First part of lecture 8
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23
inputs/prerequisites.tex
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inputs/prerequisites.tex
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\begin{theorem}[Chebyshev's inequality] % TODO Proof
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Let $X$ be a r.v.~with $\Var(x) < \infty$.
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Then $\forall \epsilon > 0 : \bP \left[ \left| X - \bE[X] \right| > \epsilon\right] \le \frac{\Var(x)}{\epsilon^2}$.
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\end{theorem}
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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 $X_i$.
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How do we prove that something happens almost surely?
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\begin{lemma}[Borel-Cantelli]
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If we have a sequence of events $(A_n)_{n \ge 1}$
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such that $\sum_{n \ge 1} \bP(A_n) < \infty$,
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then $\bP[ A_n \text{for infinitely many $n$}] = 0$
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(more precisely: $\bP[\limsup_{n \to \infty} A_n] = 0$).
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The converse also holds for independent events $A_n$.
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\end{lemma}
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Modes of covergence: $L^p$, in probability, a.s.
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117
inputs/vl3.tex
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inputs/vl3.tex
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$(\Omega, \cF, \bP)$ Probability Space, $X : ( \Omega, \cF) \to (\R, \cB(\R))$ random variable.
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Then $\Q(\cdot) = \bP [ x\in \cdot ]$ is the distribution of $X$ under $\bP$.
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\section{Independence and product measures}
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In order to define the notion of independence, we first need to construct
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product measures in order to be able to consider several random variables
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at the same time.
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The finite case of a product is straightforward:
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\begin{theorem}{Product measure (finite)}
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Let $(\Omega_1, \cF, \bP)$ and $(\Omega_2, \cF_2, \bP_2)$ be probability spaces.
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Let $\Omega \coloneqq \Omega_1 \times \Omega_2$
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and $R \coloneqq \{A_1 \times A_2 | A_1 \in \cF_1, A_2 \in \cF_2 \}$.
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Let $\cF$ be $\sigma(R)$ (the sigma algebra generated by $R$).
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Then there exists a unique probability measure $\bP$ on $\Omega$
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such that for every rectangle $R = A_1 \times A_2 \in \cR$, $\bP(A_1 \times A_2) = \bP(A_1) \times \bP(A_2)$.
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\end{theorem}
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\begin{proof}
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See Theorem 5.1.1 in the lecture notes on Stochastik.
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\end{proof}
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We now want to construct a product measure for infinite products.
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\begin{definition}[Independence]
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A collection $X_1, X_2, \ldots, X_n$ of random variables are called
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\vocab{mutually independent} if
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\[
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\forall a_1,\ldots,a_n \in \R :
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\bP[X_1 \le a_1, \ldots, x_n \le a_n]
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= \prod_{i=1}^n \bP[X_i \le a_i]
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\]
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This is equivalent to
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\[
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\forall B_1, \ldots, B_n \in \cB(\R):
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\bP[X_1 \in B_1, \ldots, X_n \in B_n]
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= \prod_{i=1}^n \bP[X_i \in B_i]
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\]
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\end{definition}
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\begin{example}
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Suppose we throw a dice twice. Let $A \coloneqq \{\text{first throw even}\}$,
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$B \coloneqq \{second throw even\}$
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and $C \coloneqq \{\text{sum even}\} $.
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Are $\One_A, \One_B, \One_C$ mutually independent random variables?
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\end{example}
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It is easy the see, that the random variables are pairwise independent,
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but not mutually independent.
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The definition of mutual independence can be rephrased as follos:
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Let $X_1, X_2, \ldots, X_n$ r.v.s. Let $\bP[(X_1,\ldots, X_n) \in \cdot ] \text{\reflectbox{$\coloneqq$}} \Q^{\otimes}(\cdot )$.
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Then $\Q^{\otimes}$ is a probability measure on $\R^n$.
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\begin{fact}
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$X_1,\ldots, X_n$ are mutually independent iff $\Q^{\otimes} = \Q_1 \otimes \ldots \otimes \Q_n$.
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\end{fact}
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By constructing an infinite product, we can thus extend the notion of independence
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to an infinite number of r.v.s.
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\begin{goal}
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Can we construct infinitely many independent random variables?
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\end{goal}
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\begin{definition}[Consistent family of random variables]
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Let $\bP_n, n \in \N$ be a family of probability measures on $(\R^n, \cB(\R^n))$.
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The family is called \vocab{consistent} if if
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$\bP_{n+1}[B_1 \times B_2 \times \ldots \times B_n \times \R] = \bP_n[B_1 \times \ldots \times B_n]$
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for all $n \in \N, B_i \in B(\R)$.
