Suppose a pair of random variable is independent from another pair, does it mean that each random variable is independent from the other?

Question

Let $(X_1, Y_1)$, and $(X_2, Y_2)$ be two pairs of random variables, and they are assumed to be independent.

Does it mean that:

1. $X_1$ is independent from $X_2$?
2. $X_1$ is independent from $Y_2$?
3. $Y_1$ is independent from $X_1$?
4. $Y_1$ is independent from $Y_2$?

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Motivation: I had this question because in a lot of applications in statistics, you would like to find a parameter such that the joint distribution

$$P_{Y_1, Y_2}(y_1, y_2| x_1, x_2; \theta)$$ is maximized. And there is often the assumption that $(X_1, Y_1)$, and $(X_2, Y_2)$ are independent and identically distributed.

Using this assumption , $$P_{Y_1, Y_2}(y_1, y_2| x_1, x_2; \theta) = P_{Y_1}(y_1| x_1; \theta)P_{Y_1}(y_2| x_2; \theta) = \prod_{i = 1}^2 P_{Y_i}(y_i| x_i; \theta)$$

However, for this calculation to work, you must be able to show,

$$P_{Y_1, Y_2}(y_1, y_2| x_1, x_2; \theta) = \dfrac{\Pr(Y_1 = y_1, Y_2 = y_2, X_1 = x_1, X_2 = x_2; \theta)}{\Pr(X_1 = x_1, X_2 = x_2; \theta)}$$

which means you must split the denominator $\Pr(X_1 = x_1, X_2 = x_2; \theta) = \Pr(X_1 = x_1; \theta)\Pr(X_2 = x_2; \theta)$. However, the independence of $X_1, X_2$ is not explicitly stated. So it is implicit in the assumption? Or is an assumption missing?

Answer 1

YES, YES, NO and YES. If two sigma algebras are independent then any sub-sigma algebras of these are also independent. This proves 1) 2) and 4).

For 3) take two independent (non-constant) random variables $U$ and $V$ and take $X_1=Y_1=U, X_2=Y_2=V$.

Answer 2

I agree that the answers to your four questions are all "YES" (provided the typo in question 3 is fixed).

However, in view of your motivation, I'd like to point out that the way to derive the factorisation equation is like this. First, in view of the independence between $(x_1, y_1)$ and $(x_2, y_2)$, we have $$ p(x_1, x_2, y_1, y_2) = p(x_1, y_1) p(x_2, y_2).$$

Next, it's generally the case that $p(x_1, x_2, y_1, y_2) = p(y_1, y_2 | x_1, x_2) p(x_1, x_2)$ and $p(x_1, y_1) = p(y_1 | x_1)p(x_1)$ and $p(x_2, y_2) = p(y_2 | x_2) p(x_2)$. So $$ p(y_1, y_2 | x_1, x_2) p(x_1, x_2) = p(y_1 | x_1)p(x_1) p(y_2 | x_2) p(x_2)$$

Using independence between $x_1$ and $x_2$ once more (this is like what you were asking for in your denominator), we have $p(x_1, x_2) = p(x_1)p(x_2)$, so $$ p(y_1, y_2 | x_1, x_2) p(x_1) p(x_2) = p(y_1 | x_1)p(x_1) p(y_2 | x_2) p(x_2).$$

Cancelling factors of $p(x_1)p(x_2)$ gives $$ p(y_1, y_2 | x_1, x_2) = p(y_1 | x_1) p(y_2 | x_2) ,$$ which is what you were after.

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Edit: To address the comment below...

Saying that $(x_1, y_1)$ and $(x_2, y_2)$ are independent is to say that $$ p(x_1, x_2, y_1, y_2) = p(x_1, y_1) p(x_2, y_2).$$

Integrating both sides with respect to $y_1$ and $y_2$, we have $$ p(x_1, x_2) = \iint dy_1 dy_2 p(x_1, x_2, y_1, y_2) = \iint dy_1 dy_2 p(x_1, y_1) p(x_2, y_2) \\ = \int dy_1 p(x_1, y_1) \times \int dy_2 p(x_2, y_2) = p(x_1) p(x_2), $$ which is what you wanted to prove.