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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\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}
\q1
(a)
$$\Pr(|Z_n-a|>\epsilon) = \Pr(Z_n<a-\epsilon) = \Pr(X_i<a-\epsilon)^n,$$
which, since $\Pr(X_i<a-\epsilon)=1-\epsilon/a,$ (for small $\epsilon$)
$\to 0$ as $n\to\infty$ for all $\epsilon>0.$
(b)
$$\Pr(|\sqrt{Z_n}-\sqrt{a}|>\epsilon) =
\Pr(\sqrt{Z_n}<\sqrt{a}-\epsilon) =
\Pr(Z_n<a-\sqrt{a}\epsilon+\epsilon^2) =
\Pr(X_i<a-\sqrt{a}\epsilon+\epsilon^2)^n =
(1-\epsilon(\sqrt{a}-\epsilon)/a)^n \to 0,$$
as $n\to\infty$ because $\epsilon<\sqrt{a},$ otherwise the probability
is zero because the domain would be outside of the uniform distribution.
% not done here
(c)
$$\Pr(U_n\leq x) = \Pr(1-Z_n\leq x/n) = \Pr(Z_n\geq 1-x/n) =
1-\Pr(Z_n < 1-x/n) = 1-\Pr(X_i < x/n)^n = 1-(1-x/n)^n,$$
for $0<x<n,$ which becomes $x>0$ as $n\to\infty,$ and limits to
$1-e^{-x},$ by definition of the exponential.
\q2
Let $Z_n$ be the normalization of sum $S_n$ of $n$ independent Bernoulli
variables, each with parameter $p$ and thus mean $p$ and variance $pq.$
$$\Pr(|Z_n|\leq x) = \Pr(-x\leq Z_n\leq x) =
\Pr(np-\sqrt{npq}x\leq S_n\leq np+\sqrt{npq}x) = \sum{n\choose k}p^k
q^{n-k}.$$
By the central limit theorem, as $n\to\infty,$
$$\Pr(|Z_n|\leq x) = \Pr(Z_n\leq x) - \Pr(Z_n\leq -x) = \int_{-x}^x
{1\over\sqrt{2\pi}}e^{-\fr12 u^2}du = 2\int_0^x
{1\over\sqrt{2\pi}}e^{-\fr12 u^2}du.$$
\q3
$X_n,$ the binomial distribution is equivalent to the sum of $n$
independent Bernoulli distributions $B_i$ with parameter $p.$
$$\E([n^{-1}X_n-p]^2) = \E(n^{-2}[(B_1-p)+(B_2-p)+\cdots+(B_n-p)]^2) =
n^{-2}(n\var(B_i)) = p(1-p)/n,$$
which $\to 0$, as $n\to\infty.$ Converging in mean square implies
converging in probability, therefore the distribution converges in
probability to $p.$
\q5
With $P_n$ the poisson distribution with parameter $n,$
$$\Pr(P_n = k) = {n^k e^{-\lambda}\over k!},$$
$$e^{-n}\left(1+n+{n^2\over 2!}+\cdots+{n^n\over n!}\right) =
\sum_{k=0}^n \Pr(P_n=k) = \Pr(P_n\leq n).$$
$P_n = X_1+X_2+\cdots+X_n$ where $X_i$ has the Poisson distribution with
parameter 1.
The normalized version of $P_n$ is $(P_n-n)/\sqrt{x},$ because the mean
and variance of $X_i$ is $1.$
As $n\to\infty,$ this distribution approaches the standard normal (by
the central limit theorem), and
$\Pr(P_n\leq n) \to \Pr(N(0,1)\leq0) = 1/2.$
\q7
$$\E([X_n+Y_n-(X+Y)]^2) = \E([X_n-X]^2) + \E([Y_n-Y]^2) +
2\E([Y_n-Y][X_n-X]).$$
As $n\to\infty,$ this approaches $2\E([Y_n-Y][X_n-x]) \leq
2\sqrt{\var(Y_n-Y)\var(X_n-X)},$ of which both variances $\to 0$ because
$Y_n \to Y$ in mean square and similar for $X_n,$ and therefore the
first expectation approaches zero and $X_n+Y_n \to X+Y.$
\q8
$$\E([X_n-X]^2)\to0\qquad\hbox{\it\ as } n\to\infty$$
Let $V=1.$ By Cauchy-Schwartz,
$$\E(X_n-X)^2\leq\E([X_n-X]^2)\Rightarrow\E([X_n-X])\to 0,$$
as $n\to\infty.$
By linearity of expectations,
$$\E(X_n)\to\E(X)\qquad\hbox{\it\ as } n\to\infty.$$
If $\Pr(Z_n=0) = 1-{1\over n}$ and $\Pr(Z_n=n) = {1\over n},$ $X_n\to
X=0$ in probability because as $n\to\infty,$ $\Pr(X>\epsilon) \to 0,$
but $\E(X) = \E(0) = 0$ and $\E(X_n) = {n\over n} = 1.$
\q11
$$\Pr(|X|\geq a) = \Pr(g(X)\geq g(a)),$$
by the fact that $g$ is symmetric and strictly increasing on $x>0.$
$$\Pr(g(X)\geq g(a)) \leq {\E(g(X))\over g(a)},$$
by Markov's inequality (since $g(X)$ is a random variable) given that
$g(x)>0,$ therefore the inequality is true.
\q14
$X_n$ converges in mean square to the random variable $X,$ so
$$\E([X_n-X]^2) \to 0\qquad\hbox{\it\ as } n\to\infty,$$
$$\E([X_n-X_m]^2) = \E([X_n-X-(X_m-X)]^2) =
\E([X_n-X]^2)-2\E([X_n-X][X_m-X])+\E([X_m-X]^2) = 2\E([X_n-X][X_m-X]),$$
as $n,m\to\infty,$ from the first assumption.
By the Cauchy-Schwartz inequality,
$$\E([X_n-X][X_m-X])^2\leq \E([X_n-X]^2)\E([X_m-X])^2 \to 0,$$
as $n\to\infty,$ so $\E([X_n-X][X_m-X]) \to 0,$ so the assumption is
proved.
$$\cov(X_n,X_m) = \E([X_n-\E(X_n)][X_m-\E(X_m)]) =
\E([X_n-\mu][X_m-\mu])$$
$X_n$ and $X_m$ are sufficiently similar to $X,$ so
$$\E([X_n-\mu][X_m-\mu]) \to \E([X-\mu]^2) = \sigma^2 =
\sqrt{\var(X_n)\var(X_m)} \Rightarrow \rho\to 1.$$
% An incomplete answer: how is it sufficiently similar?
\bye
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