Transitive property of equality of random variables Let $X_1$, $X_2$ and $X_3$ be discrete random variables. If $\Pr[X_1 = X_2] = 1$ and $\Pr[X_1 = X_3] = 1$, is it true that $\Pr[X_2 = X_3] = 1$?
Intuitively, since $\Pr[X = Y] = \sum_{z \in \mathbb{Z}} f_{XY}(z,z)$, I think that the transitive property of equality holds for random variables, but I don't know if there exists some result that confirms it.
 A: Suppose that $X$, $Y$ and $Z$ are random variables defined on the same probablity space $(\Omega,\mathcal F,P)$ such that $P(X=Y)=1$ and $P(X=Z)=1$.
Since $P(X=Y)=1$, there exists $\Omega_1\subset\Omega$ such that $X(\omega)=Y(\omega)$ for each $\omega\in\Omega_1$ and $P(\Omega_1)=1$. Similarly, there exists $\Omega_2\subset\Omega$ such that $X(\omega)=Z(\omega)$ for each $\omega\in\Omega_2$ and $P(\Omega_2)=1$.
For any $\omega\in\Omega_1\cap\Omega_2$, we have that $Y(\omega)=Z(\omega)$. Let us evaluate the probability $P(\Omega_1\cap\Omega_2)$. We have that $P(\Omega_1\cap\Omega_2)=1-P(\Omega_1^c\cup\Omega_2^c)$. But $P(\Omega_1^c\cup\Omega_2^c)=0$ since $P(\Omega_1^c)=0$ and $P(\Omega_2^c)=0$. Hence, $P(Y=Z)=1$.
A: Note that 
$$ \{X_2 = X_3\}  \supseteq \{X_1 = X_3\} \cap \{X_1 = X_2\} $$
that is, in points $\omega \in \Omega$ where $X_1(\omega) = X_3(\omega)$ and $X_1(\omega) = X_2(\omega)$ we must have $X_2(\omega) = X_3(\omega)$. Hence, we have
\begin{align*}
  \def\P{\mathbf P}\P[X_2 = X_3] &\ge \P[X_1 = X_2, X_1 = X_3]\\
       &= \P[X_1 = X_2] + \P[X_1 = X_3] - \P[\{X_1 = X_3\} \cup \{X_2 = X_3\}]\\
       &= 2 - \P[\{X_1 = X_2\} \cup \{X_1 = X_3\}]\\
       &\ge 2-1\\
       &= 1.
\end{align*}
Hence, $\P[X_2 = X_3] = 1$.
A: If $A$ and $B$ are events with $P(A)=1=P(B)$ then: $$P(A\cap B)=P(A)+P(B)-P(A\cup B)=1+1-1=1$$
Applying that on $A=\{X_1=X_2\}$ and $B=\{X_1=X_3\}$ we find that $P(X_1=X_2=X_3)=1$ and consequently $P(X_2=X_3)=1$
