Probability Theory 9 | Independence for Events [dark version]

TL;DR

Independence means conditioning on one event does not change the probability of the other: P(A|B)=P(A) and P(B|A)=P(B).

Briefing Cornell Notes

Briefing

Independence in probability is the idea that learning one event gives no information about how likely another event is. Formally, event B should not change the probability of event A: the conditional probability P(A|B) must equal the unconditional probability P(A). The same requirement must hold in the opposite direction as well, with P(B|A)=P(B). When these conditions hold, the probability of both events occurring together factors neatly into a product, P(A∩B)=P(A)P(B), which becomes the practical test and the working definition.

The discussion starts by contrasting independence with mutual exclusivity. Two disjoint events can’t occur together, but that doesn’t automatically mean they’re independent—because conditional probabilities can still shift. The key diagnostic is whether conditioning on one event leaves the other event’s likelihood unchanged. A simple visualization with a sample space split into two halves illustrates the “no change” principle: if A and B each occupy half of the sample space, and their overlap also occupies a quarter in a way consistent with the product rule, then conditioning doesn’t distort probabilities and the events are independent.

From there, the definition is sharpened using conditional probability. Independence of events A and B is equivalent to the intersection probability equaling the product of their individual probabilities. This extends beyond pairs: a whole family of events {A_i} is independent when probabilities of intersections over any finite subcollection factor into the product of the corresponding individual probabilities. The requirement is not “for all infinite intersections at once,” but rather for every finite selection of events from the family.

Concrete examples make the criterion tangible. For a die rolled twice (keeping order), the sample space has 36 equally likely outcomes. One event A is “the first throw is 6.” Another event B is “the sum of the two throws is 7.” Each event has probability 1/6, and their intersection occurs only when the first throw is 6 and the second is 1, giving probability 1/36. Since P(A∩B)=1/36 and P(A)P(B)=(1/6)(1/6)=1/36, the events are independent—even though they are defined in different ways (one by a single face, the other by a sum constraint).

A continuous example uses the unit interval as the sample space with a uniform distribution, where probabilities come from integrating a constant density over intervals. The transcript then connects independence to integration: for independent events A and B, the probability of their intersection can be written as an integral over the product of indicator functions. Independence allows that intersection integral to factor into a product of two integrals, mirroring the discrete rule P(A∩B)=P(A)P(B). That factorization is highlighted as a calculation tool that simplifies later work in probability.

Cornell Notes

Independence means one event provides no probabilistic information about another. For two events A and B, independence is captured by the condition that conditioning doesn’t change likelihood: P(A|B)=P(A) and P(B|A)=P(B), which is equivalent to the product rule P(A∩B)=P(A)P(B). The idea generalizes to a family of events {A_i}: any finite intersection probability must equal the product of the individual probabilities. The transcript demonstrates this with two die-roll events (first roll is 6; sum is 7), where P(A∩B)=1/36 matches P(A)P(B). It then extends the same factorization logic to continuous settings using uniform distributions and indicator functions, where intersection probabilities become integrals that split into products under independence.

Why does mutual exclusivity not automatically imply independence?

Mutual exclusivity means A∩B=∅, so P(A∩B)=0. Independence would require P(A∩B)=P(A)P(B). That only holds if at least one of P(A) or P(B) is also 0. In general, disjoint events can still have positive probabilities, making P(A)P(B)>0 while P(A∩B)=0, so the product rule fails.

What is the practical test for independence between two events?

Compute P(A∩B) and compare it to P(A)P(B). If they match, then conditioning doesn’t change probabilities and the events are independent. Equivalently, check P(A|B)=P(A) (and similarly P(B|A)=P(B)), but the intersection-vs-product criterion is usually the most direct.

In the two-roll die example, how do the probabilities come out?

With two ordered rolls, there are 36 equally likely outcomes. Event A: first roll is 6 → probability 6/36=1/6. Event B: sum is 7 → ordered pairs (1,6),(2,5),(3,4),(4,3),(5,2),(6,1), so probability 6/36=1/6. Intersection A∩B requires first roll 6 and second roll 1, giving probability 1/36. Since (1/6)(1/6)=1/36, A and B are independent.

How does independence look in the continuous uniform example?

Using the unit interval Ω=[0,1] with uniform density 1, probabilities of events defined by subsets of Ω are computed via integrals of the density over those subsets. Indicator functions (1 on the event, 0 outside) let the probability of an event be written as an integral. For independent events A and B, the integral over the intersection region factors into a product of two integrals, paralleling the discrete rule P(A∩B)=P(A)P(B).

What does “independent family of events” require beyond two events?

For a family {A_i}, independence requires that for every finite set of indices, the probability of the intersection equals the product of the individual probabilities. The transcript notes that only finitely many events are involved in each such product check (and the empty set can be ignored as a trivial case).

Review Questions

State the equivalence between independence and the product rule for two events.
For a family of events {A_i}, what must hold for every finite subcollection to call the family independent?
In the two-roll die example, list the outcomes in A∩B and verify the product P(A)P(B).

Key Points

1
Independence means conditioning on one event does not change the probability of the other: P(A|B)=P(A) and P(B|A)=P(B).
2
For two events, independence is equivalent to the product rule P(A∩B)=P(A)P(B).
3
Independence extends to families of events: any finite intersection probability must equal the product of the corresponding individual probabilities.
4
Disjoint (mutually exclusive) events are not automatically independent; independence would require the product P(A)P(B) to be zero as well.
5
In discrete settings, independence can be checked by counting outcomes and comparing P(A∩B) to P(A)P(B).
6
In continuous settings, indicator functions and integrals represent event probabilities, and independence lets intersection integrals factor into products.

Highlights

Independence is fundamentally a “no information” condition: knowing B doesn’t make A more or less likely.

The clean criterion is P(A∩B)=P(A)P(B), which avoids repeatedly computing conditional probabilities.

A die example with A: first roll is 6 and B: sum is 7 yields P(A∩B)=1/36, matching (1/6)(1/6).

Independence carries over to continuous probability via indicator functions, turning intersection integrals into products.

Topics

Independence
Conditional Probability
Independent Events
Discrete Probability
Continuous Probability