ThmDex – An index of mathematical definitions, results, and conjectures.

Result R4637 on D1716: Event

Formulation 0

Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that

(i)	$X, Y : \Omega \to \Xi$ are each a D202: Random variable on $P$
(ii)	$\{ X \in E \}, \{ Y \in F \} \in \mathcal{F}$ are each an D1716: Event in $P$

Then \begin{equation} \mathbb{P}(\{ X \in E \} \cap \{ X = Y \}) = \mathbb{P}(\{ Y \in E \} \cap \{ X = Y \}) \end{equation}

Formulation 1

Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that

(i)	$X, Y : \Omega \to \Xi$ are each a D202: Random variable on $P$
(ii)	$\{ X \in E \}, \{ Y \in F \} \in \mathcal{F}$ are each an D1716: Event in $P$

Then \begin{equation} \mathbb{P}( X \in E, X = Y ) = \mathbb{P}( Y \in E, X = Y ) \end{equation}

Formulation 2

Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that

(i)	$X, Y : \Omega \to \Xi$ are each a D202: Random variable on $P$
(ii)	$\{ X \in E \}, \{ Y \in F \} \in \mathcal{F}$ are each an D1716: Event in $P$

Then \begin{equation} \mathbb{P}( X \in E \text{ and } X = Y ) = \mathbb{P}( Y \in E \text{ and } X = Y ) \end{equation}

Proofs

Proof 0

Let $P = (\Omega, \mathcal{F}, \mathbb{P})$ be a D1159: Probability space such that

(i)	$X, Y : \Omega \to \Xi$ are each a D202: Random variable on $P$
(ii)	$\{ X \in E \}, \{ Y \in F \} \in \mathcal{F}$ are each an D1716: Event in $P$

We have \begin{equation} \begin{split} \mathbb{P}(\{ X \in E \} \cap \{ X = Y \}) & = \mathbb{P} \left( \{ \omega \in \Omega : X(\omega) \in E \} \cap \{ \omega \in \Omega : X(\omega) = Y(\omega) \} \right) \\ & = \mathbb{P} \left( \{ \omega \in \Omega : X(\omega) \in E \text{ and } X(\omega) = Y(\omega) \} \right) \\ & = \mathbb{P} \left( \{ \omega \in \Omega : X(\omega) = Y(\omega) \in E \text{ and } X(\omega) = Y(\omega) \} \right) \\ & = \mathbb{P} \left( \{ \omega \in \Omega : Y(\omega) \in E \text{ and } X(\omega) = Y(\omega) \} \right) \\ & = \mathbb{P} \left( \{ \omega \in \Omega : Y(\omega) \in E \} \cap \{ \omega \in \Omega : X(\omega) = Y(\omega) \} \right) \\ & = \mathbb{P}(\{ Y \in E \} \cap \{ X = Y \}) \end{split} \end{equation} $\square$