\(% 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 differentiation, we have \[ [G(H(t))]' = G'(H(t)) \cdot H'(t). \]
Hence, we have \begin {align*} \Expt (Y) & = \LEvalAt {[G(H(t))]'}{t = 1} \\ & = G'(H(1)) \cdot H'(1) \\ & = G'(1) \cdot H'(1) \\ & = \Expt (N) \cdot \Expt (X_i). \end {align*}
By differentiating twice, we have \[ [G(H(t))]'' = G''(H(t)) \cdot H'(t) \cdot H'(t) + G'(H(t)) \cdot H''(t). \]
Hence, we have \begin {align*} \Var (Y) & = \Expt (Y(Y - 1)) + \Expt (Y) - \Expt (Y)^2 \\ & = \LEvalAt {[G(H(t))]''}{t = 1} + \Expt (Y) - \Expt (Y)^2 \\ & = G''(H(1)) \cdot H'(1) \cdot H'(1) + G'(H(1)) \cdot H''(1) + \Expt (Y) - \Expt (Y)^2 \\ & = G''(1) \cdot H'(1)^2 + G'(1) \cdot H''(1) + \Expt (Y) - \Expt (Y)^2 \\ & = \Expt (N(N - 1)) \cdot \Expt (X_i)^2 + \Expt (N) \cdot \Expt (X_i (X_i - 1)) + \Expt (Y) - \Expt (Y)^2 \\ & = \left [\Var (N) + \Expt (N)^2 - \Expt (N)\right ] \cdot \Expt (X_i)^2 + \Expt (N) \cdot \left [\Var (X_i) + \Expt (X_i^2) - \Expt (X_i)\right ] \\ & \phantom {=} + \Expt (N) \cdot \Expt (X_i) - \Expt (N)^2 \cdot \Expt (X_i)^2 \\ & =\Var (N) \Expt (X_i)^2 + \Expt (N) \Var (X_i). \end {align*}
As defined, we have \(N \sim \Geometric \left (\frac {1}{2}\right )\), and hence \[ G(t) = \frac {\frac {1}{2} \cdot t}{1 - \left (1 - \frac {1}{2}\right )t} = \frac {t}{2 - t}, \] and \[ \Expt (N) = 1 / \frac {1}{2} = 2, \Var (N) = \frac {1 - \frac {1}{2}}{\left (\frac {1}{2}\right )^2} = 2. \]
We have \(X_i \sim \Binomial \left (1, \frac {1}{2}\right )\), and hence \[ H(t) = \frac {1}{2} \cdot t^0 + \frac {1}{2} \cdot t^1 = \frac {1}{2} \cdot (1 + t), \] and \[ \Expt (X_i) = 1 \cdot \frac {1}{2} = \frac {1}{2}, \Var (X_i) = 1 \cdot \frac {1}{2} \cdot \frac {1}{2} = \frac {1}{4}. \]
Hence, for \(Y = \sum _{i = 1}^{N} X_i\), we have \[ \text {p.g.f.}_{Y}(t) = G(H(t)) = \frac {\frac {1}{2} (1 + t)}{2 - \frac {1}{2} (1 + t)} = \frac {1 + t}{3 - t}, \] and by the formula for expectation and variance, we have \[ \Expt (Y) = \Expt (N) \Expt (X_i) = 2 \cdot \frac {1}{2} = 1, \] and \[ \Var (Y) = \Var (N) \cdot \Expt (X_i)^2 + \Expt (N) \cdot \Var (X_i) = 2 \cdot \left (\frac {1}{2}\right )^2 + 2 \cdot \frac {1}{4} = 1. \]
By expressing the probability generating function of \(Y\) as a power series, we notice that \begin {align*} \text {p.g.f.}_{Y}(t) & = \frac {1 + t}{3 - t} \\ & = -1 + \frac {4}{3 - t} \\ & = -1 + \frac {4}{3} \cdot \frac {1}{1 - \frac {t}{3}} \\ & = -1 + \frac {4}{3} \sum _{r = 0}^{\infty } \left (\frac {t}{3}\right )^r \\ & = -1 + \frac {4}{3} + \frac {4}{3} \sum _{r = 1}^{\infty } 3^{-r} \cdot t^r \\ & = \frac {1}{3} + \frac {4}{3} \sum _{r = 1}^{\infty } 3^{-r} \cdot t^r, \end {align*}
and hence \begin {align*} \Prob (Y = y) = \begin {cases} \frac {1}{3}, & y = 0, \\ \frac {4}{3^{y + 1}}, & \text {otherwise}. \end {cases} \end {align*}