Is this proof about independence of events rigorous?

Question

This is my friend's proof that that if random events $A$ and $B$ are independent, then $A$ and $B^\complement$ are independent, where $B^\complement$ is the complement of $B:$

Two events are independent if the occurrence of one does not affect the probability of the other. Now, $P(B^\complement)$ is given by $1 - P(B).$ Because we can get $B^\complement$ from $B,$ nothing can be inferred from $B^\complement$ that can not be inferred from $B.$ Therefore, if $B$ and $A$ are independent events, then $B^\complement$ and $A$ must also be independent.

He argues that his proof is as rigorous as proofs like this one using algebra, because neither makes logical leaps that lack rigor.

I think that his proof is not rigorous at all, because unlike the algebraic proof it seems to make assumptions not based in axioms. For example, it seems to assume that you can't make a new inference from the same information.

Who is correct?

If you are talking about a Mathematical proof then your friend is certainly wrong. He probably has different notion of 'proof'. — geetha290krm, May 22 '23 at 05:50
A proof which isn’t based on axioms and logic is by definition not rigorous. However, he has the right idea that information content is all that matters as far as independence goes. You normally prove that intuition however by going through the math. You can formalize that idea with sigma algebras, but that’s more effort than it’s worth. — Eric, May 22 '23 at 06:29
In particular, he seems to think it’s obvious that $\Pr(A|B)=\Pr(A)$ implies $\Pr(A|f(B))=\Pr(A)$ which is true, but seems to require proof to me. — Eric, May 22 '23 at 06:37
If this were a small part of a large proof, I’d probably say you don’t even need to prove it or can give a cursory summary, but if you’re trying to prove just this point, then you should do it formally. — Eric, May 22 '23 at 06:47
The definition of "rigorous" as mostly used in mathematics, is the property of sticking to a fixed set of rules in which quantities are manipulated. To prove that $A$ and $B^c$, you must prove within the bounds of mathematical logic and the definitions provided in probability, that $P(A\cap B^c) = P(A)P(B^c)$. The argument provided does not do this. It uses a definition of independence that is intuitive, but not mathematical. In fact all its arguments are not mathematical, but can be translated to mathematical statements. That much merit can certainly be afforded to it! — Sarvesh Ravichandran Iyer, May 23 '23 at 06:41
To avoid divide-by-zero issues, assume $0<P[B]<1$. The sentence "because we can get $B^c$ from $B$" seems to mean that we can know the truth or falsehood of $B^c$ by knowing the truth or falsehood of $B$. Yes, but then the remainder of the argument seems to assume that independence of $A$ and $B$ implies both $P[A|B]=P[A]$ (true by basic defn of independence) and that $P[A|B^c]=P[A]$ (only true by the fact you are trying to prove). It is better to use $$(P[AB]=P[A]P[B])\implies P[AB^c]=P[A]-P[AB]=...=P[A]P[B^c]$$ and then interpret that to justify the fact about truth/falsehood of $B$. — Michael, May 23 '23 at 20:07

score 5 · Answer 1 · edited May 23 '23 at 07:13

This is not a proof but rather an intuitive explanation of what is happening behind the curtains. While the reasoning is sound, it does not prove the statement as it lacks the unambiguous formalism to make sure of the assumption we are going from to the conclusions we arrive to. One may easily come up with wrong proofs with this, some will be easy to spot, but some not at all.

Let $u_1,u_2,v_1,v_2$ be four vectors of the plane going in the upper right direction such that the steepness of vector $v_1$ is greater than the one of $u_1$ and the steepness of vector $v_2$ is greater than the one of $u_2$. Since each vector has a greater steepness, the sum $v_1+v_2$ also has a greater steepness than $u_1+u_2$.

It sounds right, but it is false.

Is this proof about independence of events rigorous?

1 Answers1