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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?

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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?

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  • $\begingroup$ 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? $\endgroup$
    – niobium
    Commented Nov 16, 2022 at 13:03
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    $\begingroup$ @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! $\endgroup$
    – 2469158533
    Commented Nov 16, 2022 at 13:09
  • $\begingroup$ No problem, welcome to stackexchange by the way $\endgroup$
    – niobium
    Commented Nov 16, 2022 at 13:11
  • $\begingroup$ @niobium Thank you! Happy to be here :) $\endgroup$
    – 2469158533
    Commented Nov 16, 2022 at 13:18

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