Let A and B be two random variables both independent from another random variable C. If A is independent from B, is A*B also independent from C? And if A and B are no independent from each other?
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$\begingroup$ Because $(A,B)\to AB$ is a function of the random variable $(A,B),$ this question is a special case of the duplicate question. $\endgroup$– whuber ♦Commented May 10, 2020 at 11:32
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$\begingroup$ I can provide a counter-example where the independence is pairwise. Suppose $(A,B,C,AB)$ can take any of the values $(0,0,0,0)$, $(0,1,1,0)$, $(1,0,1,0)$, $(1,1,0,1)$ each with probability $\frac14$. Then you have pairwise independence of $A$ and $C$, of $B$ and $C$, and of $A$ and $B$. But $AB$ and $C$ are not independent as $P(AB=1\mid C=0)=\frac12\not= 0=P(AB=1 \mid C=1)$ $\endgroup$– HenryCommented May 10, 2020 at 11:38
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$\begingroup$ @Henry That is a sufficiently interesting and clarifying example (pointing to the possibility of another interpretation of the question) to warrant re-opening this question: thank you. $\endgroup$– whuber ♦Commented May 10, 2020 at 12:56
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1 Answer
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If all you have is pairwise independence then there is a counterexample. Suppose the following four cases each have probability $\frac14$:
A B C AB
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1
Then $A$ is independent of $C$ and $B$ is independent of $C$, and $A$ is independent of $B$.
But $AB$ and $C$ are not independent as $\mathbb P(AB=1\mid C=0)=\frac12\not= 0=\mathbb P(AB=1 \mid C=1)$
In this example $A$, $B$ and $C$ are pairwise independent as suggested by the question, but are not mutually independent. If they had been mutually independent then it would also follow that $AB$ would be independent of $C$. A slightly weaker condition is that if $A$ and $B$ were jointly independent of $C$ then it would follow that $AB$ would be independent of $C$ even if $A$ and $B$ were not independent of each other.
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1$\begingroup$ Remark: $C = A \operatorname{XOR} B$, and $AB = A \operatorname{AND} B$; this might give some insight on the structure of the counterexample. $\endgroup$ Commented May 11, 2020 at 7:15
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1$\begingroup$ Another counterexample: take $A$ and $C$ to be independent coin-flips valued in $\{-1,1\}$, and then take $B = A/C$. It’s not hard to check that $B$ is independent of $C$, and it’s immediately clear that $AB$ is not independent of $C$. $\endgroup$– PLLCommented May 11, 2020 at 10:06
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$\begingroup$ I'm having trouble understanding this answer and comments. The question specifies that A, B, and C are all random variables independent of each other. In the answer aren't you missing 4 rows: 001, 010, 100, and 111 for ABC? $\endgroup$ Commented May 11, 2020 at 16:35
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$\begingroup$ @JasonGoemaat In my example, can you see that $A$ and $C$ are independent? Can you see that $B$ and $C$ are independent? Can you see that $A$ and $B$ are independent? $\endgroup$– HenryCommented May 11, 2020 at 16:51
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$\begingroup$ I think I see what you mean in that There is an equal likelihood of C being 0 or 1 if A is 0 or A is one and if B is or or B is 1. But then isn't C not an 'random variable' because it's value is totally dependent on the values of A and B? There is a 0 percent chance of C being 0 in the case of A and B being 0? $\endgroup$ Commented May 11, 2020 at 17:01