119 lines
4.6 KiB
TeX
119 lines
4.6 KiB
TeX
\lecture{2}{2023-04-11}{Independence, Kolmogorov's consistency theorem, consistent families}
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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.
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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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$B \coloneqq \{\text{second throw even}\}$
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and $C \coloneqq \{\text{sum even}\} $.
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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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\end{example}
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\begin{definition}
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Let $(\Omega, \cF, \bP)$ be a probability space
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and $X : ( \Omega, \cF) \to (\R, \cB(\R))$ a random variable.
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Then $\Q(\cdot) \coloneqq \bP [ X \in \cdot ]$ is called the \vocab{distribution}
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of $X$ under $\bP$.
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\end{definition}
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Let $X_1, \ldots, X_n$ be random variables and $\Q^{\otimes}(\cdot ) \coloneqq \bP[(X_1,\ldots, X_n) \in \cdot ]$
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their \vocab{joint distribution}.
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Then $\Q^{\otimes}$ is a probability measure on $\R^n$.
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The definition of mutual independence can be rephrased as follows:
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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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where $\Q_i$ is the distribution of $X_i$.
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In this setting, $\Q_i$ is called the \vocab{marginal distribution} of $X_i$.
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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 random variables.
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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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\label{def:consistentfamily}
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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
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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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\label{thm:kolmogorovconsistency}
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\footnote{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}[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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