\newfam\rsfs \newfam\bbold \def\scr#1{{\fam\rsfs #1}} \def\bb#1{{\fam\bbold #1}} \let\oldcal\cal \def\cal#1{{\oldcal #1}} \font\rsfsten=rsfs10 \font\rsfssev=rsfs7 \font\rsfsfiv=rsfs5 \textfont\rsfs=\rsfsten \scriptfont\rsfs=\rsfssev \scriptscriptfont\rsfs=\rsfsfiv \font\bbten=msbm10 \font\bbsev=msbm7 \font\bbfiv=msbm5 \textfont\bbold=\bbten \scriptfont\bbold=\bbsev \scriptscriptfont\bbold=\bbfiv \def\Pr{\bb P} \def\E{\bb E} \newcount\qnum \def\q{\afterassignment\qq\qnum=} \def\qq{\qqq{\number\qnum}} \def\qqq#1{\bigskip\goodbreak\noindent{\bf#1)}\smallskip} \def\align#1{\vcenter{\halign{$##\hfil$&&$\hfil##$\cr#1}}} \tabskip=1em \q7 By continuity, $X$ has a density function, so $$\E(X) = \int_0^\infty xf_X(x) dx = x(F_X(x)-1)\big|_0^\infty - \int_0^\infty (F_X(x)-1) dx,$$ by integration by parts. Because $xF_X(x)|_0^\infty = 0,$ (since $\Pr(X\leq\infty)-1 = 0.$) $$\E(X) = \int_0^\infty 1-F_X(x)dx.$$ \q9 $$F_{X'}(x) = \bigg\{\align{ F_X(x)&\hbox{if }x<a,\cr 1&\hbox{if }x\geq a.\cr}$$ For $x\leq a,$ the distribution function is the same because the $X' \leq x$ when $X\leq x.$ For $x\geq a,$ $\Pr(X'\leq x) = \Pr(X<a) + \Pr(X\geq a) = 1.$ \q10 $$f_X(x) = \bigg\{\align{ e^{-x}&\hbox{if }x>0,\cr 0&\hbox{if }x\leq0.\cr}$$ $$Y = (X-2)/(X+1) \to YX+Y = X-2 \to (Y-1)X = -(2+Y) \to X = -(2+Y)/(Y-1).$$ $$dX/dY = -{3\over(Y-1)^2}.$$ By 5.52, since $g(x)$ is strictly decreasing, $$\align{f_Y(y)&=&-f_X\left(-{2+y\over y-1}\right) \left(-{3\over(y-1)^2}\right)\hfill\cr &=&\bigg\{\align{ e^{2+y\over y-1}{3\over(y-1)^2}&-2<y<1.\cr 0&{\rm otherwise.}\cr}\hfill\cr}$$ Note that this bound is true because ${2+y\over y-1}\in(0,\infty)\iff y\in(-2,1).$ \q14 $$f_X(x) = \bigg\{\align{ 1&x\in[0,1],\cr 0&{\rm otherwise}.\cr}$$ By theorem 5.50, with $Y = g(X) = {3X\over 1-X} \to 3X-Y(1-X) = 3X+YX-Y = 0 \to X = {Y\over 3+Y},$ (note that $g(x)$ is strictly decreasing on $(0,\infty).$) $$f_Y(y) = -f_X(g^{-1}(y))[g^{-1}]'(y) = f_X\left({y\over 3+y}\right){3\over(3+y)^2} = \bigg\{\align{ {3\over(3+y)^2}&y>0,\cr \hfil 0&{\rm otherwise}.\cr}$$ \bye