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

Mathematics Asked on January 5, 2022

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 $$Y_2$$?

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?

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.

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.

Answered by Kenny Wong on January 5, 2022

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 $$X_1=Y_1=U, X_2=Y_2=V$$.

Answered by Kavi Rama Murthy on January 5, 2022

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