Math 5010 Introduction to Probability Daniel Conus Lehigh University Davar Khoshnevisan University of Utah Firas Rassoul-Agha University of Utah Last Scribed May 8, 2013 Lecture 1 1. A few examples Example 1.1 (The Monte Hall problem; Steve Selvin, 1975). Three doors: behind one is a nice prize; behind the other two lie goats. You choose a door however you wish. The host (Monte Hall) knows where the prize is, and proceeds by showing you a door that has a goat behind it. Then, the host gives you the option to switch the door you had chosen to the remaining unopened door. Should you switch? The answer is “yes.” Here is why: If you do not switch, then you can only win if you had chosen the correct door to begin with. In other words, your winning odds are 1:3 if you don’t switch. On the other hand, if you do switch, then you win if you choose a goat right from the start (for then the host opens the other door with the goat and when you switch you have the prize). This has 2:3 odds. In other words, your chance to win doubles if you switch; therefore, you should. 2. The sample space, events, and outcomes We need a math model for describing “random” events that result from performing an “experiment.” We cannot use “frequency of occurrence” as a model because it does not have the power of “prediction.” For instance, if our definition of a fair coin is that the frequency of heads has to converge to 1/2 as the number 1 2 1 of tosses grows to infinity, then we have done things backwards. We have used our prediction to make a definition. What we should do is first define a model, then draw from it that prediction about the frequency of heads. Here is how we will do things. First, we define a state space (or sample space) that we will denote by Ω. We think of the elements of Ω as outcomes of the experiment. Then, we specify is a collection F of subsets of Ω. Each of these subsets is called an event. These events are the ones we are “allowed” to talk about the probability of. When Ω is finite, F can be taken to be the collection of all subsets of Ω. The next step is to assign a probability P(A) to every A F . We will 2 talk about this after the following examples. Example 1.2. Roll a six-sided die; what is the probability of rolling a six? First, write a sample space. Here is a natural one: Ω = f1, 2, 3, 4, 5, 6g. In this case, Ω is finite and we want F to be the collection of all subsets of Ω. That is, F = ? , f1g ,..., f6g , f1, 2g ,..., f1, 6g ,..., f1, 2, . , 6g . Example 1.3. Toss two coins; what is the probability that we get two heads? A natural sample space is Ω = (H1 , H2) , (H1 , T2) , (T1 , H2) , (T1 , T2) . Once we have readied a sample space Ω and an event-space F , we need to assign a probability to every event. This assignment cannot be made at whim; it has to satisfy some properties. 3. Rules of probability 3.1. Aside: Algebra of sets. Before we set forth the axioms [or rules] of probability, we need to recall some notation from set theory [“modern math” in high school]. Suppose A and B are two sets. We recall the following: A B denotes the union of A and B. In other words, a point x is • [ an element of A B if and only if either x is in A or in B or both; [ A B denotes the intersection of A and B. In other words, a point • \ x is in A B if and only if x is in both A and B; \ Ac denotes the “complement” of A. In other words, a point x is • in Ac if and only if x is not in A. 3. Rules of probability 3 c We write A r B for A B . The elements of A r B are points that are in A \ but not in B. Clearly, A B = B A and A B = B A. Also, A (B C) = [ [ \ \ [ [ (A B) C; therefore, we do not have to put parentheses, and simply will [ [ write A B C instead. In this way we can define A An inductively [ [ 1 [···[ without any ambiguities. Similar remarks apply to intersections. We say that A is a subset of B, and write A B, when A is inside B; ⊆ that is, all elements of A are also elements of B. It should be clear that if A B, then A B = A and A B = B. ⊆ n \ n [ Therefore, if A1 A2 An, then i=1Ai = A1 and i=1Ai = An. ⊆ ⊆ · · · ⊆ c \ c [ It should also be clear that A A = ? and A A = Ω. It is also not \ [ very hard to see that (A B)c = Ac Bc. (Not being in A or B is the same [ \ thing as not being in A and not being in B.) Similarly, (A B)c = Ac Bc. \ [ Rule 1. 