# Probability Independence check similar to linear independence?

Linear independence check can be done as: given vectors $$v_1, v_2, \dots v_n$$, if there exists $$a_1, a_2, \dots a_n$$, not all zero, such that $$a_1 v_1 + a_2 v_2 + \dots + a_n v_n = 0$$, then $$v_i$$ are not linear independent; otherwise, $$v_i$$ are linear independent.

This check is quite straight-forward as it will lead to solve a $$n\times n$$ linear equation, or computing the det of a $$n\times n$$ matrix.

Probability independence check is a bit different. It's defined as:

Given a probability triple $$(\Omega, F, P)$$, a collection of events $$\{A_1, A_2, \dots, A_n \}$$ is independent if for all $$k \in N$$, and all possible $$\{i_1, i_2, \dots, i_k\}$$, $$P(A_{i_1}\cap A_{i_2} \cap \dots \cap A_{i_k})=P(A_{i_1})P(A_{i_2}) \dots P(A_{i_k})$$

This is quite troublesome as there are so many combinations.

Is there a similar way to check the independent of a collection of events? Or even better is it possible to convert the problem to check of vectors' linear independence?

• I'm afraid you are stuck with troublesome. The fact that each situation uses the word "independent" doesn't really help with the arguments. – Ethan Bolker Jul 4 at 18:31
• @EthanBolker i'm not wishing independence always brings the same check; I'm just wishing there's an easy way to check Probability Independence. is that possible? – athos Jul 4 at 18:32
• I don't think so... – Ethan Bolker Jul 4 at 18:38

I might note that it is entirely possible that of these $$2^n$$ equations, $$2^n - 1$$ are true, but one is false.
However, in practice one almost never actually checks all these equations. Rather, some assumptions are made in setting up a probability model, as a consequence of which we can deduce the independence. Thus if the $$A_i$$ depend only on different parts of an experiment that have no physical connection to each other, then they are independent.