1 00:00:00,000 --> 00:00:01,390 2 00:00:01,390 --> 00:00:02,400 Hi. 3 00:00:02,400 --> 00:00:05,140 In this problem, we're going to use the set of probability 4 00:00:05,140 --> 00:00:07,830 axioms to derive the probability of the difference 5 00:00:07,830 --> 00:00:09,170 of two events. 6 00:00:09,170 --> 00:00:11,430 Now, before we get started, there's one thing you might 7 00:00:11,430 --> 00:00:13,730 notice that, the equation we're trying to prove is 8 00:00:13,730 --> 00:00:15,340 actually quite complicated. 9 00:00:15,340 --> 00:00:17,730 And I don't like it either, so the first thing we're going to 10 00:00:17,730 --> 00:00:21,930 do will be to find a simpler notation for the events that 11 00:00:21,930 --> 00:00:23,180 we're interested in. 12 00:00:23,180 --> 00:00:26,180 13 00:00:26,180 --> 00:00:29,780 So we start with two events, A and B, and there might be some 14 00:00:29,780 --> 00:00:31,720 intersection between the two events. 15 00:00:31,720 --> 00:00:36,460 We'll label the set of points or samples in A that are not 16 00:00:36,460 --> 00:00:45,800 in B, as a set C. So C will be A intersection B complement. 17 00:00:45,800 --> 00:00:51,680 Similarly, for all points that are in B but not in A, this 18 00:00:51,680 --> 00:00:54,770 area, we'll call it D. 19 00:00:54,770 --> 00:01:01,310 And D will be the set A complement intersection B. And 20 00:01:01,310 --> 00:01:06,130 finally, for points that are in the intersection of A and 21 00:01:06,130 --> 00:01:13,540 B, we'll call it E. So E is A intersection B. And for the 22 00:01:13,540 --> 00:01:15,830 rest of our problem, we're going to be using the notation 23 00:01:15,830 --> 00:01:21,990 C, D, and E instead of whatever's down below. 24 00:01:21,990 --> 00:01:26,010 If we use this notation, we can rewrite our objective as 25 00:01:26,010 --> 00:01:27,790 the following. 26 00:01:27,790 --> 00:01:34,220 We want to show that the probability of C union D is 27 00:01:34,220 --> 00:01:38,210 equal to the probability of the event A plus the 28 00:01:38,210 --> 00:01:46,760 probability of B minus twice the probability of E. And that 29 00:01:46,760 --> 00:01:48,010 will be our goal for the problem. 30 00:01:48,010 --> 00:01:51,140 31 00:01:51,140 --> 00:01:54,110 Now, let's take a minute to review what the axioms are, 32 00:01:54,110 --> 00:01:56,310 what the probability axioms are. 33 00:01:56,310 --> 00:01:59,000 The first one says non-negativity. 34 00:01:59,000 --> 00:02:01,870 We take any event A, then the probability of A 35 00:02:01,870 --> 00:02:04,150 must be at least 0. 36 00:02:04,150 --> 00:02:07,910 The second normalization says the probability of the entire 37 00:02:07,910 --> 00:02:12,420 space, the entire sample space omega, must be equal to 1. 38 00:02:12,420 --> 00:02:16,610 And finally, the additivity axiom, which will be the axiom 39 00:02:16,610 --> 00:02:19,550 that we're going to use for this problem says, if there 40 00:02:19,550 --> 00:02:23,780 are two events, A and B that are disjoint-- 41 00:02:23,780 --> 00:02:26,765 which means they don't have anything in common, therefore. 42 00:02:26,765 --> 00:02:29,340 the intersection is the empty set. 43 00:02:29,340 --> 00:02:32,870 Then the probability of their union will be equal to the 44 00:02:32,870 --> 00:02:37,610 probably A plus the probability of B. For the rest 45 00:02:37,610 --> 00:02:41,090 of the problem, I will refer to this axiom as add. 46 00:02:41,090 --> 00:02:44,610 So whenever we invoke this axiom, I'll write 47 00:02:44,610 --> 00:02:47,010 "add" on the board. 48 00:02:47,010 --> 00:02:48,510 Let's get started. 49 00:02:48,510 --> 00:02:52,950 First, we'll invoke the additivity axioms to argue 50 00:02:52,950 --> 00:02:58,470 that the probability of C union D is simply the sum of 51 00:02:58,470 --> 00:03:01,680 probability of C plus probability of 52 00:03:01,680 --> 00:03:04,780 D. Why is this true? 53 00:03:04,780 --> 00:03:09,610 We can apply this axiom, because the set C here and the 54 00:03:09,610 --> 00:03:14,410 set D here, they're completely disjoint from each other. 55 00:03:14,410 --> 00:03:20,400 And in a similar way, we'll also notice the following. 56 00:03:20,400 --> 00:03:30,390 We see that A is equal to the union of the set C and E. 57 00:03:30,390 --> 00:03:36,020 And also, C and E, they're disjoint with each other, 58 00:03:36,020 --> 00:03:39,970 because C and E by definition don't share any points. 59 00:03:39,970 --> 00:03:44,440 And therefore, we have probably A is equal to 60 00:03:44,440 --> 00:03:51,760 probability of C plus the probability of E. Now, in a 61 00:03:51,760 --> 00:03:57,620 similar way, the probability of event B can also be written 62 00:03:57,620 --> 00:04:03,890 as a probability of D plus the probability of E, because 63 00:04:03,890 --> 00:04:07,480 event B is the union of D and E. 64 00:04:07,480 --> 00:04:10,240 And D and E are disjoint from each other. 65 00:04:10,240 --> 00:04:14,150 So we again invoke the additivity axiom. 66 00:04:14,150 --> 00:04:18,529 Now, this should be enough to prove our final claim. 67 00:04:18,529 --> 00:04:25,660 We have the probability of C union D. By the very first 68 00:04:25,660 --> 00:04:29,810 line, we see this is simply probability of C plus the 69 00:04:29,810 --> 00:04:33,690 probability of D. 70 00:04:33,690 --> 00:04:36,620 Now, I'm going to insert two terms here to make the 71 00:04:36,620 --> 00:04:40,920 connection with a second part of the equation more obvious. 72 00:04:40,920 --> 00:04:48,990 That is, I will write probability C plus probability 73 00:04:48,990 --> 00:04:57,290 E plus probability D plus probability of E. Now, I've 74 00:04:57,290 --> 00:04:59,810 just added two terms here-- 75 00:04:59,810 --> 00:05:04,960 probability E. So to make the equality valid or subtract it 76 00:05:04,960 --> 00:05:10,100 out two times, the probability of E. 77 00:05:10,100 --> 00:05:13,510 Hence this equality is valid. 78 00:05:13,510 --> 00:05:17,370 So if we look at this equation, we see that there 79 00:05:17,370 --> 00:05:20,260 are two parts here that we've already seen 80 00:05:20,260 --> 00:05:22,810 before right here. 81 00:05:22,810 --> 00:05:30,570 The very first parenthesis is equal to the probability of A. 82 00:05:30,570 --> 00:05:34,690 And the value of the second parenthesis is equal to the 83 00:05:34,690 --> 00:05:39,750 probability of B. We just derived these here. 84 00:05:39,750 --> 00:05:47,730 And finally, we have the minus 2 probability of E. This line 85 00:05:47,730 --> 00:05:51,450 plus this line gives us the final equation. 86 00:05:51,450 --> 00:05:53,610 And that will be the answer for the problem. 87 00:05:53,610 --> 00:05:54,860