# Independence of Random Variables in measure theoretic sense

In preparation for my graduate school, I am studying probability theory myself but am confused about the independence of random variables and it has been weeks since I am stuck on this. Could you please all share your thoughts?

Definition of independence of events

So this says that two random variables are independent if the events {X $$\in A$$} and {Y$$\in B$$} are independent. But here this assumes that there is same probability measure "P" defined on domain space $$\Omega$$. Therefore, using P on these sets makes sense. How would this definition make sense when we want to show whether two Normal Random variables in R are independent? What should be the underlying ($$\Omega$$,$$\wp$$, P) in that case? Moreover, doesn't each Normal RV generate a different probability measure on R with Borel Sigma-Algebra, then does this make sense to speak about independence when we have different measures?

• Your first link gives a 404 error, and your second link is to a text-heavy image. Please fix the first link, and copy the text of the second link into your question (images are not terribly searchable, eat up bandwidth, and are inaccessible to folk who access the site using screenreaders). Commented May 27, 2021 at 19:01
• Welcome to MSE. The first link doesn't work. Anyway, please type your questions instead of posting links where possible. If you need help formatting math on this site, here's tutorial. Commented May 27, 2021 at 19:01
• These might be related to your question: stats.stackexchange.com/questions/429178/… and stats.stackexchange.com/questions/450215/…
– Joe
Commented May 27, 2021 at 19:13
• Thanks for your reply @Joe I am confused about the following question: How to prove the independence of any two normally distributed random variables in the measure-theoretic sense? Do we assume some standard probability measure on R as we assume standard sigma-algebra in the form of Borel sets? Commented May 27, 2021 at 19:25
• Not sure why this question is closed. It is a perfectly reasonable question based on a natural misunderstanding. Commented May 27, 2021 at 20:01

Imagine throwing a fair, six-sided die. We model the possible outcomes and their probabilities using a probability space, where $$\Omega=\{1,2,3,4,5,6\}$$, $$\mathcal p$$ is the power set of $$\Omega$$, and $$P(A):=\frac{\#A}{6}$$, where $$\#A$$ is the number of elements of $$A$$ (so it's a uniform probability distribution). This is our underlying probability space.
Once we have this probability space (and not a second earlier!) we can define random variables on it. This underlying space is crucial, since random variables are functions with this probability space as their domain. For instance, we could define the random variable $$X$$ which measures wether the roll yields an even number, so $$X(\omega)=1$$ if $$\omega\in\Omega$$ is even, and $$X(\omega)=0$$ otherwise. Similarly, we could define a completely different random variable $$Y$$ on the same probability space, measuring wether the roll yields a prime. These two random variables are not independent, since the chance that the roll yields an even prime is $$1/6$$ (since 2 is the only even prime), but if they were independent it would be $$P(X=1,Y=1)=P(X=1)P(Y=1)=\frac12 \cdot\frac12=\frac14$$.
• But how would find the probability of the intersection part? If X and Y are RV, then P(X$\in A$ and Y $\in B$)=P($X \in A$). P(Y $\in B$), but how do we find P for the intersection part when $\Omega$ and P are not known? The other two Probabilities will definitely come from distributions. Commented May 28, 2021 at 20:55