If X and Y are equal almost surely, then they have the same distribution

Question

If $2$ r.v are equal a.s. can we write $\mathbb P((X\in B)\triangle (Y\in B))=0$

why is this so?

I get the equivalent statement $\mathbb P([(X\in B)\triangle (Y\in B)]^C)=1$ intuitively, but I don't see how to show rigorously that either follows from $P(\omega \in \Omega | X(\omega) - Y(\omega) \in \{ 0 \}) = 1$.

It wasn't me, but probably because this appears to be a duplicate of your other question, and takes careful examination to see that you're asking something slightly different. — Math1000, Sep 14 '15 at 20:02
@Math1000 Of which of my questions does this appear to be a duplicate? :| — BCLC, Sep 14 '15 at 20:04
Oh, you edited the question, so I saw your avatar, and the title is extremely similar. — Math1000, Sep 14 '15 at 20:06

score 5 · Accepted Answer · answered Sep 14 '15 at 19:54

5

Note that the inclusion $\{X \in B\} \Delta \{Y \in B\} \subset \{X \ne Y\}$ holds, which immediately yields the result.

To see the inclusion: Assume $\omega$ is in the set on the left-hand side; Wlog $\omega \in \{X \in B\}$. Then by the definition of the symmetric difference $\omega \notin \{Y \in B\}$, i.e. we have $X(\omega) \in B$ and $Y(\omega) \notin B$. But this can only be if $X(\omega) \ne Y(\omega)$.

answered Sep 14 '15 at 19:54

Dominik

20,241

Thanks Dominik! Contrapositive didn't seem to cross my mind – BCLC Sep 14 '15 at 19:56

score 0 · Answer 2 · answered May 15 '18 at 10:44

$$\{X=Y\} \subseteq [(X\in B)\triangle (Y\in B)]^C$$

Suppose $\omega \in \{X=Y\}$ but on the contrary that $\omega \in (X\in B)\triangle (Y\in B)$.

Then we have two cases by $X$: $X(\omega) \in B$ or $X(\omega) \notin B$

If $X(\omega) \in B$, then $\omega$ belongs to $X \in B \cap Y \notin B$. Hence $Y(\omega) \notin B$. However, $Y(\omega)=X(\omega) \in B$. ↯

If $X(\omega) \notin B$, then $\omega$ belongs to $X \notin B \cap Y \in B$. Hence $Y(\omega) \in B$. However, $Y(\omega)=X(\omega) \notin B$. ↯

score 0 · Answer 3 · answered May 15 '18 at 10:50

$$\{X=Y\} \subseteq [(X\in B)\triangle (Y\in B)]^C$$

Suppose $\omega \in \{X=Y\}$. To show that $\omega \in [(X\in B)\triangle (Y\in B)]^C$.

We have to show that $\omega$ belongs to both of the following sets

$X \in B \cup Y \notin B$

$X \notin B \cup Y \in B$

For $X \in B \cup Y \notin B$, we have two cases by $X$: $X(\omega) \in B$ or $X(\omega) \notin B$

If $X(\omega) \in B$, then $\omega$ belongs to $X \in B \cap Y \notin B$.
If $X(\omega) \notin B$, then $Y(\omega) \notin B$ and hence $\omega$ belongs to $X \in B \cap Y \notin B$.

For $X \notin B \cup Y \in B$, we again have two cases by $X$: $X(\omega) \in B$ or $X(\omega) \notin B$

If $X(\omega) \in B$, then $Y(\omega) \in B$ and hence $\omega$ belongs to $X \notin B \cup Y \in B$.
If $X(\omega) \notin B$, then $\omega$ belongs to $X \notin B \cup Y \in B$.

If X and Y are equal almost surely, then they have the same distribution

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