$A_1$, $A_2$, $B_1$ and $B_2$ are random variables with

$P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$.

Are $A_1$ and $B_1$ independent? I would think not, because if e.g. $P(A_2)=0$ and $P(B_2)=0$, then the equation above would be correct for dependent $A_1$ and $B_1$.

Thanks for any help and thoughts!

  • 2
    $\begingroup$ You probably mean events and not random variables. In that case your thinking is correct. $\endgroup$ – Kavi Rama Murthy Nov 15 '18 at 12:07

As it is said in the comment, be careful not to mix random variables with events : $P(A_1)=0$ means nothing for a random variable. An event is something that might happen or not and it has a probability while a random variable is a variable that has a random value and it has not a single probability.

Events and random variables are related : you can create a random variable from an event by saying that its value is 1 when then event happens and 0 otherwise. Reciprocally, for any random variable $V$, you can consider one event for each of its possible values : $V=1$, $V=2$,...

Your equation (E) : $P(A_1, A_2, B_1, B_2) = P(A_1, A_2) P(B_1, B_2)$ is the definition of independence between the unions of variables $\{A_1,A_2\}$ and $\{B_1,B_2\}$.

But its arguments are random variables and it is a convenient notation that actually means one equation for each combinations of the related events.

eg : $P(A_1=T,A_2=F,B_1=1,B_2=F)=P(A_1=T,A_2=F)P(B_1=T,B_2=F)$

If your random variables are associated with 4 events, (E) is actually 16 equations, not only the one associated with True values.

So if all those 16 equations are true, you have independence of $\{A_1,A_2\}$ and $\{B_1,B_2\}$ that implies the independence between any subsets and so independence between $A_1$ and $B_1$.


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