# Probability of 2 discrete i.i.d. random variable to be 0.

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

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.

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

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