Unit 3, lesson 18

Independent Events

What it means for events to be independent, how the product criterion tests it, why disjoint events with positive probability are dependent, why complements keep independence, and how pairwise and mutual independence differ.

  • Video 18
  • Independence

A generator outputs one of the records 00, 01, 10 and 11, each with probability 1/41/4. Learning that the second bit is 1 leaves the chance that the first bit is 1 at exactly 1/21/2. A second generator also makes each bit equal to 1 with probability 1/21/2, yet there the same news raises that chance for the first bit to 4/54/5. Independence is a property of the whole probability model, not of how unrelated two descriptions sound, and this lesson defines it by a product rule, links it to conditional probability and separates pairwise from mutual independence.

Key terms

Independent events

Definition Two events AA and BB in the same probability model are independent if

P(A∩B)=P(A) P(B).P(A \cap B) = P(A)\,P(B).

The definition is symmetric in AA and BB and uses no conditional probability, so it applies even when one of the events has probability 00. It depends on the probabilities assigned in the model, not only on the events as sets of outcomes.

In plain words For independent events, the probability that both occur is exactly the product of their separate probabilities. When the conditional probability is defined, learning that one of them occurred leaves the probability of the other unchanged.

Example In the uniform model on Ω={00,01,10,11}\Omega = \{00, 01, 10, 11\}, let A={10,11}A = \{10, 11\} be the event that the first bit is 1 and B={01,11}B = \{01, 11\} the event that the second bit is 1. Then A∩B={11}A \cap B = \{11\}, and

P(A∩B)=14=12⋅12,P(A \cap B) = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2},

so AA and BB are independent.

Product criterion

Definition The product criterion tests independence by computing P(A)P(A), P(B)P(B) and P(A∩B)P(A \cap B) and comparing P(A∩B)P(A \cap B) with P(A)P(B)P(A)P(B). When P(B)>0P(B) > 0, it is equivalent to a conditional test,

P(A∩B)=P(A)P(B)  ⟺  P(A∣B)=P(A),\begin{gathered} P(A \cap B) = P(A)P(B) \\[0.4em] \iff P(A \mid B) = P(A), \end{gathered}

because dividing the product equation by P(B)P(B) gives the conditional equation and multiplying back reverses the step. Likewise, when P(A)>0P(A) > 0 it is equivalent to P(B∣A)=P(B)P(B \mid A) = P(B). An event with probability 00 or probability 11 satisfies the criterion with every event. If P(B)=0P(B) = 0, both sides are 00. If P(B)=1P(B) = 1, then P(A∩B)=P(A)=P(A)P(B)P(A \cap B) = P(A) = P(A)P(B).

In plain words Multiply the two probabilities and compare the product with the probability that both events occur. When the condition has positive probability, the same test can be read as “learning that one event occurred does not change the chance of the other.”

Example Draw one card from a well-shuffled standard deck of 52, and let AA be the event of a heart and BB the event of an ace. Then P(A)=1/4P(A) = 1/4, P(B)=1/13P(B) = 1/13, and the only card in both events is the ace of hearts, so

P(A∩B)=152=14⋅113.P(A \cap B) = \frac{1}{52} = \frac{1}{4} \cdot \frac{1}{13}.

The events are independent, even though both describe the same card. In conditional form, P(B∣A)=1/13=P(B)P(B \mid A) = 1/13 = P(B), since one of the 13 hearts is an ace.

Dependent events

Definition Events AA and BB are dependent when they are not independent, that is, when

P(A∩B)≠P(A) P(B).P(A \cap B) \ne P(A)\,P(B).

An event of probability 00 or 11 is independent of every event, so both probabilities of dependent events lie strictly between 00 and 11, and then P(A∣B)≠P(A)P(A \mid B) \ne P(A) and P(B∣A)≠P(B)P(B \mid A) \ne P(B). Disjoint events with P(A)>0P(A) > 0 and P(B)>0P(B) > 0 are always dependent, since P(A∩B)=0P(A \cap B) = 0 while P(A)P(B)>0P(A)P(B) > 0. Dependence is a statement about probabilities in the model and does not by itself show that one event causes the other.

