\lecture{2}{2023-04-11}{Independence, Kolmogorov's consistency theorem, consistent families}
\section{Independence and Product Measures}

In order to define the notion of independence, we first need to construct
product measures.

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 $A \coloneqq \{\text{first throw even}\}$,
    $B \coloneqq \{\text{second throw even}\}$
    and $C \coloneqq \{\text{sum even}\} $.

    It is easy the see, that the random variables are pairwise independent,
    but not mutually independent.
\end{example}

\begin{definition}
    Let $(\Omega, \cF, \bP)$ be a probability space
    and $X : ( \Omega, \cF) \to  (\R, \cB(\R))$ a random variable.
    Then $\Q(\cdot) \coloneqq \bP [ X \in \cdot ]$ is called the \vocab{distribution}
    of $X$ under $\bP$.
\end{definition}

Let $X_1, \ldots, X_n$ be random variables and $\Q^{\otimes}(\cdot ) \coloneqq \bP[(X_1,\ldots, X_n) \in \cdot ]$
their \vocab{joint distribution}.
Then $\Q^{\otimes}$ is a probability measure on $\R^n$.

The definition of mutual independence can be rephrased as follows:
\begin{fact}
    $X_1,\ldots, X_n$ are mutually independent iff $\Q^{\otimes} = \Q_1 \otimes \ldots \otimes \Q_n$,
    where $\Q_i$ is the distribution of $X_i$.
    In this setting, $\Q_i$ is called the \vocab{marginal distribution} of $X_i$.
\end{fact}
By constructing an infinite product, we can thus extend the notion of independence
to an infinite number of random variables.
\begin{goal}
    Can we construct infinitely many independent random variables?
\end{goal}

\begin{definition}[Consistent family of random variables]
    \label{def:consistentfamily}
    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
    $\bP_{n+1}[B_1 \times  B_2 \times  \ldots \times B_n \times  \R] = \bP_n[B_1 \times  \ldots \times  B_n]$
    for all $n \in \N, B_i \in  B(\R)$.

\end{definition}

\begin{theorem}[Kolmogorov extension / consistency theorem]
    \label{thm:kolmogorovconsistency}
    \footnote{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}[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}