\(% 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}}\)

2013.3.13 Question 13

1. 1. Since \(0 \leq X \leq 1\), we must have that \[ F(x) = \int _{0}^{x} f(t) \Diff t \] for \(0 \leq x \leq 1\). Hence, since \(0 \leq f(t) \leq M\) for \(0 \leq t \leq x \leq 1\), we have \[ 0 = \int _{0}^{x} 0 \Diff t \leq F(x) \leq \int _{0}^{x} M \Diff t = Mx, \] as desired.

   2. Since \(0 \leq X \leq 1\), we must have \(F(0) = 0\) and \(F(1) = 1\). Let the desired integral be \(I\), using integration by parts, we have \begin {align*} I & = \int _{0}^{1} 2g(x) F(x) f(x) \Diff x \\ & = \int _{0}^{1} 2 g(x) F(x) \Diff F(x) \\ & = \left [2g(x) F(x)^2\right ]_{0}^{1} - 2 \int _{0}^{1} F(x) \Diff (g(x) F(x)) \\ & = 2g(1) F(1)^2 - 2g(0) F(0)^2 - 2 \int _{0}^{1} g'(x) F(x)^2 \Diff x - 2\int _{0}^{1} g(x) F(x) f(x) \Diff x \\ & = 2g(1) - 2\int _{0}^{1} g'(x) F(x)^2 \Diff x - I. \end {align*}

      This means \[ 2 I = 2 g(1) - 2 \int _{0}^{1} g'(x) F(x)^2 \Diff x, \] and hence \[ I = g(1) - \int _{0}^{1} g'(x) F(x)^2 \Diff x. \]

2. 1. Since \(0 \leq Y \leq 1\), we must have \begin {align*} \int _{0}^{1} k F(y) f(y) \Diff y & = k \int _{0}^{1} F(y) \Diff F(y) \\ & = k \cdot \frac {1}{2} \cdot \left [F(y)^2\right ]_{0}^{1} \\ & = k \cdot \frac {1}{2} \cdot \left [F(1)^2 - F(0)^2\right ] \\ & = \frac {k}{2} \cdot (1^2 - 0^2) \\ & = \frac {k}{2} \\ & = 1, \end {align*}

      and hence \(k = 2\).

   2. Notice that \begin {align*} \Expt \left (Y^n\right ) & = \int _{0}^{1} 2y^n F(y) f(y) \Diff y \\ & \leq \int _{0}^{1} 2y^n My f(y) \Diff y \\ & = 2M \int _{0}^{1} y^{n + 1} f(y) \Diff y \\ & = 2M \Expt \left (X^{n + 1}\right ) \\ & = 2M \mu _{n + 1}, \end {align*}

      and that \begin {align*} \Expt \left (Y^n\right ) & = \int _{0}^{1} 2y^n F(y) f(y) \Diff y \\ & = \LEvalAt {y^n}{y = 1} - \int _{0}^{1} (y^n)' F(y)^2 \Diff y \\ & = 1 - n \int _{0}^{1} y^{n - 1} F(y)^2 \Diff y \\ & \geq 1 - n \int _{0}^{1} y^{n - 1} My F(y) \Diff y \\ & = 1 - M n \int _{0}^{1} y^n F(y) \Diff y \\ & = 1 - \frac {Mn}{n + 1} \int _{0}^{1} F(y) \Diff (y^{n + 1}) \\ & = 1 - \frac {Mn}{n + 1} \left (\left [F(y) y^{n + 1}\right ]_{0}^{1} - \int _{0}^{1} y^{n + 1} \Diff F(y)\right ) \\ & = 1 - \frac {Mn}{n + 1} \left (F(1) \cdot 1^{n + 1} - F(0) \cdot 0^{n + 1} - \int _{0}^{1} y^{n + 1} f(y) \Diff y\right ) \\ & = 1 - \frac {Mn}{n + 1} \left (1 - \Expt \left (X^{n + 1}\right )\right ) \\ & = 1 - \frac {nM}{n + 1}\mu _{n + 1} - \frac {nM}{n + 1}, \end {align*}

      as desired.

   3. Since we have for non-negative \(n\), \[ 1 + \frac {nM}{n + 1}\mu _{n + 1} - \frac {nM}{n + 1} \leq 2M \mu _{n + 1}, \] and hence for \(n \geq 1\), we have \[ 1 + \frac {(n - 1)M}{n} \mu _n - \frac {(n - 1)M}{n} \leq 2M \mu _n, \] which multiplying both sides by \(n\) gives \[ n + (n - 1)M \mu _n - (n - 1)M \leq 2Mn \mu _n, \] and rearranging gives \[ n - (n - 1) M \leq M(n + 1) \mu _n, \] hence \[ \mu _n \geq \frac {n - (n - 1)M}{M(n + 1)} = \frac {n}{(n + 1)M} - \frac {n - 1}{n + 1}, \] as desired.