8 Hewitt–Savage 0–1 Law, Symmetric Difference Basics, Conditioning
1. Basics of Symmetric Difference
Let $A, B, C$ be events. Recall the symmetric difference \(A \triangle B = (A \cap B^c) \cup (A^c \cap B).\)
(1) Triangle Property
\(P(A \triangle B) \le P(A \triangle C) + P(C \triangle B).\)
(2) Convergence in Symmetric Difference
If
\(P(A_n \triangle A) \to 0,\)
then
\(P(A_n) \to P(A)\)
and
\(P(A_n \cap A) \to P(A).\)
(3) Two sequences converging to the same event
If
\(P(A_n \triangle A) \to 0 \quad\text{and}\quad P(B_n \triangle A) \to 0,\)
then
\(P(A_n \cap B_n) \to P(A).\)
Proof sketch (from board):
$$ P(A_n \triangle A) = P(A_n) - P(A_n \cap A)
- P(A) - P(A_n \cap A) \to 0 $$ implies each term converges to the correct limit.
2. Hewitt–Savage 0–1 Law
Let ${X_k}_{k\ge 1}$ be iid random variables.
Let
\(A \in \sigma(X_1, X_2, \ldots)\)
be a permutable event:
Definition.
$A$ is permutable if
\(\pi(A) = A \quad\text{for every finite permutation }\pi.\)
Theorem (Hewitt–Savage 0–1 Law)
If $A$ is permutable, then
\(P(A) \in \{0,1\}.\)
Example
Take
\(A = \bigl\{ \limsup_{n\to\infty} S_n > 55 \bigr\},\)
where $S_n = \sum_{k=1}^n X_k$.
This event does not change if you permute finitely many $X_k$,
so $P(A) \in {0,1}$.
3. Proof Sketch of Hewitt–Savage (from lecture)
Choose approximations
\(A_n \in \sigma(X_1,\dots,X_n)
\quad\text{such that}\quad
P(A_n \triangle A) \to 0.\)
Since $A_n \to A$ in symmetric difference: \(P(A_n) \to P(A).\)
If $A$ is permutable, choose a finite permutation $\pi_n$ that moves
${1,\dots,n}$ to ${n+1,\dots,2n}$.
Then: \(P(\pi_n(A_n) \triangle A) = P(A_n \triangle A) \to 0.\)
Because $\pi_n(A_n)$ and $A_n$ depend on disjoint sets of iid variables, they are independent, and: \(P(\pi_n(A_n) \cap A_n) = P(\pi_n(A_n))P(A_n) = P(A_n)^2 \to P(A)^2.\)
But also, \(P(\pi_n(A_n) \cap A_n) \to P(A).\)
Thus, \(P(A) = P(A)^2 \quad\Rightarrow\quad P(A)\in\{0,1\}.\)
4. Beginning of Conditional Expectation (Durrett, Ch. 5)
We work on a probability space $(\Omega, \mathcal{F}_0, P)$
with $\mathcal{F} \subset \mathcal{F}_0$.
| Let $X$ be integrable, $E | X | <\infty$. |
Definition
A random variable $Y$ is the conditional expectation of $X$ given $\mathcal{F}$ if:
- $Y$ is $\mathcal{F}$-measurable.
- For all $A \in \mathcal{F}$, \(E[X \mathbf{1}_A] = E[Y \mathbf{1}_A].\)
Notation: $Y = E(X\mid \mathcal{F})$.
Useful Equivalent Form
For every bounded $\mathcal{F}$-measurable $Z$, \(E[XZ] = E[YZ].\)
5. Examples & Properties
Uniqueness (a.s.)
If $Y$ and $Y’$ both satisfy the definition, then
\(Y = Y' \quad\text{a.s.}\)
Example 1
If $X$ is already $\mathcal{F}$-measurable, then \(E(X\mid \mathcal{F}) = X.\)
Example 2
If $X$ is independent of $\mathcal{F}$, then \(E(X\mid \mathcal{F}) = E(X).\)
Proof sketch:
For any bounded $Z \in \mathcal{F}$,
\(E(XZ) = E(X)E(Z) = E(E(X)Z).\)
Thus $E(X)$ satisfies the defining property.
6. Radon–Nikodym Viewpoint
Given measure space $(\Omega,\mathcal{F}_0)$.
Let $\mu$ be sigma-finite and absolutely continuous w.r.t $P$:
\(\mu \ll P.\)
If $A \in \mathcal{F}_0$ and $P(A)=0$, then $\mu(A)=0$.
Density Example
If $(X,Y)$ has joint density $f_{X,Y}(x,y)$, then \(f_{Y\mid X=x}(y) = \frac{f_{X,Y}(x,y)}{f_X(x)}\) and \(E(Y \mid X=x) = \int_{-\infty}^{\infty} y\, f_{Y\mid X=x}(y)\,dy.\)
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