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Part CCSIR NET June 2023symmetry-of-dependence

Symmetry of dependence

Let A, B be two events in a discrete probability space with P(A) > 0 and P(B) > 0. Which of the following are necessarily true?

  1. A.If P(A | B) = 0 then P(B | A) = 0.✓ Correct
  2. B.If P(A | B) = 1 then P(B | A) = 1.
  3. C.If P(A | B) > P(A) then P(B | A) > P(B).✓ Correct
  4. D.If P(A | B) > P(B) then P(B | A) > P(A).

The idea, in plain words

Sort these relations into ones that treat the two events alike and ones that do not. Having empty overlap is even-handed, and so is one event raising the other's chance, because both are really statements about the overlap versus the product. Containment is not even-handed.

Solution

(1) P(A∩B) = 0 is symmetric.

(3) P(A|B) > P(A) ⇔ P(A∩B) > P(A)P(B), symmetric in A, B.

(2) A ⊇ B (a.s.) does not give B ⊇ A.

(4) Take B ⊂ A with P(B) small.

Why each option is right or wrong

Checked against 2 symbolic computations
  • A.Correct

    True

    because the hypothesis is secretly symmetric.

    P(A | B) = 0 means P(A ∩ B)/P(B) = 0

    hence P(A ∩ B) = 0.

    That statement makes no distinction between A and B

    so P(B | A) = P(A ∩ B)/P(A) = 0 as well.

    The trigger: rewrite every conditional statement in terms of P(A ∩ B) and see whether what remains is symmetric.

  • False, and here the symmetry genuinely breaks.

    P(A | B) = 1 means P(A ∩ B) = P(B), that is B is contained in A up to null sets.

    Then P(B | A) = P(B)/P(A)

    which is less than 1 whenever P(B) < P(A).

    Containment runs one way only.

  • C.Correct

    True, and it is the symmetric case again.

    P(A | B) > P(A) is equivalent to P(A ∩ B) > P(A)P(B)

    which is symmetric in A and B, and that in turn is equivalent to P(B | A) > P(B).

    Positive association is a mutual relation — if B raises the chance of A, then A raises the chance of B by exactly the same criterion.

  • False.

    Written out, the hypothesis is P(A∩B)>P(B)2P(A \cap B) > P(B)^{2} and the conclusion P(A∩B)>P(A)2P(A \cap B) > P(A)^{2}

    which are different statements.

    Take B ⊂ A with P(A) = 9/10 and P(B) = 1/10: then P(A | B) = 1 > 1/10 = P(B)

    so the hypothesis holds

    but P(B | A) = 1/9 (verified) and SymPy confirms 1/9 − 9/10 < 0

    so the conclusion fails.

    Compare with (C): the same shape of statement

    but comparing against the other event's probability destroys the symmetry.

2 of these 2 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Dependence misread

Independence, pairwise vs mutual, or correlation vs dependence.

Drill statements like this

Related counterexample: Pairwise independent events are mutually independent

More on this topic

The chapter behind this: Probability axioms, conditioning and independence — free to read

From Probability › Axioms, conditional probability, independence, Bayes

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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