\newfam\rsfs
\newfam\bbold
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\def\bb#1{{\fam\bbold #1}}
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\def\cal#1{{\oldcal #1}}
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\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
