, Y = r) \\ & = \Prob (X = t) \Prob (Y = t) + 2 \sum _{r = 1}^{t - 1} \Prob (X = t) \Prob (Y = r) \\ & = q^{t - 1} p q^{t - 1}p + 2 q^{t - 1} p \sum _{r = 1}^{t - 1} q^{r - 1} p \\ & = q^{2t - 2} p^2 + 2 q^{t - 1} p^2 \sum _{r = 1}^{t - 1} q^{r - 1} \\ & = q^{2t - 2} p^2 + 2 q^{t - 1} p^2 \frac {1 - q^{t - 1}}{1 - q} \\ & = q^{2t - 2} p^2 + 2 q^{t - 1} p^2 \frac {1 - q^{t - 1}}{p} \\ & = q^{2t - 2} p^2 + 2 q^{t - 1} p (1 - q^{t - 1}) \\ & = pq^{t - 1} \left (pq^{t - 1} + 2 - 2q^{t - 1}\right ) \\ & = pq^{t - 1} \left ((1 - q)q^{t - 1} + 2 - 2q^{t - 1}\right ) \\ & = pq^{t - 1} \left (2 + q^t - q^{t - 1}\right ) \end {align*}
Since \(U = \abs *{X - Y}\), we have \(U \geq 0\). For \(u \geq 1\), we have \begin {align*} \Prob (U = u) & = \Prob (\abs *{X - Y} = u) \\ & = \Prob (X - Y = \pm u) \\ & = 2 \Prob (X - Y = u) \\ & = 2 \sum _{t = 1}^{\infty } \Prob (X = u + t, Y = t) \\ & = 2 \sum _{t = 1}^{\infty } \Prob (X = u + t) \Prob (Y = t) \\ & = 2 \sum _{t = 1}^{\infty } q^{u + t - 1} p q^{t - 1} p \\ & = 2 q^u p^2 \sum _{t = 1}^{\infty } q^{2t - 2} \\ & = 2 q^u p^2 \cdot \frac {1}{1 - q^2} \\ & = 2 q^u p^2 \cdot \frac {1}{(1 + q)p} \\ & = \frac {2 q^u p}{1 + q}, \end {align*}
and for \(u = 0\), \begin {align*} \Prob (U = 0) & = \Prob (X = Y) \\ & = \sum _{t = 1}^{\infty } \Prob (X = Y = t) \\ & = \sum _{t = 1}^{\infty } \Prob (X = t) \Prob (Y = t) \\ & = \sum _{t = 1}^{\infty } q^{t - 1} p q^{t - 1} p \\ & = p^2 \sum _{t = 1}^{\infty } q^{2t - 2} \\ & = p^2 \cdot \frac {1}{1 - q^2} \\ & = \frac {p}{1 + q}. \end {align*}
Since \(W = \min \{X, Y\}\), we have \(W \geq 1\). For \(w \geq 1\), we have \begin {align*} \Prob (W = w) & = \Prob (X = Y = w) + \Prob (X = w, Y > X) + \Prob (Y = w, Y < X) \\ & = \Prob (X = w, Y = w) + 2\Prob (X = w, Y > X) \\ & = \Prob (X = w) \Prob (Y = w) + 2 \sum _{r = w + 1}^{\infty } \Prob (X = w, Y = r) \\ & = \Prob (X = w) \Prob (Y = w) + 2 \sum _{r = w + 1}^{\infty } \Prob (X = w) \Prob (Y = r) \\ & = q^{w - 1} p q^{w - 1} p + 2 \sum _{r = w + 1}^{\infty } q^{w - 1} p q^{r - 1} p \\ & = q^{2w - 2} p^2 + 2 q^{w - 2} p^2 \sum _{r = w + 1}^{\infty } q^r \\ & = q^{2w - 2} p^2 + 2 q^{w - 2} p^2 q^{w + 1} \cdot \frac {1}{1 - q} \\ & = q^{2w - 2} p^2 + 2 q^{2w - 1} p^2 \cdot \frac {1}{p} \\ & = q^{2w - 2} p^2 + 2 q^{2w - 1} p \\ & = q^{2w - 2} p \left (p + 2q\right ) \\ & = q^{2w - 2} p \left (1 + q\right ). \end {align*}
Since \(S = 2\) and \(T = 3\), the maximum of \(X\) and \(Y\) is \(3\), but they sum to \(2\), and this is impossible, so \[ \Prob (S = 2, T = 3) = 0. \]
However, \begin {align*} \Prob (S = 2) \Prob (T = 3) & = (2 - 1) q^{2 - 2} p^2 \cdot p q^{3 - 1} (2 + q^3 - q^{3 - 1}) \\ & = p^3 q^2 (2 + q^3 - q^2) \\ & \neq 0 \\ & = \Prob (S = 2, T = 3), \end {align*}
as desired.
