From b080d4aa6ccfd51732d463a23add0f94e91908ce Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Tue, 6 Oct 2020 19:13:38 -0400
Subject: did two homeworks in math

---
 houdre/hw5.tex | 111 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 111 insertions(+)
 create mode 100644 houdre/hw5.tex

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diff --git a/houdre/hw5.tex b/houdre/hw5.tex
new file mode 100644
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+++ b/houdre/hw5.tex
@@ -0,0 +1,111 @@
+\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\fr#1#2{{#1\over #2}}
+\def\var{\mathop{\rm var}\nolimits}
+\def\infint{\int_{-\infty}^\infty}
+
+\q1
+
+The mean of this distribution $\E(X)$ is
+$$\E(X) = \infint xf(x) dx.$$
+This integral evaluates to 0 by symmetry because $f(x) =
+{1\over2}ce^{-c|x|}$ is an even function, so $xf(x)$ is odd.
+
+The variance of this distribution is $\var(X) = \E(X^2) - \E(X)^2 =
+\E(X^2).$ By theorem 5.58,
+$$\E(X^2) = \infint x^2f(x) dx = \infint x^2{1\over2}ce^{-c|x|} dx
+= \int_0^\infty x^2ce^{-cx} dx
+= ce^0{2\over c^3} = 2c^{-2}.$$
+by repeated integration by parts
+
+\q2
+
+$$\Pr(X\geq w) = \sum_{k=w}^\infty {1\over k!}\lambda^ke^{-\lambda}.$$
+$$\Pr(Y\leq \lambda) = \int_0^\lambda {1\over\Gamma(w)} x^{w-1}e^{-x}dx
+= \int_0^\lambda {x^{w-1}e^{-x}\over (w-1)!} dx,$$
+because $\Gamma(w) = (w-1)!$
+
+We are going to prove this equality by induction on $w$.
+It is true for $w=1$ because the Poisson distribution will sum to
+$1-e^{-\lambda}$ (because $\Pr(X\geq0) = 1$ and $\Pr(X<1) = \Pr(X=0) =
+e^{-\lambda}\lambda^0/0!.$)
+The Gamma distribution is ${1\over\Gamma(1)}\int_0^\lambda x^0e^{-x}dx =
+-e^{-x}\big|^\lambda_0 = -e^{-\lambda} + 1.$
+
+Assuming the following equality holds for $w,$ it will be shown to
+hold for $w+1$:
+$$\sum_{k=w}^\infty {1\over k!}\lambda^ke^{-\lambda} = \int_0^\lambda
+{x^{w-1}e^{-x}\over (w-1)!} dx.$$
+$$\sum_{k=w}^\infty {1\over k!}\lambda^ke^{-\lambda}
+= {1\over w!}\lambda^we^{-\lambda} + \sum_{k=w+1}^\infty
+{\lambda^ke^{-\lambda}\over k!}.$$
+$${1\over (w-1)!}\int_0^\lambda x^{w-1}e^{-x}dx
+= {1\over (w-1)!}\left({x^we^{-x}\over w}\big|^\lambda_0 +
+\int_0^\lambda {x^we^{-x}\over w} dx\right) =
+{\lambda^we^{-\lambda}\over w!} + \int_0^\lambda {x^we^{-x}\over
+\Gamma(w+1)} dx$$
+by integration by parts.
+$$\Longrightarrow \sum_{k=n+1}^\infty {\lambda^ke^{-\lambda}\over k!} =
+\int_0^\lambda {x^we^{-x}\over \Gamma(w+1)} dx.$$
+
+QED
+
+\q3
+
+The density function $f(x)$ is proportional, so there is some constant
+$c$ such that $$1 = f(x) = c\infint g(x)dx = 2c\int_1^\infty x^{-n}dx =
+2c\left(-{x^{1-n}\over n-1}\right)\big|^\infty_1 = 2c\left({1\over
+n-1}\right) \to c = {n-1\over2}.$$
+
+The mean and variance of $X$ exist when $\E(X)$ and $\E(X^2)$ exist,
+respectively. The first exists when $n>2$ because for $n=2,$
+$$\E(X) = \int_1^\infty xx^{-n}dx = \ln(x)\big|^\infty_1,$$ and,
+similarly for $\E(X^2)$ under $n\leq3.$ $n>3$ is the condition that it
+exists because $$\int_1^\infty x^2x^{-n}dx = \ln(x)\big|^\infty_1,$$ for
+$n=3$ (and $\int\ln(x)dx$ for $n=2$).
+
+\q4
+
+The density function of $Y=|X|$ is:
+$$\{x\geq0: {2\over\sqrt{2\pi}}\exp(-{1\over2}x^2).\hbox{ 0
+otherwise.}\}$$
+
+$$\E(Y) = \int_0^\infty {2x\over \sqrt{2\pi}}\exp(-{1\over2}x^2)dx =
+{1\over\sqrt{2\pi}}\int_0^\infty\exp(-\fr u2)du,$$
+by u-substitution, becoming
+$$-{2\over\sqrt{2\pi}}e^{-u\over2}\big|_0^\infty =
+{\sqrt{2}\over\sqrt{\pi}}.$$
+
+Similarly, $\var(x) = \E(Y^2) - \E(Y)^2.$
+$$\E(Y^2) = \int_0^\infty {2x^2\over \sqrt{2\pi}}\exp(-\fr12 x^2)dx
+= -{2\over\sqrt{2\pi}}x\exp(-\fr12 x^2)\big|^\infty_0
+- {2\over\sqrt{2\pi}}\int_0^\infty -e^{-\fr12 x^2}dx,$$
+by integration by parts, turning into
+$$0 + 1,$$
+because the first evaluates to zero at both extrema and the second is
+the distribution function of the normal distribution so integrates to 1.
+
+\bye
-- 
cgit