Chapter 3 : Conditional Probability and Independence

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1 STAT/MATH 394 A - PROBABILITY I UW Autumn Quarter 2016 Néhémy Lim Chapter 3 : Conditional Probability and Independence 1 Conditional Probabilities How should we modify the probability of an event when some supplementary information concerning the result of an experiment is available? The concept of conditional probability allows to answer this question. Example 1. A professional tennis player has the following yearly records : 45 victories (W) - 17 losses (L). Before playing a match, one would then bet that he would win with probability 45/( ) Actually, we may suspect that the surface he is playing on might have some effect on his performance. To get better insight on this assumption, we have the following data on the player s records : W L Grass 7 3 Hard 27 9 Clay 11 5 Intuitively, if we know that the player is going to play a match on clay, we feel that the probability of winning should be updated to 11/(11+5) = This intuition turns out to be true and is confirmed by the definition of conditional probability. Definition 1.1 (Conditional probability). Let (Ω, A, P) be a probability space and F A an event such that > 0, we define the conditional probability that an event E A occurs given or knowing that F has occurred, and denote it by P(E F ), as follows : P(E F ) = P(E F ) (1) Proposition 1.1. Let (Ω, A, P) be a probability space and F A an event such that > 0. Function mapping P F : A R defined by P F (E) = P(E F ) is a probability measure on (Ω, A). Proof. We need to check that P F measure : satisfies the three axioms of a probability 1

2 1. For any event E A, is P F (E) 0? By definition, P F (E) = P(E F )/. Since P(E F ) 0 and > 0. So P F (E) Is P F (Ω) = 1? By definition, P F (Ω) = P(Ω F )/ = / = Let (E i ) be a countably infinite sequence of events that are mutually exclusive (i.e. E i E j = if i j), do we have P F ( E i) = P F (E i )? ( ) P F E i ( ) = P E i F by definition = P ( E i F ) = P ( (E i F )) distributivity of over = P(E i F ) since P is a probability measure and events E i are mutually exclusive events = P F (E i ) Remark : In case the sample space Ω is finite with uniform probability, that is all outcomes are equally likely to occur then, conditional on the event that the outcome lies in a subset F Ω, it may be convenient to compute conditional probabilities of the form P(E F ) by using F as the reduced sample space. Proposition 1.2. Let Ω be a finite sample space and consider the associated probability space (Ω, P(Ω), P) with uniform probability measure P. Let F A be an event such that > 0, the conditional probability of event E A given F can be computed as : P(E F ) = E F (2) F Example 2. You draw a card from a standard deck of 52 cards. If the card you draw is an ace, what is the probability that this is the ace of spades? Let A be the event that the card is an ace and S be the event that the card is a spade. Using the above proposition, the desired probability is thus : P(S A) = S A A = 1 4 2

3 Proposition 1.3 (Multiplication rule). Let E 1,..., E n be a sequence of n N, n 1 events. Then we have, ( n ) ( ) n 1 P E i = P(E 1 )P(E 2 E 1 )P(E 3 E 1 E 2 )... P E n E i (3) In particular, for n = 2, we have : P(E 1 E 2 ) = P(E 1 )P(E 2 E 1 ) Proof. To prove the multiplication rule, just apply the definition of conditional probability to its right-hand side. Example 3. Three cards are dealt successively at random and without replacement from a standard deck of 52 playing cards. What is the probability of receiving, in order, a king, a queen, and a jack? Let K 1, Q 2 and J 3 be the events of being dealt a king in the first draw, a queen in the second draw and a jack in the third draw, respectively. We want to find P(K 1 Q 2 J 3 ). According to the multiplication rule, we have : P(K 1 Q 2 J 3 ) = P(K 1 )P(Q 2 K 1 )P(J 3 K 1 Q 2 ) = For any two events E and F such that P(E) > 0 and > 0, we have that P(E F ) = P(E F ) so P(E F ) = P(E F ). Similarly, we have that P(F E) = P(E F ) P(E). By replacing P(E F ) by P(E F ) in the last expression, we obtain the so-called first Bayes s formula that is stated as follows : Property 1.1 (First Bayes s formula). Let E and F be two events such that P(E) > 0 and > 0. We have that P(F E) = P(E F ) P(E) (4) Example 4. According to a study, 0.5% of the population exhibit symptoms of video game addiction. The study highlights that the proportion of addicts is 2% among teenagers. We also know that there are 15% of teenagers in the population. What is the probability that a randomly selected addict is a teenager? Let us denote A the event that a person is an addict and T the event that the person is a teenager. Then the desired probability is : P(T A) = P(A T )P(T ) P(A) = = 0.6 In Chapter 2, one of the main results is called the law of total probability. This result can be rewritten using conditional probabilities instead of probabilities of intersection thanks to the multiplication rule. 3

