# Are the products of dependent and independent random variables independent?

Let A, B, C, and D be four random variables such that A and B are independent, and C and D are dependent. It is unknown whether A and C are independent nor whether B and D are independent. Let E and F represent the products E = AC and F = BD. Are E and F necessarily independent?

If not, say we add the knowledge that A and C are independent and B and D are independent; now are E and F necessarily independent?

• Simple counterexample: let A=D and B=C, with A and B independent, whence C and D are independent. Now E=AC=AB and F=BD=BA are identical: as far from independent as you can get. This shows that trying to dodge the problem by ruling out degenerate RVs or discrete RVs is not going to work. As a generalization, consider A,B,C mutually independent and D=A, entailing E=AC and F=AB. Evidently E and F aren't going to be independent in general. This shows that five of the six pairs from A,B,C,D can be independent, but the single dependence between A and D can make E and F dependent.
– whuber
Commented Oct 24, 2011 at 16:52
• @whuber If all six of the pairs from A,B,C,D are independent, that is, we have pairwise independence, are AC and BD independent? Your last sentence seems to invite the inference that if all six pairs were independent, then AC and BD would be independent, and I am fairly sure that this is not necessarily so. Could you clarify this point? Commented Oct 24, 2011 at 17:17
• @Dilip You are correct: pairwise independence of A,B,C,D still does not imply independence of AC and BD. For instance, it's easy to construct examples where A,B,C are pairwise independent but A,B,C themselves are not independent. The example I offered is so simple and general that it appears to provide insight into why AC and BD need not be independent, regardless of issues about multiway dependencies (or lack thereof).
– whuber
Commented Oct 24, 2011 at 17:58

First, define dependent to mean not independent, that is, the joint distribution is not the product of the marginal distributions. Note also that all constant variables are independent of everything.

Though this may look like cheating, if $A = B = 1$ and $C = D \in \{0,1\}$, with the constraint that their common distribution is not degenerate, then $A$ and $B$ are independent, $C$ and $D$ are not, and since $E = C$ and $F = D$, then $E$ and $F$ are not independent either. Furthermore, $A$ and $C$ are independent and $B$ and $D$ are independent by degeneracy of the distributions of $A$ and $B$.

• To add to your example, if $A = B = 0$ instead, then $E = F = 0$ are independent, and thus the answer to the OP's question "Are $E$ and $F$ necessarily independent?" is that $E$ and $F$ are not necessarily independent; they may be dependent or independent as illustrated by the examples. Commented Oct 25, 2011 at 12:51

In both cases, the answer is No, $E = AC$ and $F = BD$ are not necessarily independent. Your "added knowledge" makes $A, B, C, D$ pairwise independent but that is not sufficient to guarantee that $AC$ and $BD$ are mutually independent. If $A, B, C, D$ are _mutually independent, then $AC$ and $BD$ are mutually independent events.

Given that C and D are dependent, E and F are definitely NOT independent. Rather they are dependent as well.

Even if A & C is independent, and B & D is independent, the above dependency will remain same.

• Could you add to your answer a proof that $E$ and $F$ are always dependent? I believe that all that can be said is that $E$ and $F$ are not necessarily independent, that is, in general, $E$ and $F$ are dependent, but there may be instances where $E$ and $F$ are independent. Borrowing and editing the example given by @NRH, if $A=B=0$ instead of $A=B=1$, then $E = F = 0$ are independent. But maybe you have a proof that except for the case of constant variables $A$ and $B$, $E$ and $F$ are always dependent? Commented Oct 25, 2011 at 12:47
• I thought that can be done easily. Given that - C and D are dependent, we case see - C = f(D). Hence, we can say - E = AC = Af(D) We already have F = BD hence we can extend to prove E = g(F) = A/B*(f(D)/D). I think, we need to prove the relationship between the conditional probability functions - which can be done if above variables are defined more concretely. Commented Oct 25, 2011 at 14:59
• No, this just proves it for the case when $C = f(D)$. Your original statement "Given that C and D are dependent, E and F are definitely NOT independent." is not completely correct, as the example of $A=B=0$ being degenerate RVs shows. So, I am asking you for a proof showing that if $A$ and $B$ are independent non-degenerate RVs, and $C$ and $D$ are dependent RVs, then $AC$ and $BD$ are necessarily dependent RVs. This would make a slightly modified version your original statement completely correct: definitely not independent provided $A$ and $B$ are non-degenerate. Commented Oct 25, 2011 at 15:53