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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?

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  • $\begingroup$ I'm afraid you are stuck with troublesome. The fact that each situation uses the word "independent" doesn't really help with the arguments. $\endgroup$ – Ethan Bolker Jul 4 at 18:31
  • $\begingroup$ @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? $\endgroup$ – athos Jul 4 at 18:32
  • $\begingroup$ I don't think so... $\endgroup$ – Ethan Bolker Jul 4 at 18:38
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As noted by Ethan Bolker, there is no relation between the two meanings of "independent", and no shortcut in general.

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.

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