1 00:00:02,110 --> 00:00:04,830 We will now continue and derive some additional 2 00:00:04,830 --> 00:00:08,560 properties of probability laws which are, again, consequences 3 00:00:08,560 --> 00:00:11,680 of the axioms that we have introduced. 4 00:00:11,680 --> 00:00:14,140 The first property is the following. 5 00:00:14,140 --> 00:00:17,490 If we have two sets and one set is 6 00:00:17,490 --> 00:00:20,170 smaller than the other-- 7 00:00:20,170 --> 00:00:22,270 so we have a picture as follows. 8 00:00:22,270 --> 00:00:24,120 We have our sample space. 9 00:00:24,120 --> 00:00:27,080 And we have a certain set, A. 10 00:00:27,080 --> 00:00:32,780 And then we have a certain set, B, which is even bigger. 11 00:00:32,780 --> 00:00:38,230 So the set B is the bigger blue set. 12 00:00:38,230 --> 00:00:44,890 So if B is a set which is larger than A, then, 13 00:00:44,890 --> 00:00:48,200 naturally, the probability that the outcome falls inside 14 00:00:48,200 --> 00:00:51,290 B should be at least as big as the probability that the 15 00:00:51,290 --> 00:00:53,380 outcome falls inside A. 16 00:00:53,380 --> 00:00:56,120 How do we prove this formally? 17 00:00:56,120 --> 00:01:02,410 The set B can be expressed as a union of two pieces. 18 00:01:02,410 --> 00:01:06,290 One piece is the set A itself. 19 00:01:06,290 --> 00:01:13,300 The second piece is whatever elements of B there are, that 20 00:01:13,300 --> 00:01:18,030 do not belong in A. What are these elements? 21 00:01:18,030 --> 00:01:23,789 They are elements that belong to B. And they do not belong 22 00:01:23,789 --> 00:01:27,720 to A, which means that they belong to the complement of A. 23 00:01:27,720 --> 00:01:32,050 So we have expressed the set B as the union of two pieces. 24 00:01:32,050 --> 00:01:36,930 Now this piece is A. This piece here is outside A. So 25 00:01:36,930 --> 00:01:38,979 these two pieces are disjoint. 26 00:01:38,979 --> 00:01:42,190 And so we can apply the additivity axiom, and write 27 00:01:42,190 --> 00:01:47,470 that the probability of B is equal to the probability of A 28 00:01:47,470 --> 00:01:50,600 plus the probability of the other set. 29 00:01:53,570 --> 00:01:57,960 And since probabilities are non-negative, this expression 30 00:01:57,960 --> 00:02:02,860 here is at least as large as the probability of A. And this 31 00:02:02,860 --> 00:02:04,710 concludes the proof of the property that 32 00:02:04,710 --> 00:02:05,640 we wanted to show. 33 00:02:05,640 --> 00:02:08,978 Indeed, the probability of A is less than or equal to the 34 00:02:08,978 --> 00:02:12,770 probability of B. 35 00:02:12,770 --> 00:02:16,200 The next property we will show is the following. 36 00:02:16,200 --> 00:02:19,960 It allows us to write the probability of the union of 37 00:02:19,960 --> 00:02:24,040 two sets for the case now, where the two sets are not 38 00:02:24,040 --> 00:02:26,520 necessarily disjoint. 39 00:02:26,520 --> 00:02:31,460 So the picture is as follows. 40 00:02:31,460 --> 00:02:40,440 We have our two sets, A and B. These sets are 41 00:02:40,440 --> 00:02:43,370 not necessarily disjoint. 42 00:02:43,370 --> 00:02:45,950 And we want to say something about the probability of the 43 00:02:45,950 --> 00:02:50,880 union of A and B. 44 00:02:50,880 --> 00:02:54,770 Now the union of A and B consists of three pieces. 45 00:02:54,770 --> 00:02:59,440 One piece is this one here. 46 00:02:59,440 --> 00:03:06,100 And that piece consists of those elements of A that do 47 00:03:06,100 --> 00:03:12,080 not belong to B. So they belong to B complement. 48 00:03:12,080 --> 00:03:15,340 This set has a certain probability, let's call it 49 00:03:15,340 --> 00:03:19,840 little a and indicate it on this diagram. 50 00:03:19,840 --> 00:03:23,410 So a is the probability of this piece. 51 00:03:23,410 --> 00:03:26,300 Another piece is this one here, which is the 52 00:03:26,300 --> 00:03:30,329 intersection of A and B. It has a certain probability that 53 00:03:30,329 --> 00:03:32,370 we denote by little b. 54 00:03:32,370 --> 00:03:36,510 This is the probability of A intersection B. 55 00:03:36,510 --> 00:03:40,730 And finally, there's another piece, which is out here. 56 00:03:40,730 --> 00:03:43,950 And that piece has a certain probability c. 57 00:03:43,950 --> 00:03:46,280 It is the probability of that set. 58 00:03:46,280 --> 00:03:48,010 And what is that set? 59 00:03:48,010 --> 00:03:50,030 That set is the following. 60 00:03:50,030 --> 00:03:53,850 It's that part of B that consists of elements that do 61 00:03:53,850 --> 00:03:58,520 not belong in A. So it's B intersection with the 62 00:03:58,520 --> 00:04:02,470 complement of A. 63 00:04:02,470 --> 00:04:06,250 Now let's express the two sides of this equality here in 64 00:04:06,250 --> 00:04:09,280 terms of little a, little b, and little c, and see whether 65 00:04:09,280 --> 00:04:11,190 we get the same thing. 