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imporved typesetting of renewal theorem

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Josia Pietsch 2023-07-12 12:38:07 +02:00
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inputs/lecture_06.tex
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@ -172,21 +172,31 @@ In order to prove \autoref{thm2}, we need the following:
\begin{proof}
By SLLN, $\frac{S_n}{n} \xrightarrow{a.s.} m$ as $n \to \infty$.
Note that $N_t \uparrow \infty$ a.s.~as $t \to \infty (\ast\ast)$, since
$\{N_t \ge n\} = \{X_1 + \ldots+ X_n \le t\}$ thus $N_t \uparrow \infty$ as $t \uparrow \infty$.
Note that
\begin{equation}
N_t \uparrow \infty \text{ a.s.~as } t \to \infty, \label{eqn:renewalnt}
\end{equation}
since $\{N_t \ge n\} = \{X_1 + \ldots+ X_n \le t\}$.
\begin{claim}
$\bP[\frac{S_n}{n} \xrightarrow{n \to \infty} m , N_t \xrightarrow{t \to \infty} \infty] = 1$.
$\bP[\frac{S_n}{n} \xrightarrow{n \to \infty} m
\land N_t \xrightarrow{t \to \infty} \infty] = 1$.
\end{claim}
\begin{subproof}
Let $A \coloneqq \{\omega: \frac{S_n(\omega)}{n} \xrightarrow{n \to \infty} m\}$ and $B \coloneqq \{\omega : N_t(\omega \xrightarrow{t \to \infty} \infty\}$.
By the SLLN, we have $\bP(A^C) = 0$ and $\ast\ast \implies \bP(B^C) = 0$.
Let $A \coloneqq \{\omega: \frac{S_n(\omega)}{n} \xrightarrow{n \to \infty} m\}$
and $B \coloneqq \{\omega : N_t(\omega) \xrightarrow{t \to \infty} \infty\}$.
By the SLLN we have $\bP(A^C) = 0$
and by \eqref{eqn:renewalnt} it holds that $\bP(B^C) = 0$.
\end{subproof}
Equivalently, $\bP\left[ \frac{S_{N_t}}{N_t} \xrightarrow{t \to \infty} m, \frac{S_{N_t + 1}}{N_t + 1} \xrightarrow{t \to \infty} m \right] = 1$.
Equivalently,
$\bP\left[ \frac{S_{N_t}}{N_t} \xrightarrow{t \to \infty} m
\land \frac{S_{N_t + 1}}{N_t + 1} \xrightarrow{t \to \infty} m \right] = 1$.
By definition, we have $S_{N_t} \le t \le S_{N_t + t}$.
Then $\frac{S_{N_t}}{N_t} \le \frac{t}{N_t} \le S_{N_t + 1}{N_t} \le \frac{S_{N_t + 1}}{N_t + 1} \cdot \frac{N_t + 1}{N_t}$.
By definition, we have $S_{N_t} \le t \le S_{N_t + 1}$.
Thus
\[\frac{S_{N_t}}{N_t} \le \frac{t}{N_t} \le \frac {S_{N_t + 1}}{N_t}
\le \frac{S_{N_t + 1}}{N_t + 1} \cdot \frac{N_t + 1}{N_t}.\]
Hence $\frac{t}{N_t} \to m$.
\end{proof}