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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
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