# Logical disjunction OR between two independent random variables

Consider the Bernouilli experiment of tossing a coin $$2$$ consecutive times, with the probability of getting "heads" of $$p=0,8$$

The base space can be described as follows $$\Omega=\{HH,TT,HT,TH\}$$

And the two random variables $$X_1$$ and $$X_2$$ can be assigned the following values:

$$X_1(HH) = 0$$$$X_1(HT) = 0$$ $$X_1(TT) = 1$$ $$X_1(TH) = 1$$ $$X_2(TT) = 1$$ $$X_2(HT) = 1$$ $$X_2(HH) = 0$$ $$X_2(TH) = 0$$

While it is well documented what is the probability of $$\mathbb P(X_1=i ,X_2=j)$$ , the comma meaning a logical conjunction $$AND$$, I am having a hard time finding what would be the probability of $$\mathbb P (X_1=1\lor X_2=j)$$ . Does it correspond to a set union? Does it has to do with the independence of the two random variables?

## 1 Answer

Did you mean to write $$\cup$$ rather than $$\lor$$? The latter is typically used for logical propositions (e.g: $$P \lor Q$$), and the former for set unions. They're not exactly the same thing. Similarly, for $$P(X_1 = i, X_2 = j)$$ we're talking about $$P(X_1 =i \cap X_2 =j)$$.

So $$P(X_1 = 1 \cup X_2 = j)$$ refers to the probability that $$X_1 = 1$$ or $$X_2 = j$$. Generally, this is the same thing as adding up their individual probabilities and then taking away $$P(X_1 \cap X_2 )$$ (since we're overcounting by that amount when we add the individual probabilities): $$P(X_1 = 1 \cup X_2 = j) = P(X_1 = 1) + P(X_2 = j) - P(X_1 = 1 \cap X_2 = j)$$

In this case, you are correct that $$X_1$$ and $$X_2$$ are independent, since knowing something about one doesn't tell us anything about the other. So this further becomes $$P(X_1 = 1 \cup X_2 = j) = P(X_1 = 1) + P(X_2 = j) - P(X_1 = 1) P (X_2 = j)$$

So if $$j=1$$, we have $$P(X_1 = 1 \cup X_2 = 1)$$ which is $$0.8 + 0.2 - 0.16 = 0.84$$. Does that make sense?

• Yes I meant $\cup$ instead of $\lor$ Did you mean they are independent (and not mutually exclusive)? According to this website two events (or R.V.) are independent if $P(X_1=i,X_2=j)=P(X_1=i)P(X_2=j)$ which is the case I think in my example? Commented Nov 16, 2022 at 13:03
• @niobium In your old post I didn’t see the entire probability space, but yes from what I can see here they’re now independent since knowing something about $X_1$ doesn’t tell us anything about $X_2$. Let me see if I can edit my answer to reflect this—it would change! Commented Nov 16, 2022 at 13:09
• No problem, welcome to stackexchange by the way Commented Nov 16, 2022 at 13:11
• @niobium Thank you! Happy to be here :) Commented Nov 16, 2022 at 13:18