From a31c07408cfd10b82e3228ccb0b0c9dcd5a33a4b Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Tue, 20 Oct 2020 11:34:37 -0400
Subject: hw 5 part ii /houdre

---
 houdre/hw5ii.tex | 45 ++++++++++++++++++++++++++++++++++++++++++---
 1 file changed, 42 insertions(+), 3 deletions(-)

diff --git a/houdre/hw5ii.tex b/houdre/hw5ii.tex
index 3e99232..51b3208 100644
--- a/houdre/hw5ii.tex
+++ b/houdre/hw5ii.tex
@@ -23,16 +23,55 @@
 \def\q{\afterassignment\qq\qnum=}
 \def\qq{\qqq{\number\qnum}}
 \def\qqq#1{\bigskip\goodbreak\noindent{\bf#1)}\smallskip}
-\def\fr#1#2{{#1\over #2}}
-\def\var{\mathop{\rm var}\nolimits}
-\def\infint{\int_{-\infty}^\infty}
+\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
-- 
cgit