$X$ is independent of the event A then $\int_A 1_B(X(\omega)) dP(\omega) = P(A) P(X\in B)$

I started like this :

$$ \int_A 1_B(X(\omega)) dP(\omega) = \int 1_A(\omega) 1_B(X(\omega)) dP(\omega) = \int 1_A(\omega) 1_{X^{-1}B}(\omega) dP(\omega) = \int 1_{A \cap X^{-1}B}(\omega) dP(\omega) = \int_{ A \cap X^{-1}B} dP(\omega) = P ( A \cap X^{-1}B )$$

I'm stuck at this point, I'm trying to indroduce the independence ( $P(A\cap B ) = P(A)P(B) )$ but I don't see how I can do that as I have the event $X^{-1}(B)$ as the second event.

could anyone help with the rest?

  • $\begingroup$ Have you used independence yet? $\endgroup$ – AnonymousCoward Dec 4 '17 at 10:55
  • $\begingroup$ I didn't use independence yet. This is where I wanna introduce it ($P(A \cap B) = P(A)P(B)$ if $A$ and $B$ are idependent events $\endgroup$ – user30614 Dec 4 '17 at 10:57

$$ \int_{A} 1_{\{w : X^{-1} \in B\}} dP = \int 1_A 1_{\{w : X^{-1} \in B\}} dP = \int 1_{A\cap {\{w : X^{-1} \in B\}}} dP = P (A \cap {\{w : X^{-1} \in B\}}) = P(A)P(\{w : X^{-1} \in B\}) = P(A)P(X \in B) $$

  • $\begingroup$ Thank you. I guess the thing I don't get is : What does it mean to have $ X $ which is a random variable being independent from an event $ A $? an event indépendent from another OR a random variable independent from another makes sense to me but not $ X $ and $ A $ $\endgroup$ – user30614 Dec 4 '17 at 12:12
  • $\begingroup$ I think about it as two sets $\{w \in \Omega : X(w) \in B\}$ and $A$ is a set of someother omegas, depending how the event $A$ is defined. So basically you have two sets of omegas, which you want to show is independent, one set satisfying properties of A and the other satisfying that the image under X lies in B. Does it make sense? @user30614 $\endgroup$ – Olba12 Dec 4 '17 at 12:29
  • $\begingroup$ it does thank you !!!! $\endgroup$ – user30614 Dec 4 '17 at 14:11
  • $\begingroup$ @user30614 If you prefer: $1_A$ and $X$ are both random variables. $\endgroup$ – Henrik Dec 4 '17 at 16:50
  • $\begingroup$ ok perfect, thank you, that is clear now $\endgroup$ – user30614 Dec 5 '17 at 9:55

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