\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
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\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\align#1{\vcenter{\halign{$\displaystyle##\hfil$\tabskip1em&&
    $\hfil\displaystyle##$\cr#1}}}
\def\fr#1#2{{#1\over #2}}
\def\var{\mathop{\rm var}\nolimits}
\def\cov{\mathop{\rm cov}\nolimits}
\def\infint{\int_{-\infty}^\infty}
\def\pa#1#2{\partial#1/\partial#2}

\q2

$$\E\left(\fr1{n-1}\sum_{i=1}^n(X_i-\overline X)^2\right)
= \E\left(\fr1{n-1}\sum_{i=1}^n(X_i-\mu - (\overline X -
\mu))^2\right)$$$$
= \E\left(\fr1{n-1}\sum_{i=1}^n((X_i-\mu)^2-2(X_i-\mu)(\overline X-\mu)
+ (\overline X-\mu)^2)\right)$$$$
= \E\left(\fr1{n-1}\right)
= \E\left(\fr1{n-1}(\sum_{i=1}^n(X_i-\mu)^2-n(\overline X-\mu)^2)\right)
= \fr1{n-1}(n\sigma^2 - n\E(\overline X-\mu)^2)$$$$
= \fr1{n-1}(n\sigma^2 - n\E(({1\over n}(X_1-\mu)+\cdots+(X_n-\mu))^2)
= \fr1{n-1}(n\sigma^2 - n\E(({1\over n}(X_1-\mu)+\cdots+(X_n-\mu))^2)
$$$$
= \fr1{n-1}(n\sigma^2 - {1\over n}
(\E((X_1-\mu)^2)+\cdots+\E((X_n-\mu)^2) + \E((X_1-\mu)(X_2-\mu))+\cdots)
= \fr1{n-1}(n\sigma^2 - \sigma^2) = \sigma^2.$$
Note $\E((X_i-\mu)(X_j-\mu)) = \E(X_i-\mu)\E(X_j-\mu) = 0,$ when $i\neq
j,$ by independence.

\q3

$1 = \E(S_n/S_n) = n\E(X_1/S_n) \to \E(X_1/S_n) = 1/n,$ using linearity
and symmetry.
$\E(S_m/S_n) = m\E(X_1/S_n) = m/n.$
However, it is not necessary that $\E(X_k/S_n) \neq \E(X_1/S_n)$ for
$k>n,$ so the equality doesn't necessarily hold for $m>n.$

\q8

The mgf is the product of the sum's components' mgf, or in other words
$M_{X_1}(x)^n$ for $N=n.$
$M_U(ln(s)) = \E(e^{\ln(s)X}) = \E(s^X) = G_U(s),$ and the general mgf
is $G_N(M_{X_1}(x)) = M_N(\ln(M_{X_1}(x))).$

\q10

Expanding $\cov(X,Y) = \E(XY)-\E(X)\E(Y)$ gives
$$\cov(X,Y) = \E((X_1+\cdots+X_n)(Y_i+\cdots+Y_n))-\E(X_1+\cdots+X_n)
\E(Y_1+\cdots+Y_n),$$
becomes, by linearity,
$$\cov(X,Y) = \E(X_1Y_1)+\cdots+\E(X_1Y_n)\cdots+\E(X_nY_n) - (\E X_1 +
\cdots + \E X_n)(\E Y_1 + \cdots + \E Y_n)$$
$$= \E(X_1Y_1) + \E(X_2Y_2) + \cdots + \E(X_nY_n),$$
by independence for $X_i,Y_j:i\neq j.$

By the earlier theorem, since each game is independent and identical,
$\cov(X,Y) = n\cov(X_1,Y_1).$
$$\cov(X_1,Y_1) = \E(XY) - \E(X)\E(Y) = 2pq - (2p^2 + 2pq)(2q^2 + 2pq) =
2pq - 4p^2q^2 - 4p^3q - 4p^2q^2 - 4pq^3$$$$= 2pq - 4pq(2pq+p^2+q^2) = -2pq.$$
So $\cov(X,Y) = -2npq.$

\q14

The coupon collecting time to all coupons is the sum of geometric
distributions with parameters $p=c/c,(c-1)/c\ldots1/c.$
The moment generating function of a geometric distribution is
$$\sum_x^\infty e^{tx}\Pr(X=x) = \sum_x^\infty e^{tx}pq^x =
\sum_x^\infty p(e^tq)^x = {p\over 1-e^tq}.$$

The moment generating function is therefore the product of this series
of geometric moment generating functions:
$${c!\over c^c}{1\over(1-e^t)(1-2e^t)\cdots(1-ce^t)}$$

\q17

$$f(x) = {1\over2}e^{-|x|} = {1\over2\pi}\infint e^{-itx}\phi(t)dt =
{1\over2\pi}\infint e^{-itx}{1\over1+t^2}dt =
{1\over2\pi}\infint -e^{itx}{1\over1+t^2}dt =
{1\over2}e^{-|x|},$$
based on 7.80, using $\phi(t) = {1\over1+t^2}.$

\q18

The characteristic function of the Cauchy distribution is $e^{-|t|}.$
$$M_{A_n}(t) = \E(e^{itn^{-1}(X_1+\cdots+X_n)}) =
\root n\of{M_{X_1}(t)^n} = M_{X_1}(t).$$

\q20

%deeply unsure

By applying the partition theorem to $M(t) = \E(e^{tX}),$ we trivially
get $$M(t) = \sum_{k=1}^\infty \Pr(N=k)\E(e^{tX}|N=k).$$

Given that $\Pr(N=k) = {1\over(e-1)k!}$ and 
Given these specific distributions,
$$\Pr(X\leq x) = \Pr(U_1\leq x)\cdots\Pr(U_k\leq x) = x^n \to
f_X(x) = nx^{n-1}$$$$ \to \E(X^j) = \int_0^1 x^jnx^{n-1}dx = \int_0^1
nx^{n+j}/(j+n)\to \E(e^{tX}|N=k) = \sum_{j=0}^\infty {1\over
j!}t^jnx^{n+j}/(j+n)$$$$
\to M(t) = \sum_{k=1}^\infty {1\over(e-1)k!}\sum_{j=0}^\infty {1\over
j!}t^jnx^{n+j}/(j+n).$$

using $\E(e^{tX}) = \sum_{n=0}^\infty {t^j\over j!}\E(X^j).$

The difference of $R$ and $X$ must be exponentially distributed because
the quotient of their moment generating functions is similar to an
exponential moment generating function.

\bye