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I'm trying to come up with an exercise that would help with the conceptual understanding of random variables and their independence. I came up with the question presented below. It is however always a pleasant surprise to discover a hole in your own understanding (of mathematics in general, in this case), that prevents you from answering your own question.

I therefore came here for help. Do you think the question below is meaningful in the first place and if it is, how might I prove or disprove the claim presented in it?

The question

Let $X : \Omega \to A \subseteq \mathbb R$ and $Y : \Omega \to B \subseteq \mathbb R$ be random variables. Do the events $E_1, E_2 \subset \Omega$ have to be independent for the events $X(E_1)$ and $Y(E_2)$ to be independent?

The question in another format

Is the implication $$ \text{''$E_1$ and $E_2$ are independent''} \quad\Leftarrow\quad \text{''$X(E_1)$ and $Y(E_2)$ are independent''} $$ true?

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    $\begingroup$ I'm a bit confused by the question... you refer to $X(E_1)$ and $Y(E_2)$ as events, but they're subsets of $\mathbb R$, not $\Omega$. What's the probability measure on $\mathbb R$ that you would use to evaluate their independence? $\endgroup$ Commented Mar 2, 2020 at 15:09
  • $\begingroup$ I dunno. Maybe if I tried considering the case of marbles of different colors being taken out of a bag without replacement, set $E_1$ as blue marble was taken on the first try and $E_2$ as blumarble was lifted on the second try, when there are 3 orange and 2 blue marbles in the bag, it would be pretty easy to calculate the different probabilities in the sample space. In my head, the image of an event simply forms another sample point in the codomain of the random variable, which is why I called $X(E_1)$ an event. $\endgroup$
    – sesodesa
    Commented Mar 2, 2020 at 15:24
  • $\begingroup$ I mean, it's definitely true that the variables push the sets $E_1, E_2$ forward onto the common codomain ($\mathbb R$). But if you want to talk about those images as events, then you're now regarding $\mathbb R$ as a sample space, and it needs to be equipped with a probability measure of its own. So in particular, you want to be able to evaluate $\mathbb P_* (X(E_1) \cap Y(E_2)) = \mathbb P_*(X(E_1)) \mathbb P_* (Y(E_2))$ with some probability measure $\mathbb P_*$ on $\mathbb R$. $\endgroup$ Commented Mar 2, 2020 at 15:29
  • $\begingroup$ To be clear, I'm not saying the question doesn't make sense (yet) -- I just don't see what you're trying to go for (yet). In your example, I know what the events $E_1, E_2$ are. What are the variables $X, Y$? $\endgroup$ Commented Mar 2, 2020 at 15:32
  • $\begingroup$ That is what I still need to figure out. Could $X$ return the cumulative number of blue marbles lifted after each event, as in after $E_1$ $X$ might return $0$ or $1$, and after the second lift something between $0$ and $2$. But then $Y$ is still to be determined. Point is, I don't have an answer (yet). $\endgroup$
    – sesodesa
    Commented Mar 2, 2020 at 16:11

1 Answer 1

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In the spirit of providing an answer:

Do you think the question below is meaningful in the first place and if it is, how might I prove or disprove the claim presented in it?

At the moment, I don't think the question is meaningful. I don't know how to evaluate the independence of $X(E_1)$ and $Y(E_2)$, both of which are subsets of the codomain and not $\Omega$.

Whenever we discuss independence of things that seem not to live in $\Omega$, the trick is always a sneaky pullback to $\Omega$. For instance, I can ask whether the events $\{X \in A\}$ and $\{Y \in B\}$ are independent precisely because they are secret shorthand for $\{\omega \in \Omega: X(\omega) \in A\}$ and $\{\omega \in \Omega: Y(\omega) \in B\}$. But here you seem to have explicitly bucked that trend by deliberately pushing forward into $\mathbb R$ without an obvious way to pull back into $\Omega$ to use the probability measure.

There's no obvious way to pull back, either, since a pullback implicitly requires some notion of the mapping of how you got to the codomain. If I push a set forward through $X$, I can pull it back through $X$, and I can do the same for $Y$, but which pullback am I to use when I consider the intersection of those two? That would seem to be the essential thing I'm tasked with doing, after all.

Let's make this more concrete. Suppose $\Omega = (0, 1)$ with Borel sets / measure. Let $X(\omega) = 4 \omega + 1$ and $Y(\omega) = 5 - 4\omega$. Note that the images of $X, Y$ are $(1, 5)$ and $(2, 6)$, respectively. Consider the sets $A, B \subset \Omega$ defined as $A = (1/4, 3/4)$ and $B = (0, 1/2)$. These sets are independent because $\mathbb P(A) = 1/2$, $\mathbb P(B) = 1/2$, $\mathbb P(A \cap B) = 1/4$. I can say this because I know what the probability measure is, and it lives on the space $\Omega = (0, 1)$.

Now, let's consider the images of those sets; we have $X(A) = (2, 4)$ and $Y(B) = (3, 5)$. Are these sets independent? I have no idea, because I have no idea what it means to compute the probability of the intersection $(3, 4)$. You might be tempted to use Lebesgue / Borel measure as a default, but that's an ad hoc solution with no real meaning or interpretation here. Moreover, there's no reason at all this intersection had to be finite in the first place, and if it wasn't, Lebesgue measure wouldn't have anything constructive to contribute to this conversation anyway.

Another thing you might try is to pull the intersection back via the maps $X, Y$ and intersect back on $\Omega$ -- but the pullback of $(3, 4)$ under $X$ is $(1/2, 3/4)$ and the pullback of $(3, 4)$ under $Y$ is $(1/4, 1/2)$. These two are disjoint, which would seem to be an obstruction to usefully evaluating independence here.

I wondered at first if I just misunderstood your intent -- but then you clarified it, and it seems I did not. So I'll stick with my answer of: as written, this question seems not to make sense.

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