diff --git a/inputs/prerequisites.tex b/inputs/prerequisites.tex new file mode 100644 index 0000000..9ca735b --- /dev/null +++ b/inputs/prerequisites.tex @@ -0,0 +1,23 @@ +\begin{theorem}[Chebyshev's inequality] % TODO Proof + Let $X$ be a r.v.~with $\Var(x) < \infty$. + Then $\forall \epsilon > 0 : \bP \left[ \left| X - \bE[X] \right| > \epsilon\right] \le \frac{\Var(x)}{\epsilon^2}$. + +\end{theorem} + + +We used Chebyshev's inequality. Linearity of $\bE$, $\Var(cX) = c^2\Var(X)$ and $\Var(X_1 +\ldots + X_n) = \Var(X_1) + \ldots + \Var(X_n)$ for independent $X_i$. + + +How do we prove that something happens almost surely? +\begin{lemma}[Borel-Cantelli] + If we have a sequence of events $(A_n)_{n \ge 1}$ + such that $\sum_{n \ge 1} \bP(A_n) < \infty$, + then $\bP[ A_n \text{for infinitely many $n$}] = 0$ + (more precisely: $\bP[\limsup_{n \to \infty} A_n] = 0$). + + The converse also holds for independent events $A_n$. + +\end{lemma} + + +Modes of covergence: $L^p$, in probability, a.s. diff --git a/inputs/vl3.tex b/inputs/vl3.tex new file mode 100644 index 0000000..c5c9fd0 --- /dev/null +++ b/inputs/vl3.tex @@ -0,0 +1,117 @@ + +$(\Omega, \cF, \bP)$ Probability Space, $X : ( \Omega, \cF) \to (\R, \cB(\R))$ random variable. +Then $\Q(\cdot) = \bP [ x\in \cdot ]$ is the distribution of $X$ under $\bP$. + + +\section{Independence and product measures} + +In order to define the notion of independence, we first need to construct +product measures in order to be able to consider several random variables +at the same time. + +The finite case of a product is straightforward: +\begin{theorem}{Product measure (finite)} +Let $(\Omega_1, \cF, \bP)$ and $(\Omega_2, \cF_2, \bP_2)$ be probability spaces. +Let $\Omega \coloneqq \Omega_1 \times \Omega_2$ +and $R \coloneqq \{A_1 \times A_2 | A_1 \in \cF_1, A_2 \in \cF_2 \}$. + +Let $\cF$ be $\sigma(R)$ (the sigma algebra generated by $R$). +Then there exists a unique probability measure $\bP$ on $\Omega$ +such that for every rectangle $R = A_1 \times A_2 \in \cR$, $\bP(A_1 \times A_2) = \bP(A_1) \times \bP(A_2)$. +\end{theorem} +\begin{proof} +See Theorem 5.1.1 in the lecture notes on Stochastik. +\end{proof} + +We now want to construct a product measure for infinite products. + + +\begin{definition}[Independence] + A collection $X_1, X_2, \ldots, X_n$ of random variables are called + \vocab{mutually independent} if + \[ + \forall a_1,\ldots,a_n \in \R : + \bP[X_1 \le a_1, \ldots, x_n \le a_n] + = \prod_{i=1}^n \bP[X_i \le a_i] + \] + This is equivalent to + \[ + \forall B_1, \ldots, B_n \in \cB(\R): + \bP[X_1 \in B_1, \ldots, X_n \in B_n] + = \prod_{i=1}^n \bP[X_i \in B_i] + \] +\end{definition} + +\begin{example} + Suppose we throw a dice twice. Let $A \coloneqq \{\text{first throw even}\}$, + $B \coloneqq \{second throw even\}$ + and $C \coloneqq \{\text{sum even}\} $. + + Are $\One_A, \One_B, \One_C$ mutually independent random variables? +\end{example} +It is easy the see, that the