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\end{definition}
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\begin{theorem}[Kolmogorov extension / consistency theorem]
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Informally:
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``Probability measures are determined by finite-dimensional marginals
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(as long as these marginals are nice)''
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Let $\bP_n, n \in \N$ be probability measures on $(\R^n, \cB(\R^n))$
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which are \vocab{consistent},
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then there exists a unique probability measure $\bP^{\otimes}$
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on $(\R^\infty, B(R^\infty))$ (where $B(R^{\infty}$ has to be defined),
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such that
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\[
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\forall n \in \N, B_1,\ldots, B_n \in B(\R):
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\bP^\otimes [\cX : X_i \in B_i \forall 1 \le i \le n]
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= \bP_n[B_1 \times \ldots \times B_n]
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\]
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\end{theorem}
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\begin{remark}
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Kolmogorov's theorem can be strengthened to the case of arbitrary
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index sets. However this requires a different notion of consistency.
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\end{remark}
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\begin{example}of a consistent family:
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Let $F_1, \ldots, F_n$ be probability distribution functions
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and let $\bP_n$ be the probability measure on $\R^n$ defined
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by
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\[
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\bP_n[(a_1,b_1] \times \ldots (a_n, b_n]]
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\coloneqq (F_1(b_1) - F_1(a_1)) \cdot \ldots \cdot (F_n(b_n) - F_n(a_n)).
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\]
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It is easy to see that each $\bP_n$ is a probability measure.
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Define $X_i(\omega) = \omega_i$ where $\omega = (\omega_1, .., \omega_n)$.
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Then $X_1, \ldots, X_n$ are mutually independent with $F_i$ being
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the distribution function of $X_i$.
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In the case of $F_1 = \ldots = F_n$, then $X_1,\ldots, X_n$ are i.i.d.
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\end{example}
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inputs/vl4.tex
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inputs/vl4.tex
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\begin{notation}
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Let $\cB_n$ denote $\cB(\R^n)$.
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\end{notation}
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\begin{goal}
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Suppose we have a probability measure $\mu_n$ on $(\R^n, \cB(\R^n))$
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for each $n$.
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We want to show that there exists a unique probability measure $\bP^{\otimes}$
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on $(\R^\infty, \cB_\infty)$ (where $\cB_{\infty}$ still needs to be defined),
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such that $\bP^{\otimes}\left( \prod_{n \in \N} B_n \right) = \prod_{n \in \N} \mu_n(B_n)$
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for all $\{B_n\}_{n \in \N}$, $B_n \in \cB_1$.
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\end{goal}
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% $\bP_n = \mu_1 \otimes \ldots \otimes \mu_n$.
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\begin{remark}
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$\prod_{n \in \N} \mu_n(B_n)$ converges, since $0 \le \mu_n(B_n) \le 1$
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for all $n$.
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\end{remark}
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First we need to define $\cB_{\infty}$.
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This $\sigma$-algebra must contain all sets $\prod_{n \in \N} B_n$
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for all $B_n \in \cB_1$. We simply define $\cB_{\infty}$ to be the
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$\sigma$-algebra
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Let $\cB_\infty \coloneqq \sigma \left( \{\prod_{n \in \N} B_n | B_n \in \cB(\R)\} \right)$.
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\begin{question}
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What is there in $\cB_\infty$?
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Can we identify sets in $\cB_\infty$ for which we can define the product measure
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easily?
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\end{question}
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Let $\cF_n \coloneqq \{ C \times \R^{\infty} | C \in \cB_n\}$.
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It is easy to see that $\cF_n \subseteq \cF_{n+1}$
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and using that $\cB_n$ is a $\sigma$-algebra, we can show that $\cF_n$
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is also a $\sigma$-algebra.
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Now, for any $C \subseteq \R^n$ let $C^\ast \coloneqq C \times \R^{\infty}$.
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Thus $\cF_n = \{C^\ast : C \in \cB_n\}$.
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Define $\lambda_n : \cF_n : \to [0,1]$ by $\lambda_n(C^\ast) \coloneqq (\mu_1 \otimes \ldots \otimes \mu_n)(C)$.
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It is easy to see that $\lambda_{n+1} \defon{\cF_n} = \lambda_n$ (\vocab{consistency}).
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Recall the following theorem from measure theory:
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\begin{theorem}[Caratheodory's extension theorem] % 2.3.3 in the notes
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\label{caratheodory}
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Suppose $\cA$ is an algebra (i.e.~closed under finite union)
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und $\Omega \neq \emptyset$.
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Suppose $\bP$ is countably additive on $\cA$ (i.e.~if $(A_n)_{n}$
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are pairwise disjoint and $\bigcup_{n \in \N} A_n \subseteq \cA $
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then $\bP\left( \bigcup_{n \in \N} A_n \right) = \sum_{n \in \N} \bP(A_n)$).
