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Conditional Properties of Probability 📂Probability Theory

Conditional Properties of Probability

Theorem

Let’s say we have a probability space (Ω,F,P)( \Omega , \mathcal{F} , P) and a sub-sigma field G⊂F\mathcal{G} \subset \mathcal{F}.

  • [1] For all B∈GB \in \mathcal{G}, there is 0≤P(B∣G)≤10 \le P(B | \mathcal{G}) \le 1.
  • [2] Continuity of probability: For a nested sequence {Bn}n∈N⊂G\left\{ B_{n} \right\}_{n \in \mathbb{N}} \subset \mathcal{G}, lim⁡n→∞Bn=B  ⟹  P(Bn∣G)→P(B∣G) a.s. \lim_{n \to \infty} B_{n} = B \implies P ( B_{n} | \mathcal{G} ) \to P ( B | \mathcal{G} ) \text{ a.s.}
  • [3] If {Bn}n∈N\left\{ B_{n} \right\}_{n \in \mathbb{N}} is a partition of Ω\Omega, P(⨆n∈NBn∣G)=∑n∈NP(Bn∣G) P \left( \bigsqcup_{n \in \mathbb{N}} B_{n} | \mathcal{G} \right)= \sum_{n \in \mathbb{N}} P \left( B_{n} | \mathcal{G} \right)
  • That a sequence of events {Bn}n∈N⊂G\left\{ B_{n} \right\}_{n \in \mathbb{N}} \subset \mathcal{G} is nested means it has one of the following two properties. ∀n∈N,Bn⊂Bn+1  ⟺  Bn⊂Bn+1⊂⋯∀n∈N,Bn⊂Bn−1  ⟺  Bn⊂Bn−1⊂⋯ \forall n \in \mathbb{N}, B_{n} \subset B_{n+1} \iff B_{n} \subset B_{n+1} \subset \cdots \\ \forall n \in \mathbb{N}, B_{n} \subset B_{n-1} \iff B_{n} \subset B_{n-1} \subset \cdots
  • A nested sequence may have the following properties for some event B∈GB \in \mathcal{G}. ∀n∈N,Bn⊂Bn+1∧⋃n∈NBn=B  ⟹  lim⁡n→∞Bn=B∀n∈N,Bn⊂Bn−1∧⋂n∈NBn=B  ⟹  lim⁡n→∞Bn=B \forall n \in \mathbb{N}, B_{n} \subset B_{n+1} \land \bigcup_{n \in \mathbb{N}} B_{n} = B \implies \lim_{n \to \infty} B_{n} = B \\ \forall n \in \mathbb{N}, B_{n} \subset B_{n-1} \land \bigcap_{n \in \mathbb{N}} B_{n} = B \implies \lim_{n \to \infty} B_{n} = B
  • ⨆\bigsqcup denotes the symbol for the union of mutually exclusive sets.

Proof

[1]

Since PP is a probability, according to the definition of conditional probability and conditional expectation, for all A∈GA \in \mathcal{G} ∫A0dP≤∫AP(B∣G)dP=∫AE(1B∣G)dP=∫A1BdP≤∫A1dP \begin{align*} \int_{A} 0 dP \le & \int_{A} P(B | \mathcal{G}) dP \\ =& \int_{A} E ( \mathbb{1}_{B} | \mathcal{G} ) dP \\ =& \int_{A} \mathbb{1}_{B} dP \\ \le & \int_{A} 1 dP \end{align*} ∀A∈F,∫Afdm=0  ⟺  f=0 a.e.\displaystyle \forall A \in \mathcal{F}, \int_{A} f dm = 0 \iff f = 0 \text{ a.e.} hence 0≤P(B∣G)≤10 \le P(B | \mathcal{G}) \le 1

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[2]

We only need to show that it holds when ∀n∈N,Bn⊂Bn+1\forall n \in \mathbb{N}, B_{n} \subset B_{n+1}, by setting it as Bn:=Ω∖AnB_{n} := \Omega \setminus A_{n}, we can also show it holds when ∀n∈N,An⊂An−1\forall n \in \mathbb{N}, A_{n} \subset A_{n-1}. Assuming ∀n∈N,Bn⊂Bn+1\forall n \in \mathbb{N}, B_{n} \subset B_{n+1}, according to the definition of conditional probability and the Monotone Convergence Theorem for conditional probability, lim⁡n→NP(Bn∣G)=lim⁡n→∞E(1Bn∣G)=E(lim⁡n→∞1Bn∣G)=E(1B∣G)=P(B∣G) \begin{align*} \lim_{n \to \mathbb{N}} P(B_{n} | \mathcal{G}) \color{red}{=}& \lim_{n \to \infty} E ( \mathbb{1}_{B_{n}} | \mathcal{G} ) \\ \color{blue}{=}& E \left( \lim_{n \to \infty} \mathbb{1}_{B_{n}} | \mathcal{G} \right) \\ =& E \left( \mathbb{1}_{B} | \mathcal{G} \right) \\ \color{red}{=}& P(B | \mathcal{G} ) \end{align*}

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[3]

If {Bn}n∈N\left\{ B_{n} \right\}_{n \in \mathbb{N}} is a partition of Ω\Omega, then for all n∈Nn \in \mathbb{N}, ⨆k=1nBk⊂⨆k=1n+1Bk\displaystyle \bigsqcup_{k=1}^{n} B_{k} \subset \bigsqcup_{k=1}^{n+1} B_{k} hence, according to the continuity of probability in [2], P(⨆n=1∞Bn∣G)=P(lim⁡n→∞⨆k=1nBk∣G)=lim⁡n→∞P(⨆k=1nBk∣G)=lim⁡n→∞E(1⨆k=1nBk∣G)=lim⁡n→∞∑k=1nE(1Bk∣G)=lim⁡n→∞∑k=1nP(Bk∣G)=∑n=1∞P(Bn∣G) \begin{align*} P \left( \bigsqcup_{n=1}^{\infty} B_{n} | \mathcal{G} \right) =& P \left( \lim_{n \to \infty} \bigsqcup_{k=1}^{n} B_{k} | \mathcal{G} \right) \\ =& \lim_{n \to \infty} P \left( \bigsqcup_{k=1}^{n} B_{k} | \mathcal{G} \right) \\ =& \lim_{n \to \infty} E \left( \mathbb{1}_{\bigsqcup_{k=1}^{n} B_{k}} | \mathcal{G} \right) \\ =& \lim_{n \to \infty} \sum_{k=1}^{n} E \left( \mathbb{1}_{B_{k}} | \mathcal{G} \right) \\ =& \lim_{n \to \infty} \sum_{k=1}^{n} P \left( B_{k} | \mathcal{G} \right) \\ =& \sum_{n=1}^{\infty} P \left( B_{n} | \mathcal{G} \right) \end{align*}

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