random variables are pairwise independent, + but not mutually independent. + +The definition of mutual independence can be rephrased as follos: +Let $X_1, X_2, \ldots, X_n$ r.v.s. Let $\bP[(X_1,\ldots, X_n) \in \cdot ] \text{\reflectbox{$\coloneqq$}} \Q^{\otimes}(\cdot )$. +Then $\Q^{\otimes}$ is a probability measure on $\R^n$. + +\begin{fact} + $X_1,\ldots, X_n$ are mutually independent iff $\Q^{\otimes} = \Q_1 \otimes \ldots \otimes \Q_n$. +\end{fact} +By constructing an infinite product, we can thus extend the notion of independence +to an infinite number of r.v.s. + +\begin{goal} + Can we construct infinitely many independent random variables? +\end{goal} + +\begin{definition}[Consistent family of random variables] + Let $\bP_n, n \in \N$ be a family of probability measures on $(\R^n, \cB(\R^n))$. + The family is called \vocab{consistent} if if + $\bP_{n+1}[B_1 \times B_2 \times \ldots \times B_n \times \R] = \bP_n[B_1 \times \ldots \times B_n]$ + for all $n \in \N, B_i \in B(\R)$. + +\end{definition} + +\begin{theorem}[Kolmogorov extension / consistency theorem] + Informally: + ``Probability measures are determined by finite-dimensional marginals + (as long as these marginals are nice)'' + + Let $\bP_n, n \in \N$ be probability measures on $(\R^n, \cB(\R^n))$ + which are \vocab{consistent}, + then there exists a unique probability measure $\bP^{\otimes}$ + on $(\R^\infty, B(R^\infty))$ (where $B(R^{\infty}$ has to be defined), + such that + \[ + \forall n \in \N, B_1,\ldots, B_n \in B(\R): + \bP^\otimes [\cX : X_i \in B_i \forall 1 \le i \le n] + = \bP_n[B_1 \times \ldots \times B_n] + \] +\end{theorem} + +\begin{remark} + Kolmogorov's theorem can be strengthened to the case of arbitrary + index sets. However this requires a different notion of consistency. +\end{remark} + +\begin{example}of a consistent family: + + Let $F_1, \ldots, F_n$ be probability distribution functions + and let $\bP_n$ be the probability measure on $\R^n$ defined + by + \[ + \bP_n[(a_1,b_1] \times \ldots (a_n, b_n]] + \coloneqq (F_1(b_1) - F_1(a_1)) \cdot \ldots \cdot (F_n(b_n) - F_n(a_n)). + \] + + It is easy to see that each $\bP_n$ is a probability measure. + + Define $X_i(\omega) = \omega_i$ where $\omega = (\omega_1, .., \omega_n)$. + Then $X_1, \ldots, X_n$ are mutually independent with $F_i$ being + the distribution function of $X_i$. + In the case of $F_1 = \ldots = F_n$, then $X_1,\ldots, X_n$ are i.i.d. +\end{example} + + diff --git a/inputs/vl4.tex b/inputs/vl4.tex new file mode 100644 index 0000000..3f27476 --- /dev/null +++ b/inputs/vl4.tex @@ -0,0 +1,119 @@ +\begin{notation} + Let $\cB_n$ denote $\cB(\R^n)$. +\end{notation} +\begin{goal} + Suppose we have a probability measure $\mu_n$ on $(\R^n, \cB(\R^n))$ + for each $n$. + We want to show that there exists a unique probability measure $\bP^{\otimes}$ + on $(\R^\infty, \cB_\infty)$ (where $\cB_{\infty}$ still needs to be defined), + such that $\bP^{\otimes}\left( \prod_{n \in \N} B_n \right) = \prod_{n \in \N} \mu_n(B_n)$ + for all $\{B_n\}_{n \in \N}$, $B_n \in \cB_1$. +\end{goal} +% $\bP_n = \mu_1 \otimes \ldots \otimes \mu_n$. +\begin{remark} + $\prod_{n \in \N} \mu_n(B_n)$ converges, since $0 \le \mu_n(B_n) \le 1$ + for all $n$. +\end{remark} + +First we need to define $\cB_{\infty}$. +This $\sigma$-algebra must contain all sets $\prod_{n \in \N} B_n$ +for all $B_n \in \cB_1$. We simply define $\cB_{\infty}$ to be the +$\sigma$-algebra +Let $\cB_\infty \coloneqq \sigma \left( \{\prod_{n \in \N} B_n | B_n \in \cB(\R)\} \right)$. +\begin{question} + What is there in $\cB_\infty$? + Can we identify sets in $\cB_\infty$ for which we can define the product measure + easily? +\end{question} +Let $\cF_n \coloneqq \{ C \times \R^{\infty} | C \in \cB_n\}$. +It is easy to see that $\cF_n \subseteq \cF_{n+1}$ +and using that $\cB_n$ is a $\sigma$-algebra, we can show that $\cF_n$ +is also a $\sigma$-algebra. +Now, for any $C \subseteq \R^n$ let $C^\ast \coloneqq C \times \R^{\infty}$. + +Thus $\cF_n = \{C^\ast : C \in \cB_n\}$. +Define $\lambda_n : \cF_n : \to [0,1]$ by $\lambda_n(C^\ast) \coloneqq (\mu_1 \otimes \ldots \otimes \mu_n)(C)$. +It is easy to see that $\lambda_{n+1} \defon{\cF_n} = \lambda_n$ (\vocab{consistency}). + + +Recall the following theorem from measure theory: +\begin{theorem}[Caratheodory's extension theorem] % 2.3.3 in the notes + \label{caratheodory} + Suppose $\cA$ is an algebra (i.e.~closed under finite union) + und $\Omega \neq \emptyset$. + Suppose $\bP$ is countably additive on $\cA$ (i.e.~if $(A_n)_{n}$ + are pairwise disjoint and $\bigcup_{n \in \N} A_n \subseteq \cA $ + then $\bP\left( \bigcup_{n \in \N} A_n \right) = \sum_{n \in \N} \bP(A_n)$). + Then $\bP$ extends uniquely to a probability measure on $(\Omega, \cF)$, + where $\cF = \sigma(\cA)$. +\end{theorem} + +Define $\cF = \bigcup_{n \in \N} \cF_n$. Check that $\cF$ is an algebra. +We'll show that if we define $\lambda: \cF \to [0,1]$ with +$\lambda(A) = \lambda_n(A)$ for any $n$ where this is well defined, +then $\lambda$ is countably additive on $\cF$. +Using \autoref{caratheodory} $\lambda$ will extend uniquely to a probability measure on $\sigma(\cF)$. + +We want to prove: +\begin{enumerate}[(1)] + \item $\sigma(\cF) = \cB_\infty$, + \item $\lambda$ as defined above is countably additive on $\F$. +\end{enumerate} +\begin{proof}[Proof of (1)] + Consider an infinite dimensional box $\prod_{n \in \N} B_n$. + We have + \[ + \left( \prod_{n=1}^N B_n \right)^\ast \in \cF_n \subseteq \cF + \] + thus + \[ + \prod_{n \in \N} B_n = \bigcap_{N \in \N} \left( \prod_{n=1}^N B_n \right)^\ast \in \sigma(\cF). + \] + Since $\sigma(\cF)$ is a $\sigma$-algebra, $\cB_\infty \subseteq \sigma(\cF)$. This proves ``$\supseteq$''. + For the other direction we'll show $\cF_n \subseteq \cB_\infty$ for all $n$. + Let $\cC \coloneqq \{ Q \in \cB_n | Q^\ast \in \cB_\infty\}$. + For $B_1,\ldots,B_n \in \cB$, $B_1 \times \ldots \times B_n \in \cB_n$ + and $(B_1 \times \ldots \times B_n)^\ast \in \cB_\infty$. + We have $B_1 \times \ldots \times B_n \in \cC$. + And $\cC$ is a $\sigma$-algebra, because: + \begin{itemize} + \item $\cB_n$ is a $\sigma$-algebra + \item $\cB_\infty$ is a $\sigma$-algebra, + \item $\phi^\ast \phi$, $(\R^n \setminus Q)^\ast = \R^{\infty} \setminus Q^\ast$, $\bigcup_{i \in I} Q_i^\ast = (\bigcup_{i \in I} Q_i)^\ast$. + \end{itemize} + Thus $\cC \subseteq \cB_n$ is a $\sigma$-algebra and contains all rectangles, hence $\cC = \cB_n$. + Hence $\cF_n \subseteq \cB_\infty$ for all $n$, + thus $\cF \subseteq \cB_\infty$. Since $\cB_\infty$ is a $\sigma$-algebra, + $\sigma(\cF) \subseteq \cB_\infty$. +\end{proof} +We are going to use the following +\begin{fact} + \label{fact:finaddtocountadd} + Suppose $\cA$ is an algebra on $\Omega \neq \emptyset$, + and suppose $\bP: \cA \to [0,1]$ is a finitely additive + probability measure. + Suppose whenever $\{B_n\}_n$ is a sequence of sets from $\cA $ + decreasing to $\emptyset$ it is the case that + $\bP(B_n) \to 0$. Then $\bP$ must be countably additive. +\end{fact} +\begin{proof} + Exercise + +\end{proof} +\begin{proof}[Proof of (2)] + Let's prove that $\lambda$ is finitely additive. + $\lambda(\R^\infty) = \lambda_1(\R^\infty) = 1$. + $\lambda(\emptyset) = \lambda_1(\emptyset) = 0$. + Suppose $A_1, A_2 \in \cF$ are disjoint. + Then pick some $n$ such that $A_1, A_2 \in \cF_n$. + Take $C_1, C_2 \in \cB_n$ such that $C_1^\ast = A_1$ + and $C_2^\ast = A_2$. + Then $C_1$ and $C_2$ are disjoint and $A_1 \cup A_2 = (C_1 \cup C_2)^\ast$. + $\lambda(A_1 \cup A_2) = \lambda_n(A_1 \cup A_2) = (\mu_1 \otimes \ldots \otimes \mu_n)(C_1 \cup C_2) = \lambda_n(C_1) + \lambda_n(C_2)$ + by the definition of the finite product measure. + + In order to use \autoref{fact:finaddtocountadd}, + we need to show that if $B_n \in \cF$ with $B_n \to \emptyset \implies \lambda(B_n) \to 0$. + %TODO +\end{proof} + diff --git a/inputs/vl6.tex b/inputs/vl6.tex new file mode 100644 index 0000000..bd2acab --- /dev/null +++ b/inputs/vl6.tex @@ -0,0 +1,62 @@ +We want to show laws of large numbers: +The LHS is random and represents ``sane'' averaging. +The RHS is constant, which we can explicitly compute from the distribution of the RHS. + +We fix a probability space $(\Omega, \cF, \bP)$ once and for all. + +\begin{theorem} + \label{lln} + Let $X_1, X_2,\ldots$ be i.i.d.~random variables on $(\R, \cB(\R))$ + and $m = \bE[X_i] < \infty$ + and $\sigma^{2} = \Var(X_i) = \bE[ (X_i - \bE(X_i))^2] = \bE[X_i^2] - \bE[X_i]^2 < \infty$. + + Then + \begin{enumerate}[(a)] + \item $\frac{X_1 + \ldots + X_n}{n} \xrightarrow{n \to \infty} m$ + in probability (\vocab{weak law of large numbers}, WLLN), + \item $\frac{X_1 + \ldots + X_n}{n} \xrightarrow{n \to \infty} m$ + almost surely (\vocab{strong law of large numbers}, SLLN). + \end{enumerate} +\end{theorem} +\begin{refproof}{lln} + \begin{enumerate}[(a)] + \item Given $\epsilon > 0$, we need to show