In plain words For dependent events, the probability that both occur is larger or smaller than the product of their separate probabilities, so learning whether one of them occurred changes the chance of the other. Mutually exclusive events with positive probabilities are an extreme case, because one rules the other out.

Example Keep the events AA and BB about the first and second bit, but give the records the probabilities 0.40.4 for 00, 0.10.1 for 01, 0.10.1 for 10 and 0.40.4 for 11. Each bit is still 1 with probability 0.50.5, so P(A)P(B)=0.25P(A)P(B) = 0.25. The intersection A∩BA \cap B is the record 11 alone, which now has probability 0.40.4, so

P(A∩B)=0.4≠0.25,P(A \cap B) = 0.4 \ne 0.25,

and AA and BB are now dependent, with P(A∣B)=0.4/0.5=0.8P(A \mid B) = 0.4/0.5 = 0.8. The probabilities of the single events are the same as in the uniform model, so they alone cannot settle independence. The joint probabilities decide. Disjoint events get their own treatment in Disjoint Events and Independence.

Independence of complements

Definition If AA and BB are independent, then so are AA and BcB^{c}, AcA^{c} and BB, and AcA^{c} and BcB^{c}. For the first pair, AA is the disjoint union of A∩BA \cap B and A∩BcA \cap B^{c}, so

P(A∩Bc)=P(A)−P(A∩B)=P(A)−P(A)P(B)=P(A)(1−P(B))=P(A)P(Bc).\begin{aligned} &P(A \cap B^{c}) \\[0.4em] &\quad = P(A) - P(A \cap B) \\[0.4em] &\quad = P(A) - P(A)P(B) \\[0.4em] &\quad = P(A)\bigl(1 - P(B)\bigr) \\[0.4em] &\quad = P(A)P(B^{c}). \end{aligned}

Swapping the roles of AA and BB gives the second pair, and applying the result to the pair AcA^{c} and BB gives the third. The statement is about complementing a member of an independent pair. An event AA is independent of its own complement exactly when P(A)P(A) is 00 or 11, since P(A∩Ac)=0P(A \cap A^{c}) = 0 while P(A)P(Ac)=P(A)(1−P(A))P(A)P(A^{c}) = P(A)\bigl(1 - P(A)\bigr).

In plain words The events BB and BcB^{c} carry the same information, since each occurs exactly when the other does not. So replacing either event of an independent pair by its complement keeps the product rule intact.

Example For the heart AA and the ace BB from one card, the event A∩BcA \cap B^{c} is a heart that is not an ace, which holds 12 cards. So

P(A∩Bc)=1252=14⋅1213,P(A \cap B^{c}) = \frac{12}{52} = \frac{1}{4} \cdot \frac{12}{13},

which is P(A)P(Bc)P(A)P(B^{c}). Likewise, 36 cards are neither hearts nor aces, and 36/52=34⋅121336/52 = \frac{3}{4} \cdot \frac{12}{13}.

Pairwise independence

Definition Events A1,…,AnA_1, \dots, A_n are pairwise independent if every two of them are independent, that is, if P(Ai∩Aj)=P(Ai)P(Aj)P(A_i \cap A_j) = P(A_i)P(A_j) for all i≠ji \ne j.

In plain words Check the events two at a time. Each pair on its own passes the product test.

Example In the uniform two-bit model, add the event C={00,11}C = \{00, 11\} that the two bits are equal, so P(C)=1/2P(C) = 1/2. Each pair among AA, BB and CC meets only in the record 11. For example,

P(A∩C)=14=12⋅12,P(A \cap C) = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2},

and the same holds for A∩BA \cap B and B∩CB \cap C. The three events are pairwise independent.