66 00:04:11,190 --> 00:04:17,800 So the probability of A union B. A union B consists of these 67 00:04:17,800 --> 00:04:22,000 three pieces that have probabilities little a, little 68 00:04:22,000 --> 00:04:24,810 b, and little c, respectively. 69 00:04:24,810 --> 00:04:27,710 And by the additivity axiom, the probability of the union 70 00:04:27,710 --> 00:04:30,000 of A and B is the sum of the probabilities 71 00:04:30,000 --> 00:04:32,200 of these three pieces. 72 00:04:32,200 --> 00:04:36,480 Let's look now at the right hand side of that equation and 73 00:04:36,480 --> 00:04:38,800 see whether we get the same thing. 74 00:04:38,800 --> 00:04:43,070 The probability of A plus the probability of B, minus the 75 00:04:43,070 --> 00:04:49,150 probability of A intersection B is equal to the following. 76 00:04:49,150 --> 00:04:52,480 A consists of two pieces that have probabilities little a 77 00:04:52,480 --> 00:04:55,540 and little b. 78 00:04:55,540 --> 00:04:58,240 The set B consists of two pieces that have probabilities 79 00:04:58,240 --> 00:04:59,765 little b and little c. 80 00:05:02,630 --> 00:05:04,800 And then we subtract the probability of the 81 00:05:04,800 --> 00:05:07,960 intersection, which is b. 82 00:05:07,960 --> 00:05:13,740 And we notice that we can cancel here one b 83 00:05:13,740 --> 00:05:15,210 with another b. 84 00:05:15,210 --> 00:05:23,290 And what we are left with is a plus b plus c. 85 00:05:23,290 --> 00:05:25,210 So this checks. 86 00:05:25,210 --> 00:05:30,470 And indeed we have this equality here. 87 00:05:30,470 --> 00:05:33,330 We have verified that it is true. 88 00:05:33,330 --> 00:05:37,720 One particular consequence of the equality that we derived 89 00:05:37,720 --> 00:05:39,140 is the following. 90 00:05:39,140 --> 00:05:48,409 Since this term here is always non-negative, this means that 91 00:05:48,409 --> 00:05:53,390 the probability of A union B is always less than or equal 92 00:05:53,390 --> 00:05:57,430 to the probability of A plus the probability of B. This 93 00:05:57,430 --> 00:06:00,760 inequality here is quite useful whenever we want to 94 00:06:00,760 --> 00:06:02,610 argue that a certain probability is 95 00:06:02,610 --> 00:06:03,850 smaller than something. 96 00:06:03,850 --> 00:06:05,460 And it has a name. 97 00:06:05,460 --> 00:06:07,255 It's called the union bound. 98 00:06:13,020 --> 00:06:17,650 We finally consider one last consequence of our axioms. 99 00:06:17,650 --> 00:06:21,530 And namely, we are going to derive an expression, a way of 100 00:06:21,530 --> 00:06:26,470 calculating the probability of the union of three sets, not 101 00:06:26,470 --> 00:06:28,830 necessarily disjoint. 102 00:06:28,830 --> 00:06:31,890 So we have our sample space. 103 00:06:31,890 --> 00:06:37,652 And within the sample space there are three sets-- set A, 104 00:06:37,652 --> 00:06:46,630 set B, and set C. 105 00:06:46,630 --> 00:06:50,170 We are going to use a set theoretic relation. 106 00:06:50,170 --> 00:06:55,840 We are going to express the union of these three sets as 107 00:06:55,840 --> 00:07:00,370 the union of three disjoint pieces. 108 00:07:00,370 --> 00:07:03,900 What are these disjoint pieces? 109 00:07:03,900 --> 00:07:07,123 One piece is the set A itself. 110 00:07:13,050 --> 00:07:19,500 The second piece is going to be that part of B which is 111 00:07:19,500 --> 00:07:26,740 outside A. So this is the intersection of B with the 112 00:07:26,740 --> 00:07:30,860 complement of A. 113 00:07:30,860 --> 00:07:38,590 The third piece is going to be whatever is left in order to 114 00:07:38,590 --> 00:07:40,400 form the union of the three sets. 115 00:07:40,400 --> 00:07:45,980 What is left is that part of C that does not belong to A and 116 00:07:45,980 --> 00:07:53,020 that does not belong to B. So that part is C intersection 117 00:07:53,020 --> 00:07:58,170 with A complement and B complement. 118 00:07:58,170 --> 00:08:02,380 Now this set here, of course, is the same as that set 119 00:08:02,380 --> 00:08:05,230 because intersection of two sets is the same no matter in 120 00:08:05,230 --> 00:08:07,430 which order we take the two sets. 121 00:08:07,430 --> 00:08:11,410 And similarly, the set that we have here is the same one that 122 00:08:11,410 --> 00:08:13,900 appears in that expression. 123 00:08:13,900 --> 00:08:16,840 Now we notice that these three pieces, the red, the blue, and 124 00:08:16,840 --> 00:08:19,560 the green, are disjoint from each other. 125 00:08:19,560 --> 00:08:25,820 So by the additivity axiom, the probability of this union 126 00:08:25,820 --> 00:08:31,080 here is going to be the sum of the probabilities 127 00:08:31,080 --> 00:08:33,200 of the three pieces. 128 00:08:33,200 --> 00:08:35,470 And that's exactly the expression 129 00:08:35,470 --> 00:08:36,720 the we have up here.