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Then $\bP$ extends uniquely to a probability measure on $(\Omega, \cF)$,
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where $\cF = \sigma(\cA)$.
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\end{theorem}
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Define $\cF = \bigcup_{n \in \N} \cF_n$. Check that $\cF$ is an algebra.
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We'll show that if we define $\lambda: \cF \to [0,1]$ with
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$\lambda(A) = \lambda_n(A)$ for any $n$ where this is well defined,
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then $\lambda$ is countably additive on $\cF$.
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Using \autoref{caratheodory} $\lambda$ will extend uniquely to a probability measure on $\sigma(\cF)$.
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We want to prove:
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\begin{enumerate}[(1)]
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\item $\sigma(\cF) = \cB_\infty$,
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\item $\lambda$ as defined above is countably additive on $\F$.
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\end{enumerate}
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\begin{proof}[Proof of (1)]
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Consider an infinite dimensional box $\prod_{n \in \N} B_n$.
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We have
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\[
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\left( \prod_{n=1}^N B_n \right)^\ast \in \cF_n \subseteq \cF
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\]
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thus
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\[
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\prod_{n \in \N} B_n = \bigcap_{N \in \N} \left( \prod_{n=1}^N B_n \right)^\ast \in \sigma(\cF).
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\]
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Since $\sigma(\cF)$ is a $\sigma$-algebra, $\cB_\infty \subseteq \sigma(\cF)$. This proves ``$\supseteq$''.
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For the other direction we'll show $\cF_n \subseteq \cB_\infty$ for all $n$.
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Let $\cC \coloneqq \{ Q \in \cB_n | Q^\ast \in \cB_\infty\}$.
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For $B_1,\ldots,B_n \in \cB$, $B_1 \times \ldots \times B_n \in \cB_n$
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and $(B_1 \times \ldots \times B_n)^\ast \in \cB_\infty$.
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We have $B_1 \times \ldots \times B_n \in \cC$.
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And $\cC$ is a $\sigma$-algebra, because:
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\begin{itemize}
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\item $\cB_n$ is a $\sigma$-algebra
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\item $\cB_\infty$ is a $\sigma$-algebra,
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\item $\phi^\ast \phi$, $(\R^n \setminus Q)^\ast = \R^{\infty} \setminus Q^\ast$, $\bigcup_{i \in I} Q_i^\ast = (\bigcup_{i \in I} Q_i)^\ast$.
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\end{itemize}
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Thus $\cC \subseteq \cB_n$ is a $\sigma$-algebra and contains all rectangles, hence $\cC = \cB_n$.
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Hence $\cF_n \subseteq \cB_\infty$ for all $n$,
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thus $\cF \subseteq \cB_\infty$. Since $\cB_\infty$ is a $\sigma$-algebra,
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$\sigma(\cF) \subseteq \cB_\infty$.
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\end{proof}
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We are going to use the following
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\begin{fact}
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\label{fact:finaddtocountadd}
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Suppose $\cA$ is an algebra on $\Omega \neq \emptyset$,
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and suppose $\bP: \cA \to [0,1]$ is a finitely additive
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probability measure.
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Suppose whenever $\{B_n\}_n$ is a sequence of sets from $\cA $
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decreasing to $\emptyset$ it is the case that
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$\bP(B_n) \to 0$. Then $\bP$ must be countably additive.
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\end{fact}
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\begin{proof}
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Exercise
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\end{proof}
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\begin{proof}[Proof of (2)]
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Let's prove that $\lambda$ is finitely additive.
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$\lambda(\R^\infty) = \lambda_1(\R^\infty) = 1$.
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$\lambda(\emptyset) = \lambda_1(\emptyset) = 0$.
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Suppose $A_1, A_2 \in \cF$ are disjoint.
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Then pick some $n$ such that $A_1, A_2 \in \cF_n$.
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Take $C_1, C_2 \in \cB_n$ such that $C_1^\ast = A_1$
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and $C_2^\ast = A_2$.
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Then $C_1$ and $C_2$ are disjoint and $A_1 \cup A_2 = (C_1 \cup C_2)^\ast$.
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$\lambda(A_1 \cup A_2) = \lambda_n(A_1 \cup A_2) = (\mu_1 \otimes \ldots \otimes \mu_n)(C_1 \cup C_2) = \lambda_n(C_1) + \lambda_n(C_2)$
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by the definition of the finite product measure.
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In order to use \autoref{fact:finaddtocountadd},
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we need to show that if $B_n \in \cF$ with $B_n \to \emptyset \implies \lambda(B_n) \to 0$.
