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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}
+
+\q1
+
+Let $A \setminus B := \{x \in A : x\not\in B\}.$
+$(A\setminus B) \cap B = \empty,$ by definition, so they are disjoint
+sets. Therefore, if $x \in A \land x \in B,$ $x\not\in A \setminus B$
+and $x\not\in B\setminus A,$ and also the two setminuses do not
+intersect with eachother, making them disjoint sets.
+$$\Pr(A\cup B) = \Pr((A\setminus B)\cup(A\cap B)\cup(B\setminus A)) =
+\Pr(A\setminus B) + \Pr(A\cap B) + \Pr(B\setminus A) + \Pr(A\cap B) -
+\Pr(A\cap B)$$$$ = \Pr((A\setminus B)\cup(A\cap B)) + \Pr((B\setminus
+A)\cup(A\cap B)) - \Pr(A\cap B) = \Pr(A) + \Pr(B) - \Pr(A\cap B).$$
+
+\q2
+
+$$\Pr(A\cap B) + \Pr(A\cap B^C) = \Pr(A\cap B) + \Pr(\{x\in A:x\not\in
+B\}) = \Pr((A\cap B) \cup (A\setminus B)) = \Pr(A),$$
+by disjointedness and set arithmetic.
+
+\q3
+
+$$\Pr(A_1 \cup (A_2^C \cap A_3^C)) = \Pr(A_1) + \Pr(A_2^C \cap A_3^C) -
+\Pr(A_1\cap A_2^C\cap A_3^C) = 1/6 + \Pr(A_2^C)\Pr(A_3^C) -
+\Pr(A_1)\Pr(A_2^C)\Pr(A_3^C)$$$$ = 1/6 + 25/36 - 25/216 = 161/216.$$
+
+\q4
+
+\noindent{\it (a)}
+
+Pairwise exclusive implies mutually exclusive:
+$$A_1\cap A_2\cap A_3 = A_1\cap (A_2\cap A_3) = A_1\cap\empty =
+\empty,$$
+and see {\it (b)} for a description of why these probabilities are
+therefore impossible.
+
+\noindent{\it (b)}
+
+No, if they were, all three sets would be disjoint, giving
+$$\Pr(A_1\cup A_2\cup A_3) = \Pr(A_1) + \Pr(A_2) + \Pr(A_3) = 1/2 + 1/4 +
+1/3 > 1,$$
+violating a property of the probability function ($0 \leq \Pr X \leq 1$)
+
+\q5
+
+\noindent{\it (a)}
+$$\E(Y) = \E(2 - 3X) = \E(2) - 3\E(X) = 2 - 3*2 = -4.$$
+
+\noindent{\it (b)}
+$$\var Y = \E(Y^2) - \E(Y)^2 = \E((2-3X)^2) - 16 = \E(4 - 12X + 9X^2) -
+16 = 4 - 12\E(X) + 9\E(X^2) - 16 = 18,$$
+by linearity of expectation.
+
+\q6
+
+$$P_X(k) = \left\{\vbox{\halign{$#$\hfil&\hskip3em $#$\hfil\cr{\lambda^k
+e^{-\lambda}\over k!}&k \geq 0\cr 0&{\rm otherwise}\cr}}\right..$$
+
+\noindent{\it (a)}
+
+$$\E(e^{tX}) = \sum_{x=0}^\infty e^{tx}P_X(x)
+= \sum_{x=0}^\infty {e^{tx}\lambda^xe^{-\lambda}\over x!}
+= e^{-\lambda}\sum_{x=0}^\infty {e^{(t+\ln \lambda)x}\over x!}
+= e^{-\lambda}\sum_{x=0}^\infty {(\lambda e^t)^x\over x!}
+= e^{-\lambda}e^{\lambda e^t},
+$$
+by definition of the $e^ax$ Taylor series.
+
+\noindent{\it (b)}
+
+The third-order moment $\E(X^3)$ will be the third derivative of the mgf
+$\E(e^{tX}).$ at $t=0,$ so we get
+$$e^{-\lambda}(\lambda e^t)^3 e^{\lambda e^t} = \lambda^3.$$
+
+\q7
+
+\noindent{\it (a)}
+
+The pmf is $1/16$ for $X = 0$ and $X=4,$ $4/16$ for $X=1$ and $X=3,$ and
+$6/16$ for $X=2,$ or in other terms, $P_X(k) = {{4\choose k}\over 2^4}.$
+
+\noindent{\it (b)}
+
+This can be immediately computed as $$\Pr(\hbox{$X$ is odd}) = \Pr(X = 1)
++ \Pr(X = 3) = 1/2,$$ by disjointedness of those events.
+
+\q8
+
+Chebyshev's inequality gives us
+$$\Pr(|X-\E X| < .1) \geq 1-{\sigma^2\over .1^2} \geq .95$$
+The maximum value of $\sigma = \var X$ is $.05*.1^2 = \sigma^2 \to
+\sigma = .1\sqrt{.05} \approx .022.$
+
+\q9
+
+{\it (a)}
+
+On $\{x \geq 0\} = \{y \geq 1\},$
+$$F_X(x) = 1 - e^{-\lambda x}.$$
+$$F_Y(y) = F_X(e^y) = 1 - e^{-\lambda e^y}.$$
+$$f_Y(y) = \lambda e^y e^{-\lambda e^y} = \lambda e^{y-\lambda e^y}.$$
+
+{\it (b)}
+
+$\lambda < 1$ gives convergence.
+
+\bye