that + \[ + \bP\left[ \left| \frac{X_1 + \ldots + X_n}{n}\right| > \epsilon\right] \to 0 \] + as $n \to 0$. + + Let $S_n \coloneqq X_1 + \ldots + X_n$. + Then $\bE[S_n] = \bE[X_1] + \ldots + \bE[X_n] = nm$. + We have + \begin{IEEEeqnarray*}{rCl} + \bP\left[ \left| \frac{X_1 + \ldots + X_n}{n}\right| > \epsilon\right] &=& \bP\left[\left|\frac{S_n}{n}-m\right| > \epsilon\right]\\ + &\overset{\text{Chebyshev}}{\le }& \frac{\Var\left( \frac{S_n}{n} \right) }{\epsilon^2} = \frac{1}{n} \frac{\Var(X_1)}{\epsilon^2} \xrightarrow{n \to \infty} 0 + \end{IEEEeqnarray*} + since + \[\Var(\frac{S_n}{n}) = \frac{1}{n^2} \Var(S_n) = \frac{1}{n^2} n \Var(X_i).\] + \end{enumerate} +\end{refproof} +For the proof of (b) we need the following general result: +\begin{theorem} + \label{thm2} + Let $X_1, X_2, \ldots$ be independent (but not necessarily identically distributed) random variables with $\bE[X_i] = 0$ for all $i$ + and $\sum_{i=1}^n \Var(X_i) < \infty$. + Then $\sum_{n \ge 1} X_n$ converges almost surely. +\end{theorem} + +\begin{proof} + + +\end{proof} +\begin{question} + Does the converse hold? I.e.~does $\sum_{n \ge 1} X_n < \infty$ a.s.~ + then $\sum_{n \ge 1} \Var(X_n) < \infty$. +\end{question} +This does not hold. Consider for example $X_n = \frac{1}{n^2} \delta_n + \frac{1}{n^2} \delta_{-n} + (1-\frac{2}{n^2}) \delta_0$. + +\begin{refproof}{lln} + \begin{enumerate} + \item[(b)] + \end{enumerate} +\end{refproof} + diff --git a/inputs/vl7.tex b/inputs/vl7.tex new file mode 100644 index 0000000..b6f8b95 --- /dev/null +++ b/inputs/vl7.tex @@ -0,0 +1,97 @@ +\begin{refproof}{lln} + We want to deduce the SLLN (\autoref{lln}) from \autoref{thm2}. + W.l.o.g.~let us assume that $\bE[X_i] = 0$ (otherwise define $X'_i \coloneqq X_i - \bE[X_i]$). + We will show that $\frac{S_n}{n} \xrightarrow{a.s.} 0$. + Define $Y_i \coloneqq \frac{X_i}{i}$. + Then the $Y_i$ are independent and we have $\bE[Y_i] = 0$ + and $\Var(Y_i) = \frac{\sigma^2}{i^2}$. + Thus $\sum_{i=1}^\infty \Var(Y_i) < \infty$. + From \autoref{thm2} we obtain that $\sum_{i=1}^\infty Y_i < \infty$ a.s. + \begin{claim} + Let $(a_n)$ be a sequence in $\R$ such that $\sum_{n=1}^{\infty} \frac{a_n}{n}$, then $\frac{a_1 + \ldots + a_n}{n} \to 0$. + \end{claim} + \begin{subproof} + Let $S_m \coloneqq \sum_{n=1}^\infty \frac{a_n}{n}$. + By assumption, there exists $S \in \R$ + such that $S_m \to S$ as $m \to \infty$. + Note that $j \cdot (S_{j} - S_{j-1}) = a_j$. + Define $S_0 \coloneqq 0$. + Then $a_1 + \ldots + a_n = (S_1 - S_0) + 2(S_2 - S_1) + 3(S_3 - S_2) + + \ldots + n (S_n - S_{n-1})$. + Thus $a_1 + \ldots + a_n = n S_n - (S1 $ % TODO + + \end{subproof} + The claim implies SLLN. + +\end{refproof} + +We need the following inequality: +\begin{theorem}[Kolmogorov's inequality] + If $X_1,\ldots, X_n$ are independent with $\bE[X_i] = 0$ + and $\Var(X_i) = \sigma_i^2$, then + \[ + \bP\left[\max_{1 \le i \le n} \left| \sum_{j=1}^{i} X_j \right| > \epsilon \right] \le \frac{1}{\epsilon ^2} \sum_{i=1}^m \sigma_i^2 % TODO + \] +\end{theorem} +\begin{proof} + Let $A_1 \coloneqq \{\omega : |X_1(\omega)| > \epsilon\}, \ldots, + A_i := \{\omega: |X_1(\omega)| \le \epsilon, |X_1(\omega) + X_2(\omega)| \le \epsilon, \ldots, |X_1(\omega) + \ldots + X_{i-1}(\omega)| \le \epsilon, + |X_1(\omega) + \ldots + X_i(\omega)| > \epsilon\}$. + We are interested in $\bigcup_{1 \le i \le n} A_i$. + + We have + \begin{IEEEeqnarray*}{rCl} + \int_{A_i} (\underbrace{X_1 + \ldots + X_i}_C + \underbrace{X_{i+1} + \ldots + X_n}_D)^2 d \bP &=& \int_{A_i} C^2 d\bP + \underbrace{\int_{A_i} D^2 d \bP}_{\ge 0} + 2 \int_{A_i} CD d\bP\\ + &\ge & \int_{A_i} \underbrace{C^2}_{\ge \epsilon^2} d \bP + 2 \int \underbrace{\One_{A_i} (X_1 + \ldots + X_i)}_E \underbrace{(X_{i+1} + \ldots + X_n)}_D d \bP\\ + &\ge& \int_{A_i} \epsilon^2 d\bP + \end{IEEEeqnarray*} + (By the independence of $X_1,\ldots, X_n$ and therefore that of $E$ and $D$ and $\bE(X_{i+1}) = \ldots = \bE(X_n) = 0$ we have $\int D E d\bP = 0$.) + + % TODO + +\end{proof} + +\begin{refproof}{thm2} + % TODO + +\end{refproof} + + + +\paragraph{Application of SLLN} + +\begin{theorem}[Renewal theorem] + Let $X_1,X_2,\ldots$ i.i.d.~random variables with $X_i \ge 0$, $\bE[X_i] = m > 0$. The $X_i$ model waiting times. + Let $S_n \coloneqq \sum_{i=1}^n X_i$. + For all $t > 0$ let \[ + N_t \coloneqq \sup \{n : S_n \le t\}. + \] + Then $\frac{N_t}{t} \xrightarrow{a.s.} \frac{1}{m}$ as $t \to \infty$. +\end{theorem} + + The $X_i$ can be thought of as waiting times. + $S_i$ models how long you have to wait for $i$ events to occur. + +\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 $t \uparrow \infty$. + + \begin{claim} + $\bP[\frac{S_n}{n} \xrightarrow{n \to \infty} m , N_t \xrightarrow{t \to \infty} \infty] = 1$. + \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 $\ast\ast \implies \bP(B^C) = 0$. + \end{subproof} + + Equivalently, $\bP\left[ \frac{S_{N_t}}{N_t} \xrightarrow{t \to \infty} m, \frac{S_{N_t + 1}}{N_t + 1} \xrightarrow{t \to \infty} m \right] = 1$. + + By definition, we have $S_{N_t} \le t \le S_{N_t + t}$. + Then $\frac{S_{N_t}}{N_t} \le \frac{t}{N_t} \le S_{N_t + 1}{N_t} \le \frac{S_{N_t + 1}}{N_t + 1} \cdot \frac{N_t + 1}{N_t}$. + Hence $\frac{t}{N_t} \to m$. + +\end{proof} + + + diff --git a/inputs/vl8.tex b/inputs/vl8.tex new file mode 100644 index 0000000..d714310 --- /dev/null +++ b/inputs/vl8.tex @@ -0,0 +1,91 @@ +% TODO \begin{goal} +% TODO We want to drop our assumptions on finite mean or variance +% TODO and say something about the behaviour of $ \sum_{n \ge 1} X_n$ +% TODO when the $X_n$ are independent. +% TODO \end{goal} +\begin{theorem}[Theorem 3, Kolmogorov's three-series theorem] % Theorem 3 + \label{thm3} + Let $X_n$ be a family of independent random variables. + \begin{enumerate}[(a)] + \item Suppose for some $C \ge 0$, the following three series + of numbers converge: + \begin{itemize} + \item $\sum_{n \ge 1} \bP(|X_n| > C)$, + \item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n