0 6 P(A) 6 1 for every event A. Rule 2. P(Ω) = 1. “Something will happen with probability one.” Rule 3 (Addition rule). If A and B are disjoint events [i.e., A B = ?], \ then the probability that at least one of the two occurs is the sum of the individual probabilities. More precisely put, P(A B) = P(A) + P(B). [ 4 1 Homework Problems Exercise 1.1. You ask a friend to choose an integer N between 0 and 9. Let A = fN 6 5g, B = f3 6 N 6 7g and C = fN is even and > 0g. List the points that belong to the following events: (a) A B C \ \ (b) A (B Cc) [ \ (c) (A B) Cc [ \ (d) (A B) ((A C)c) \ \ [ Exercise 1.2. Let A, B and C be events in a sample space Ω. Prove the following identities: (a) (A B) C = A (B C) [ [ [ [ (b) A (B C) = (A B) (A C) \ [ \ [ \ (c) (A B)c = Ac Bc [ \ (d) (A B)c = Ac Bc \ [ Exercise 1.3. Let A, B and C be arbitrary events in a sample space Ω. Express each of the following events in terms of A, B and C using inter- sections, unions and complements. (a) A and B occur, but not C; (b) A is the only one to occur; (c) at least two of the events A, B, C occur; (d) at least one of the events A, B, C occurs; (e) exactly two of the events A, B, C occur; (f) exactly one of the events A, B, C occurs; (g) not more than one of the events A, B, C occur. Exercise 1.4. Two sets are disjoint if their intersection is empty. If A and B are disjoint events in a sample space Ω, are Ac and Bc disjoint? Are A C \ and B C disjoint ? What about A C and B C? \ [ [ Exercise 1.5. We roll a die 3 times. Give a probability space (Ω, F,P) for this experiment. Exercise 1.6. A urn contains three chips : one black, one green, and one red. We draw one chip at random. Give a sample space Ω and a collection of events F for this experiment. n Exercise 1.7. If An An-1 A1, show that i=1Ai = An and n ⊂ ⊂ · · · ⊂ \ Ai = A . [i=1 1 Lecture 2 1. Algebra of events, continued n We say that A1, , An are disjoint if Ai = ?. We say they are pair- ··· \i=1 wise disjoint if Ai Aj = ?, for all i = j. \ 6 Example 2.1. The sets f1, 2g, f2, 3g, and f1, 3g are disjoint but not pair-wise disjoint. Example 2.2. If A and B are disjoint, then A C and B C are disjoint only [ [ when C = ?. To see this, we write (A C) (B C) = (A B) C = ? C = C. [ \ [ \ [ [ On the other hand, A C and B C are obviously disjoint. \ \ Example 2.3. If A, B, C, and D are some events, then the event “B and at least A or C, but not D” is written as B (A C) r D or, equivalently, \ [ B (A C) Dc. Similarly, the event “A but not B, or C and D” is written \ [ \ (A Bc) (C D). \ [ \ Example 2.4. Now, to be more concrete, let A = f1, 3, 7, 13g, B = f2, 3, 4, 13, 15g, C = f1, 2, 3, 4, 17g, D = f13, 17, 30g. Then, A C = f1, 2, 3, 4, 7, 13, 17g, [ B (A C) = f2, 3, 4, 13g, and B (A C) r D = f2, 3, 4g. Similarly, \ [ \ [ A Bc = f1, 7g, C D = f17g, and (A Bc) (C D) = f1, 7, 17g. \ \ \ [ \ Example 2.5. We want to write the solutions to jx-5j+jx-3j > jxj as a union of disjoint intervals. For this, we first need to figure out what the absolute values are equal to. There are four cases. If x 6 0, then the inequality becomes 5 - x + 3 - x > -x, that is 8 > x, which is always satisfied (when x 6 -3). Next, is the case 0 6 x 6 3, and then we have 5 - x + 3 - x > x, which means 8 > 3x, and so 8=3 < x 6 3 is not allowed.