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If $X_1$ and $X_2$ are 2 random i.i.d. variables. What is the probability of $P(X_1 = X_2)$. \begin{equation} P(X_1=X_2) = \sum_{i \in \mathbb{Z}} P_{X_1,X_2}(i,i) = \sum_{i \in \mathbb{Z}} P_{X_1}(i)^2. \end{equation}

Similar question with same proof- Probability of iid random variables to be equal? But there doesn't seem to be a convincing answer for the discrete case for at least me in the thread. Or I couldn't understand it properly.

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    $\begingroup$ There must exist some number $i$ such that $P_{X_1}(i) > 0$, therefore the sum can never be zero. $\endgroup$ Commented Dec 23, 2019 at 13:03
  • $\begingroup$ Exactly as stated above. If $X_1 = X_2 \equiv 0$ is the constant zero random variable then it is probability 1 $\endgroup$
    – fGDu94
    Commented Dec 23, 2019 at 13:06

1 Answer 1

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We have

$ p\left( {{X_1} = {X_2}} \right) = \\ \sum\limits_{x \in {\Omega _{{X_1}}} \cap {\Omega _{{X_2}}}} {p\left( {{X_1} = {X_2} = x} \right)} = \quad \quad \quad \quad \quad \quad {\text{by total probability}} \\ \sum\limits_{x \in {\Omega _{{X_1}}} \cap {\Omega _{{X_2}}}} {p\left( {{X_1} = x,{X_2} = x} \right)} = \quad \quad \quad \quad \quad {\text{by transitivity of equality}} \\ \sum\limits_{x \in {\Omega _{{X_1}}} \cap {\Omega _{{X_2}}}} {p\left( {\left. {{X_1} = x} \right|{X_2} = x} \right)} p\left( {{X_2} = x} \right) = \quad {\text{by the product rule}} \\ \sum\limits_{x \in {\Omega _{{X_1}}} \cap {\Omega _{{X_2}}}} {p\left( {{X_1} = x} \right)} p\left( {{X_2} = x} \right) = \quad \quad \quad \quad {\text{if }}{X_1}{\text{ and }}{X_2}{\text{ are independent}} \\ {\sum\limits_{x \in {\Omega _{{X_1}}} \cap {\Omega _{{X_2}}}} {p\left( {{X_1} = x} \right)} ^2}\quad \quad \quad \quad \quad \quad \quad \quad \;\;{\text{if }}{X_1}{\text{ and }}{X_2}{\text{ have same probability distribution}} $

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  • $\begingroup$ How to proceed forward now? $\endgroup$
    – A Q
    Commented Dec 23, 2019 at 13:48
  • $\begingroup$ @AQ To proceed forward to what please? $\endgroup$ Commented Dec 23, 2019 at 15:40

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