d\bP}_{\text{\vocab{truncated mean}}}$, + \item $\sum_{n \ge 1} \underbrace{\int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2}_{\text{\vocab{truncated variance} }}$. + \end{itemize} + Then $\sum_{n \ge 1} X_n$ converges almost surely. + \item Suppose $\sum_{n \ge 1} X_n$ converges almost surely. + Then all three series above converge for every $C > 0$. + \end{enumerate} +\end{theorem} +For the proof we'll need a slight generalization of \autoref{thm2}: +\begin{theorem}[Theorem 4] % Theorem 4 + \label{thm4} + Let $\{X_n\}_n$ be independent and \vocab{uniformly bounded} + (i.e. $\exists M < \infty : \sup_n \sup_\omega |X_n(\omega)| \le M$). + Then $\sum_{n \ge 1} X_n$ converges almost surely + $\iff$ $\sum_{n \ge 1} \bE(X_n)$ and $\sum_{n \ge 1} \Var(X_n)$ + converge. +\end{theorem} +\begin{refproof}{thm3} + Assume, that we have already proved \autoref{thm4}. + We prove part (a) first. + Put $Y_n = X_n \cdot \One_{\{|X_n| \le C\}}$. + Since the $X_n$ are independent, the $Y_n$ are independent as well. + Furthermore, the $Y_n$ are uniformly bounded. + By our assumption, the series + $\sum_{n \ge 1} \int_{|X_n| \le C} X_n d\bP = \sum_{n \ge 1} \bE[Y_n]$ + and $\sum_{n \ge 1} \int_{|X_n| \le C} X_n^2 d\bP - \left( \int_{|X_n| \le C} X_n d\bP \right)^2 = \sum_{n \ge 1} \Var(Y_n)$ + converges. + By \autoref{thm4} it follows that $\sum_{n \ge 1} Y_n < \infty$ + almost surely. + Let $A_n \coloneqq \{\omega : |X_n(\omega)| > C\}$. + Since the first series $\sum_{n \ge 1} \bP(A_n) < \infty$, + by Borel-Cantelli, $\bP[\text{infinitely many $A_n$ occcur}] = 0$. + + + For the proof of (b), suppose $\sum_{n\ge 1} X_n(\omega) < \infty$ + for almost every $\omega$. + Fix an arbitrary $C > 0$. + Define + \[ + Y_n(\omega) \coloneqq \begin{cases} + X_n(\omega) & \text{if} |X_n(\omega)| \le C,\\ + C &\text{if } |X_n(\omega)| > C. + \end{cases} + \] + Then the $Y_n$ are independent and $\sum_{n \ge 1} Y_n(\omega) < \infty$ + almost surely and the $Y_n$ are uniformly bounded. + By \autoref{thm4} $\sum_{n \ge 1} \bE[Y_n]$ and $\sum_{n \ge 1} \Var(Y_n)$ + converge. + Define + \[ + Z_n(\omega) \coloneqq \begin{cases} + X_n(\omega) &\text{if } |X_n| \le C,\\ + -C &\text{if } |X_n| > C. + \end{cases} + \] + Then the $Z_n$ are independent, uniformly bounded and $\sum_{n \ge 1} Z_n(\omega) < \infty$ + almost surely. + By \autoref{thm4} we have + $\sums_{n \ge 1} \bE(Z_n) < \infty$ + and $\sums_{n \ge 1} \Var(Z_n) < \infty$. + + We have + \[ + \bE(Y_n) &=& \int_{|X_n| \le C} X_n d \bP + C \bP(|X_n| \ge C)\\ + \bE(Z_n) &=& \int_{|X_n| \le C} X_n d \bP - C \bP(|X_n| \ge C)\\ + \] + Since $\bE(Y_n) + \bE(Z_n) = 2 \int_{|X_n| \le C} X_n d\bP$ + the second series converges, + and since + $\bE(Y_n) - \bE(Z_n)$ converges, the first series converges. + For the third series, we look at + $\sum_{n \ge 1} \Var(Y_n)$ and + $\sum_{n \ge 1} \Var(Z_n)$ to conclude that this series converges + as well. +\end{refproof} + + +