4 Property 1.2 (Law of total probability). Let E A be an event and F 1,..., F n be n N, n 1 events that form a partition of the sample space Ω (i.e. n F i = Ω and F i F j = for i j), we have that n P(E) = P (E F i ) P(F i ) (5) The law of total probability entails the following result known as the second Bayes s formula : Property 1.3 (Second Bayes s formula). Let E A be an event and F 1,..., F n be nn, n 1 events that form a partition of the sample space Ω (i.e. n F i = Ω and F i F j = for i j). For a given i = 1... n, we have that P(F i E) = P(E F i )P(F i ) n j=1 P(E F j)p(f j ) (6) Example 5. A company called Fruit designs smartphones called e-phones. The items are produced on three factories. The three factories account for 20%, 30%, and 50% of the output, respectively. The fraction of defective e-phones produced is this: for the first factory, 5%; for the second factory, 3%; for the third factory, 1%. If an e-phone is chosen at random from the total output and is found to be defective, what is the probability that it was produced by the third factory? Let F i be the event that a given item is produced on the ith factory, for i = 1, 2, 3 and let us denote D the event that the item is defective. We are given that P(F 1 ) = 0.2, P(F 2 ) = 0.3 and P(F 3 ) = 0.5. In addition, P(D F 1 ) = 0.05, P(D F 2 ) = 0.03 and P(D F 3 ) = We want to find P(F 3 D). The second Bayes s formula is applied as follows: P(F 3 D) = P(D F 3 )P(F 3 ) P(D F 1 )P(F 1 ) + P(D F 2 )P(F 2 ) + P(D F 3 )P(F 3 ) = Independent Events Definition 2.1 (Independence). Let (Ω, A, P) be a probability space and E, F A two events such that P(E) > 0 and > 0, events E and F are said to be independent if one of the following equivalent identities holds : (a) P(E F ) = P(E) 4

5 (b) P(E F ) = P(E) (c) P(F E) = Example 6. A card is selected at random from an ordinary deck of 52 playing cards. If A is the event that the selected card is an ace and S is the event that it is a spade, then A and S are independent. Indeed, P(A) = 4 52 = 1 13, P(S) = = and P(A S) = 52 = P(A)P(S) Property 2.1. If events E and F are independent, then so are : E and F c E c and F E c and F c Proof. Let E and F be two independent events and let us prove that E and F c are independent, that is P(E F c ) = P(E)P(F c ). We have : P(E F c ) = P(E) P(E F ) (Law of total probability) = P(E) P(E) (since E and F are independent) = P(E)(1 ) = P(E)P(F c ) The proofs for the rest of the property are left as an exercise. Example 7. You roll one die followed by another die. Let A be the event that the sum of the die is 9. Let B be the event that the first die lands on an even number. Let C be the event that the second die lands on an odd number. Show that A and B are independent A and C are independent B and C are independent P(A B C) P(A)P(B)P(C) In a situation like this we say that A, B and C are pairwise independent but not mutually independent. Definition 2.2 (Independence of n events). Let E 1,..., E n be a sequence of n N, n 1 events, we say that events (E i )...n are pairwise independent if E i and E j are independent, that is P(E i E j ) = P(E i )P(E j ) for all i j. We say that events (E i )...n are mutually independent or independent if for all subsets I {1,..., n} ( ) P E i = P (E i ) (7) i I i I 5

6 For instance, for n = 3, if E 1, E 2 and E 3 are mutually independent, this means P(E 1 E 2 ) = P(E 1 )P(E 2 ) P(E 1 E 3 ) = P(E 1 )P(E 3 ) P(E 2 E 3 ) = P(E 2 )P(E 3 ) P(E 1 E 2 E 3 ) = P(E 1 )P(E 2 )P(E 3 ) Example 8. A person has two coins. The first coin comes up heads with probability 1/4 and tails with probability 3/4. The second coin comes up heads with probability 3/4 and tails with probability 1/4. One of the two coins is selected and then flipped twice. Each coin has the same initial probability of being selected. 1. Show that the event that the first flip is heads is not independent of the event that the second flip is heads. 2. Let F be the event that the first coin is selected. Let H 1 be the event that the first flip is heads and let H 2 be the event that the second flip is heads. Show that P(H 1 H 2 F ) = P(H 1 F )P(H 2 F ) Definition 2.3. Let E, F and G be three events such that P(G) > 0. We say that E and F are conditionally independent given G (or that conditioned on G the events E and F are independent) if P(E F G) = P(E G)P(F G) (8) It is easy to check that if E and F are conditionally independent given G then so are E and F c. It is also easy to extend this to three or more events being conditionally independent given G. 6