0
$\begingroup$

$X$ 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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.