Mutual independence

Definition Events A1,…,AnA_1, \dots, A_n are mutually independent if every subcollection of at least two of them satisfies the product rule. For all indices i1<i2<⋯<iki_1 < i_2 < \cdots < i_k with k≥2k \ge 2,

P(Ai1∩⋯∩Aik)=P(Ai1)⋯P(Aik).\begin{aligned} &P(A_{i_1} \cap \cdots \cap A_{i_k}) \\[0.4em] &\quad = P(A_{i_1}) \cdots P(A_{i_k}). \end{aligned}

For three events AA, BB and CC this means the three pair identities together with the triple identity

P(A∩B∩C)=P(A) P(B) P(C).\begin{aligned} &P(A \cap B \cap C) \\[0.4em] &\quad = P(A)\,P(B)\,P(C). \end{aligned}

Mutual independence implies pairwise independence, but pairwise independence does not imply mutual independence. The triple identity alone is not enough either. If C=∅C = \varnothing, it holds for any AA and BB, even dependent ones.

In plain words Under mutual independence, the probability of any one of the events stays the same when you learn whether some or all of the others occurred, as long as that information has positive probability. Pairwise independence guarantees this only for news about a single other event.

Example In the uniform two-bit model, the events AA, BB and CC meet only in the record 11, so

P(A∩B∩C)=14≠18.P(A \cap B \cap C) = \frac{1}{4} \ne \frac{1}{8}.

They are not mutually independent. Indeed, once both bits are known to be 1 they are certainly equal, so P(C∣A∩B)=1P(C \mid A \cap B) = 1 although P(C)=1/2P(C) = 1/2. For contrast, take three bits with all 8 records equally likely, and let AiA_i be the event that bit ii is 1. Each event holds 4 records, each pair intersection holds 2 and the triple intersection holds 1, so the probabilities 12\frac{1}{2}, 14\frac{1}{4} and 18\frac{1}{8} satisfy every product rule, and these three events are mutually independent.

Check yourself

  1. Roll one fair six-sided die. Let AA be the event of an even number, BB the event of a multiple of 3 and DD the event of a number greater than 3. Are AA and BB independent? Are AA and DD?
Show the answer

The events are A={2,4,6}A = \{2, 4, 6\}, B={3,6}B = \{3, 6\} and D={4,5,6}D = \{4, 5, 6\}. Since A∩B={6}A \cap B = \{6\},

P(A∩B)=16=12⋅13,P(A \cap B) = \frac{1}{6} = \frac{1}{2} \cdot \frac{1}{3},

so AA and BB are independent, even though both describe one roll. Since A∩D={4,6}A \cap D = \{4, 6\}, the probability P(A∩D)=1/3P(A \cap D) = 1/3 differs from P(A)P(D)=1/4P(A)P(D) = 1/4, so AA and DD are dependent. In conditional form, P(A∣D)=2/3P(A \mid D) = 2/3 while P(A)=1/2P(A) = 1/2.

  1. For the same die, let EE be the event that the roll is 1 or 2, and FF the event that it is 5 or 6. A classmate says that EE and FF are independent because they cannot happen together. Is that right?
Show the answer

No. The events are disjoint, so P(E∩F)=0P(E \cap F) = 0, but P(E)P(F)=13⋅13=19P(E)P(F) = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9}. The product criterion fails, so EE and FF are dependent. Learning that EE occurred makes FF impossible, since P(F∣E)=0P(F \mid E) = 0 while P(F)=1/3P(F) = 1/3. Disjoint events are independent exactly when at least one of them has probability 00.

  1. Toss a fair coin twice, with the four outcomes HH, HT, TH and TT equally likely. Let AA be heads on the first toss, BB heads on the second and CC exactly one head. Show that AA, BB and CC are pairwise independent but not mutually independent.
Show the answer

Each event holds two of the four outcomes, so each has probability 12\frac{1}{2}. The pair intersections are A∩B={HH}A \cap B = \{\text{HH}\}, A∩C={HT}A \cap C = \{\text{HT}\} and B∩C={TH}B \cap C = \{\text{TH}\}, each with probability 14=12⋅12\frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2}, so the events are pairwise independent. The triple intersection is empty, because heads on both tosses is not exactly one head. So

P(A∩B∩C)=0≠18,P(A \cap B \cap C) = 0 \ne \frac{1}{8},

and the three events are not mutually independent.

How this lesson connects