ThmDex – An index of mathematical definitions, results, and conjectures.
Result R3954 on D3357: Identically distributed random collection
Two almost surely equal random variables are identically distributed
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) \begin{equation} \mathbb{P}(X = Y) = 1 \end{equation}
Then \begin{equation} X \overset{d}{=} 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) \begin{equation} \mathbb{P}(X = Y) = 1 \end{equation}
(iii) $\{ X \in E \}, \{ Y \in E \} \in \mathcal{F}$ are each an D1716: Event in $P$
Then \begin{equation} \mathbb{P}(X \in E) = \mathbb{P}(Y \in E) \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 $P$
Then \begin{equation} X \overset{a.s.}{=} Y \quad \implies \quad X \overset{d}{=} Y \end{equation}
Remarks
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Remark 0
The converse statement is not true, as is shown in R3960: Identically distributed random variables need not be almost surely equal.
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) \begin{equation} \mathbb{P}(X = Y) = 1 \end{equation}
Let $E \in \mathcal{F}_{\Xi}$ be an event in $\Xi$. Since $\mathbb{P}(X = Y) = 1$ and since $\{ X \neq Y \} = \{ X = Y \}^{\complement} = \Omega \setminus \{ X = Y \}$, then result R3719: Probability of complement event implies \begin{equation} \mathbb{P}(X \neq Y) = 1 - \mathbb{P}(X = Y) = 1 - 1 = 0 \end{equation} Thus, applying results
(i) R2060: Probability of set difference
(ii) R3958: Intersecting subtrahend with minuend preserves set difference
(iii) R3934: Set intersection distributes over set difference
(iv) R4637:

yields \begin{equation} \begin{split} \mathbb{P}(X \in E) & = \mathbb{P}(X \in E) - 0 \\ & = \mathbb{P}(X \in E) - \mathbb{P}(X \neq Y) \\ & = \mathbb{P}( \{ X \in E \} \setminus \{ X \neq Y \} ) \\ & = \mathbb{P}[ \{ X \in E \} \setminus (\{ X \in E \} \cap \{ X \neq Y \}) ] \\ & = \mathbb{P}[ (\Omega \cap \{ X \in E \}) \setminus (\{ X \in E \} \cap \{ X \neq Y \}) ] \\ & = \mathbb{P}[ \{ X \in E \} \cap (\Omega \setminus \{ X \neq Y \}) ] \\ & = \mathbb{P}( \{ X \in E \} \cap \{ X = Y \} ) \\ & = \mathbb{P}( \{ Y \in E \} \cap \{ X = Y \} ) \\ & = \mathbb{P}( \{ Y \in E \} \cap \Omega \setminus \{ X \neq Y \} ) \\ & = \mathbb{P}[ (\Omega \cap \{ Y \in E \}) \setminus (\{ Y \in E \} \cap \{ X \neq Y \}) ] \\ & = \mathbb{P}[ \{ Y \in E \} \setminus (\{ Y \in E \} \cap \{ X \neq Y \}) ] \\ & = \mathbb{P}( \{ Y \in E \} \setminus \{ X \neq Y \} ) \\ & = \mathbb{P}(Y \in E) - \mathbb{P}(X \neq Y) \\ & = \mathbb{P}(Y \in E) - 0 \\ & = \mathbb{P}(Y \in E) \end{split} \end{equation} We may also opt to write the above in the equivalent form \begin{equation} \begin{split} \mathbb{P}(X \in E) & = \mathbb{P}(X \in E) - 0 \\ & = \mathbb{P}(X \in E) - \mathbb{P}(X \neq Y) \\ & = \mathbb{P}(X \in E \text{ and not } X \neq Y) \\ & = \mathbb{P}(X \in E \text{ and } X = Y) \\ & = \mathbb{P}(Y \in E \text{ and } X = Y) \\ & = \mathbb{P}(Y \in E \text{ and not } X \neq Y) \\ & = \mathbb{P}(Y \in E) - \mathbb{P}(X \neq Y) \\ & = \mathbb{P}(Y \in E) - 0 \\ & = \mathbb{P}(Y \in E) \end{split} \end{equation} Since $E \in \mathcal{F}_{\Xi}$ was arbitrary, we may conclude $X \overset{d}{=} Y$. $\square$