\(U\) and \(W\) are independent. We split this into two cases of \(U\) to consider:
When \(U = u \neq 0\), \begin {align*} \Prob (U = u, W = w) & = \Prob (X = w, Y = w + u) + \Prob (X = w + u, Y = w) \\ & = 2 \Prob (X = w, Y = w + u) \\ & = 2 \Prob (X = w) \Prob (Y = w + u) \\ & = 2 q^{w - 1} p q^{w + u - 1} p \\ & = 2 q^{2w + u - 2} p^2, \end {align*}
and \[ \Prob (U = u) \Prob (W = w) = \frac {2 q^u p}{1 + q} \cdot q^{2w - 2} p (1 + q) = 2 q^{2w + u - 2} p^2, \] and so \[ \Prob (U = u, W = w) = \Prob (U = u) \Prob (W = w). \]
Hence, we can see that \(U\) and \(W\) are independent.
\(U\) and \(S\) are not independent. Consider the case where \(S = 3\) and \(U = 0\). The event \(S = 3, U = 0\) is not possible since \(S = X + Y\) and \(U = \abs *{X - Y}\) must have the same odd-even parity, giving \[ \Prob (S = 3, U = 0) = 0. \]
On the other hand, \[ \Prob (S = 3) \Prob (U = 0) = 2 q p^2 \cdot \frac {p}{1 + q} = \frac {2qp^3}{1 + p} \neq 0. \]
This means \[ \Prob (S = 3, U = 0) \neq \Prob (S = 3) \Prob (U = 0), \] and hence \(U\) and \(S\) are not independent.
\(U\) and \(T\) are not independent. Consider the case where \(U = 1\) and \(T = 1\). The event \(U = 1, T = 1\) implies that \(X = Y \pm 1\), and that the maximum of \(X\) and \(Y\) is \(1\), and hence \(X = Y = 1\), which is impossible. Hence, \[ \Prob (U = 1, T = 1) = 0. \]
On the other hand, \[ \Prob (U = 1) \Prob (T = 1) = \frac {2qp}{1 + q} \cdot p(2 + q - 1) = 2p^2 q \neq 0. \]
This means \[ \Prob (U = 1, T = 1) \neq \Prob (U = 1) \Prob (T = 1), \] and hence \(U\) and \(T\) are not independent.
\(W\) and \(S\) are not independent. Consider the case where \(W = 2\) and \(S = 2\). On one hand, since \(\min \{X, Y\} = 2\), and \(S = X + Y = \max \{X, Y\} + \min \{X, Y\} = 2\), this means \(\max \{X, Y\} = 0\) which is impossible, and hence \[ \Prob (W = 2, S = 2) = 0. \]
On the other hand, \[ \Prob (W = 2) \Prob (S = 2) = q^2 p (1 + q) (1) q^{0} p^2 = p^3 q^2 (1 + q) \neq 0. \]
This means \[ \Prob (W = 2, S = 2) = \Prob (W = 2) \Prob (S = 2). \]
\(W\) and \(T\) are not independent. Consider the case where \(T = 1\) and \(W = 2\). Since \(T = \max \{X, Y\} = 1\) and \(W = \min \{X, Y\} = 2\), the event \(T = 1, W = 2\) is not possible, hence \[ \Prob (T = 1, W = 2) = 0. \]
On the other hand, \[ \Prob (T = 1) \Prob (W = 2) = p (2 + q - 1) q^2 p(1 + q) = p^2 q^2 (1 + q)^2 \neq 0. \]
This means \[ \Prob (T = 1, W = 2) \neq \Prob (T = 1) \Prob (W = 2), \] and hence \(W\) and \(T\) are not independent.