I’m wrestling with the idea of “independent events,” and my brain keeps doing cartwheels when I peek at extra information. Example: I flip two fair coins. Let A be “the first coin is heads” and B be “the second coin is heads.” That feels independent. But then I condition on something like C: “both coins show the same face,” or D: “there is exactly one head.” Are A and B still independent under C? Under D? My intuition flip-flops because knowing the coins match seems to glue them together, but I’m not sure if that officially breaks independence or just makes it feel that way.
I’m also mixing this up with “mutually exclusive” in my head, which I know is different, but the terms keep tangoing together. How do I properly check independence in these conditional setups, and is there a rule-of-thumb for when independence survives conditioning and when it crumbles?
Any help appreciated!
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3 Responses
Independence need not survive conditioning: to check, compute whether P(A∩B | E) = P(A | E)P(B | E); “mutually exclusive” just means A∩B=∅, which (unless one has probability 0) implies dependence.
Example: with two fair coins, under C = “same face” the space is {HH,TT}, so P(A|C)=P(B|C)=1/2 but P(A∩B|C)=1/2≠1/4; under D = “exactly one head” the space is {HT,TH}, so P(A|D)=P(B|D)=1/2 but P(A∩B|D)=0≠1/4, hence A,B are not independent (conditioning on an event independent of both, e.g., a third unrelated toss, would preserve independence).
Ohh, I love this question. Your brain’s “cartwheels” are spot on: conditioning can totally glue events together that were independent before. Let’s unpack it carefully and also keep mutual exclusivity in its own lane.
First, the baseline
– A = “first coin is heads”
– B = “second coin is heads”
– Two fair, independent coin flips; sample space: HH, HT, TH, TT, each with probability 1/4.
Unconditionally:
– P(A) = 1/2, P(B) = 1/2, and P(A∩B) = P(HH) = 1/4 = (1/2)(1/2).
So A and B are independent.
Now condition on C: “both coins show the same face” (i.e., HH or TT)
– Under C, the possible outcomes are HH and TT, each with probability 1/2.
– P(A|C) = P(first is H | HH or TT) = 1/2 (only true in HH).
– P(B|C) = 1/2 (same reason).
– P(A∩B|C) = P(HH | HH or TT) = 1/2.
Check independence under C:
– Product: P(A|C)P(B|C) = (1/2)(1/2) = 1/4.
– Joint: P(A∩B|C) = 1/2.
They’re not equal, so A and B are not independent given C. In fact, they’re positively linked: knowing one is heads makes the other more likely to be heads under C, because we’re in the “same face” world.
Now condition on D: “there is exactly one head” (i.e., HT or TH)
– Under D, the possible outcomes are HT and TH, each with probability 1/2.
– P(A|D) = P(first is H | HT or TH) = 1/2 (true only in HT).
– P(B|D) = 1/2 (true only in TH).
– P(A∩B|D) = 0 (you can’t have both heads if there’s exactly one head).
Check independence under D:
– Product: P(A|D)P(B|D) = (1/2)(1/2) = 1/4.
– Joint: P(A∩B|D) = 0.
Again not equal, so A and B are not independent given D. Now they’re negatively linked: if one is heads, the other must be tails in this conditioned world.
A quick intuition snapshot
– Conditioning on “same face” (C) ties the coins together positively.
– Conditioning on “exactly one head” (D) ties them together negatively.
– The act of conditioning on a relationship between the two flips is exactly what creates dependence. We constrained the pair, and constraints create correlations.
Mutually exclusive vs independent (keeping them untangled)
– Mutually exclusive means A∩B is impossible (probability 0).
– Independent means P(A∩B) = P(A)P(B).
– If both P(A) and P(B) are positive, then they cannot be both mutually exclusive and independent: independence would force P(A∩B) > 0, contradicting mutual exclusivity.
– Under D, A and B become mutually exclusive within the conditioned world (they can’t both be true when exactly one head occurs). So they can’t be independent there unless one has zero conditional probability, which they don’t.
How to check independence under conditioning (the recipe)
– Compute P(A|C), P(B|C), and P(A∩B|C).
– Compare P(A∩B|C) to P(A|C)P(B|C). Equality means independence given C; otherwise, dependence.
– Algebraically, the test is equivalent to P(A∩B∩C)P(C) = P(A∩C)P(B∩C).
When does independence survive conditioning?
– Rule-of-thumb: Independence usually breaks if you condition on something that “looks at” A and B together (like totals, parity, same/different, exactly one success, etc.). These constraints glue them.
– Independence is preserved if you condition on information that is independent of the whole pair (A,B). Think of conditioning on a totally separate experiment that tells you nothing about either coin.
A tiny extra worked example showing survival
– Flip a third, completely independent coin. Let E = “the third coin is heads.”
– Since E is independent of both A and B jointly, conditioning on E doesn’t change their distribution:
– P(A|E) = P(A) = 1/2, P(B|E) = 1/2, and P(A∩B|E) = P(A∩B) = 1/4.
– So P(A∩B|E) = P(A|E)P(B|E). A and B remain independent given E.
Big-picture summary
– In your coin example, A and B are independent unconditionally.
– They are not independent given C (“same face”), and not independent given D (“exactly one head”).
– Conditioning on relationships between A and B tends to create dependence (sometimes positive, sometimes negative).
– Conditioning on something external and independent of the pair preserves independence.
– To check: compute the three conditional probabilities and compare P(A∩B|C) with P(A|C)P(B|C).
I get genuinely giddy about how a single extra nugget of information can flip independence on its head-like a logic switch! It’s a neat reminder that “what you know” shapes how random variables interact.
Great question-this is exactly the kind of thing that makes our intuition wobble a bit. Before conditioning, A = “first is heads” and B = “second is heads” are independent because P(A ∩ B) = 1/4 = (1/2)(1/2). Now condition on C = “both coins match.” Given C, the only outcomes are HH and TT, each with probability 1/2. So P(A|C) = 1/2 and P(B|C) = 1/2, but P(A ∩ B|C) = P(HH|C) = 1/2, which is not (1/2)(1/2). So A and B are not independent under C-in fact, under C they’re the same event: either both happen or both don’t. For D = “exactly one head,” the only outcomes are HT and TH, again each with probability 1/2. Then P(A|D) = 1/2 and P(B|D) = 1/2, but P(A ∩ B|D) = 0, so again not independent-this time they’re perfectly negatively linked (if one is true, the other isn’t). A quick way to check in any setup is to compare P(A ∩ B | E) with P(A|E)P(B|E). Rule of thumb: conditioning on information that depends on both A and B (like “they match” or “their total is 1”) tends to glue them together and create dependence; conditioning on something about just one coin, or on something external to both, tends to preserve independence. And about “mutually exclusive” versus “independent”: mutually exclusive means they can’t both happen, while independent means learning one doesn’t change the chance of the other-aside from trivial zero-probability cases, events with positive probability can’t be both. Analogy: think of a seesaw-if I tell you the total height of the two riders is fixed, knowing one rider’s height instantly tells you the other’s; that “total” condition ties them together just like C or D ties the coins.