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%TODO
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\end{proof}
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inputs/vl6.tex
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inputs/vl6.tex
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We want to show laws of large numbers:
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The LHS is random and represents ``sane'' averaging.
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The RHS is constant, which we can explicitly compute from the distribution of the RHS.
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We fix a probability space $(\Omega, \cF, \bP)$ once and for all.
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\begin{theorem}
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\label{lln}
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Let $X_1, X_2,\ldots$ be i.i.d.~random variables on $(\R, \cB(\R))$
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and $m = \bE[X_i] < \infty$
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and $\sigma^{2} = \Var(X_i) = \bE[ (X_i - \bE(X_i))^2] = \bE[X_i^2] - \bE[X_i]^2 < \infty$.
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Then
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\begin{enumerate}[(a)]
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\item $\frac{X_1 + \ldots + X_n}{n} \xrightarrow{n \to \infty} m$
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in probability (\vocab{weak law of large numbers}, WLLN),
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\item $\frac{X_1 + \ldots + X_n}{n} \xrightarrow{n \to \infty} m$
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almost surely (\vocab{strong law of large numbers}, SLLN).
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\end{enumerate}
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\end{theorem}
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\begin{refproof}{lln}
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\begin{enumerate}[(a)]
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\item Given $\epsilon > 0$, we need to show that
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\[
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\bP\left[ \left| \frac{X_1 + \ldots + X_n}{n}\right| > \epsilon\right] \to 0 \]
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as $n \to 0$.
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Let $S_n \coloneqq X_1 + \ldots + X_n$.
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Then $\bE[S_n] = \bE[X_1] + \ldots + \bE[X_n] = nm$.
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We have
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\begin{IEEEeqnarray*}{rCl}
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\bP\left[ \left| \frac{X_1 + \ldots + X_n}{n}\right| > \epsilon\right] &=& \bP\left[\left|\frac{S_n}{n}-m\right| > \epsilon\right]\\
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&\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
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\end{IEEEeqnarray*}
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since
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\[\Var(\frac{S_n}{n}) = \frac{1}{n^2} \Var(S_n) = \frac{1}{n^2} n \Var(X_i).\]
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\end{enumerate}
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\end{refproof}
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For the proof of (b) we need the following general result:
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\begin{theorem}
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\label{thm2}
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Let $X_1, X_2, \ldots$ be independent (but not necessarily identically distributed) random variables with $\bE[X_i] = 0$ for all $i$
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and $\sum_{i=1}^n \Var(X_i) < \infty$.
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Then $\sum_{n \ge 1} X_n$ converges almost surely.
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\end{theorem}
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\begin{proof}
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\end{proof}
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\begin{question}
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Does the converse hold? I.e.~does $\sum_{n \ge 1} X_n < \infty$ a.s.~
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then $\sum_{n \ge 1} \Var(X_n) < \infty$.
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\end{question}
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This does not hold. Consider for example $X_n = \frac{1}{n^2} \delta_n + \frac{1}{n^2} \delta_{-n} + (1-\frac{2}{n^2}) \delta_0$.
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\begin{refproof}{lln}
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\begin{enumerate}
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\item[(b)]
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\end{enumerate}
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\end{refproof}
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inputs/vl7.tex
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\begin{refproof}{lln}
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We want to deduce the SLLN (\autoref{lln}) from \autoref{thm2}.
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W.l.o.g.~let us assume that $\bE[X_i] = 0$ (otherwise define $X'_i \coloneqq X_i - \bE[X_i]$).
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We will show that $\frac{S_n}{n} \xrightarrow{a.s.} 0$.
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Define $Y_i \coloneqq \frac{X_i}{i}$.
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Then the $Y_i$ are independent and we have $\bE[Y_i] = 0$
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and $\Var(Y_i) = \frac{\sigma^2}{i^2}$.
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Thus $\sum_{i=1}^\infty \Var(Y_i) < \infty$.
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From \autoref{thm2} we obtain that $\sum_{i=1}^\infty Y_i < \infty$ a.s.
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\begin{claim}
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Let $(a_n)$ be a sequence in $\R$ such that $\sum_{n=1}^{\infty} \frac{a_n}{n}$, then $\frac{a_1 + \ldots + a_n}{n} \to 0$.
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\end{claim}
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\begin{subproof}
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Let $S_m \coloneqq \sum_{n=1}^\infty \frac{a_n}{n}$.
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By assumption, there exists $S \in \R$
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such that $S_m \to S$ as $m \to \infty$.
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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}
|
||||
|
||||
|
||||
|
||||
91
inputs/vl8.tex
Normal file
91
inputs/vl8.tex
Normal file
|
|
@ -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}
|
||||
|
||||
|
||||
|
||||
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Reference in a new issue