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\newcommand{\Ukn}[1]{U_{#1,n}}
\newcommand{\Si}[1]{S_{#1}}
%\newcommand{\Si}[1]{S(#1)}
\newcommand{\SiFrac}[1]{\frac{\Si{#1}}{\Si{n+1}}}

\begin{document}

\begin{center}
Jeremy Morris
\end{center}
\begin{center}
Math 6070 Midterm (deadline: 23 April)
\end{center}

\begin{enumerate}
\item \textit{Let $U_{1}, \ldots, U_{n}$ be independent identically distributed random variables, uniform on $[0,1]$.  Let $U_{1,n} \leq U_{2,n} \leq \cdots \leq U_{n,n}$ denote the order statistics.  Let $k_{n}$ satisfy}
\[k_{n} \rightarrow \infty \textit{ and } k_{n}/n \rightarrow p \in (0,1) \quad (n \rightarrow \infty).\]  
\textit{Show that there are $a_{n},b_{n}$ such that}
\[(a_{n}\Ukn{k_{n}} - b_{n}) \overset{d}{\rightarrow} N(0,1).\]

We know that
\[(\Ukn{1},\Ukn{2},\ldots,\Ukn{n}) \overset{d}{=} \left( \SiFrac{1}, \SiFrac{2}, \ldots, \SiFrac{n} \right)\]
where
\begin{align*}
\Si{k} & = \sum_{i=1}^{k}Y_{i}.
\end{align*}
and the $Y_{i}$ are indpendent identically distributed exponential random variables with parameter $1$.
I will show that
\begin{align}
\sqrt{n}\left( \SiFrac{k_{n}} - \frac{k_{n}}{n+1}\right) \overset{d}{\rightarrow} N(0,1). \label{eq1}
\end{align}
Then we can do some algebraic manipulation so that (\ref{eq1}) can be rewritten
\begin{align*}
\sqrt{n}\left( \frac{(n+1)\Si{k_{n}} - k_{n}\Si{n+1}}{(n+1)\Si{n+1}} \right)
\end{align*}
which becomes
\begin{align*}
\sqrt{n}\left( \frac{(n+1)(\Si{k_{n}} - k_{n}) - k_{n}(\Si{n+1} - (n+1))}{(n+1)\Si{n+1}}\right)
\end{align*}
and finally
\begin{align*}
\frac{(n+1)n}{(n+1)\Si{n+1}} \left( \frac{1}{\sqrt{n}}(\Si{k_{n}} - k_{n}) - \frac{1}{\sqrt{n}}\frac{k_{n}}{n+1}(\Si{n+1} - (n+1)) \right).
\end{align*}

The first term in this statement converges in probability to one since
\begin{align*}
E\Si{n+1} & = n+1.
\end{align*}
Then, by application of Slutsky's theorem, we need to show that the remaining term converges in distribution to the normal distribution.  The remaining term is 
\begin{align}
\frac{1}{\sqrt{n}}(\Si{k_{n}} - k_{n}) - \frac{1}{\sqrt{n}}\frac{k_{n}}{n+1}(\Si{n+1} - (n+1)) \label{eq2}
\end{align}
where $\Si{k_{n}}$ and $\Si{n+1}$ are not independent.  In order to create a sum of independent random variables, we rewrite (\ref{eq2}) so that we have
\begin{align*}
\frac{1}{\sqrt{n}}(\Si{k_{n}} - k_{n}) - \frac{1}{\sqrt{n}}\frac{k_{n}}{n+1}(\Si{n+1} - (n+1) - \Si{k_{n}} + \Si{k_{n}} + k_{n} - k_{n})
\end{align*}
which can be rewritten
\begin{align}
\frac{1}{\sqrt{n}}(\Si{k_{n}} - k_{n}) - \frac{1}{\sqrt{n}}\frac{k_{n}}{n+1}\Bigl[\Bigl(\Si{k_{n}} - k_{n}\Bigr) + \Bigl(\Si{n+1}  - \Si{k_{n}} - ((n+1) - k_{n}))\Bigr)\Bigr]. \label{eq3}
\end{align}
Then $\Si{k_{n}}$ and $(\Si{n+1} - \Si{k_{n}})$ are independent so that
\begin{align*}
\Bigl(\frac{1}{\sqrt{k_{n}}}(\Si{k_{n}} - k_{n}),\frac{1}{\sqrt{n - k_{n}}}(\Si{n+1} - \Si{k_{n}} - ( (n+1) - k_{n} )\Bigr) \overset{d}{\rightarrow} (N_{1},N_{2})
\end{align*}
where $N_{1}$ and $N_{2}$ are independent standard normal random variables.  Then as $n \rightarrow \infty$, (\ref{eq3}) becomes
\begin{align*}
\sqrt{p}N_{1} - p\Bigl[ \sqrt{p}N_{1} + \sqrt{1-p}N_{2} \Bigr],
\end{align*}
which is a linear combination of independent normal random variables with mean $0$ and variance $p(1-p)$.  This shows that
\[(a_{n}U_{k_{n},n} - b_{n}) \overset{d}{\rightarrow} N(0,1).\]
where $a_{n} = \sqrt{n}$ and $b_{n} = \sqrt{n}k_{n}/(n+1)$.

\end{enumerate}

\end{document}
