houdre hws template

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-\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 $$1 = \infint f(x)dx = c\infint g(x)dx = 2c\int_1^\infty
-x^{-n}dx = 2c\left(-{x^{1-n}\over n-1}\right)\big|^\infty_1 =
-2c\left({1\over n-1}\right) \to c = {n-1\over2}.$$
-$$f(x) = {n-1\over2}g(x),$$
-and this will look like a vertically stretched $1/x^2$ graph.
-
-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