1 00:00:00,750 --> 00:00:03,290 In this segment, we start a discussion of multiple 2 00:00:03,290 --> 00:00:05,710 continuous random variables. 3 00:00:05,710 --> 00:00:09,760 Here are some objects that we're already familiar with. 4 00:00:09,760 --> 00:00:14,090 But exactly as in the discrete case, if we are dealing with 5 00:00:14,090 --> 00:00:17,930 two random variables, it is not enough to know their 6 00:00:17,930 --> 00:00:20,250 individual PDFs. 7 00:00:20,250 --> 00:00:23,230 We also need to model the relation between the two 8 00:00:23,230 --> 00:00:27,780 random variables, and this is done through a joint PDF, 9 00:00:27,780 --> 00:00:32,800 which is the continuous analog of the joint PMF. 10 00:00:32,800 --> 00:00:37,180 We will use this notation to indicate joint PDFs where we 11 00:00:37,180 --> 00:00:41,650 use f to indicate that we're dealing with a density. 12 00:00:41,650 --> 00:00:45,520 So what remains to be done is to actually define this object 13 00:00:45,520 --> 00:00:47,800 and see how we use it. 14 00:00:47,800 --> 00:00:52,440 Let us start by recalling that joint PMFs were defined in 15 00:00:52,440 --> 00:00:55,910 terms of the probability that the pair of random variables X 16 00:00:55,910 --> 00:01:01,350 and Y take certain specific values little x and little y. 17 00:01:01,350 --> 00:01:05,970 Regarding joint PDFs, we start by saying that it has to be 18 00:01:05,970 --> 00:01:08,050 non-negative. 19 00:01:08,050 --> 00:01:10,880 However, a more precise interpretation in terms of 20 00:01:10,880 --> 00:01:14,490 probabilities has to wait a little bit. 21 00:01:14,490 --> 00:01:19,270 Joint PDFs will be used to calculate probabilities. 22 00:01:19,270 --> 00:01:21,300 And this will be done in analogy with 23 00:01:21,300 --> 00:01:22,740 the discrete setting. 24 00:01:22,740 --> 00:01:25,820 In the discrete setting, the probability that the pair of 25 00:01:25,820 --> 00:01:30,660 random variables falls inside a certain set is just the sum 26 00:01:30,660 --> 00:01:35,390 of the probabilities of all of the possible pairs inside that 27 00:01:35,390 --> 00:01:37,310 particular set. 28 00:01:37,310 --> 00:01:39,880 For the continuous case, we introduce 29 00:01:39,880 --> 00:01:41,990 an analogous formula. 30 00:01:41,990 --> 00:01:46,120 We use the joint density instead of the joint PMF. 31 00:01:46,120 --> 00:01:50,700 And instead of having a summation, we now integrate. 32 00:01:50,700 --> 00:01:55,450 As in the discrete setting, we have one total unit of 33 00:01:55,450 --> 00:01:56,700 probability. 34 00:01:58,630 --> 00:02:04,090 The joint PDF tells us how this unit of probability is 35 00:02:04,090 --> 00:02:06,460 spread over the entire continuous 36 00:02:06,460 --> 00:02:08,360 two-dimensional plane. 37 00:02:08,360 --> 00:02:13,080 And we use it, we use the joint PDF, to calculate the 38 00:02:13,080 --> 00:02:19,180 probability of a certain set by finding the volume under 39 00:02:19,180 --> 00:02:24,270 the joint PDF that lies on top of that set. 40 00:02:24,270 --> 00:02:27,490 This is what this integral really represents. 41 00:02:27,490 --> 00:02:32,170 We integrate over a particular two-dimensional set, and we 42 00:02:32,170 --> 00:02:36,510 take this value that we integrate. 43 00:02:36,510 --> 00:02:40,360 And we can think of this as the height of an object that's 44 00:02:40,360 --> 00:02:43,400 sitting on top of that set. 45 00:02:43,400 --> 00:02:46,990 Now, this relation here, this calculation of probabilities, 46 00:02:46,990 --> 00:02:51,140 is not something that we are supposed to prove. 47 00:02:51,140 --> 00:02:54,160 This is, rather, the definition of 48 00:02:54,160 --> 00:02:57,450 what a joint PDF does. 49 00:02:57,450 --> 00:03:01,920 A legitimate joint PDF is any function of two variables, 50 00:03:01,920 --> 00:03:06,820 which is non-negative and which integrates to 1. 51 00:03:06,820 --> 00:03:11,670 And we will say that two random variables are jointly 52 00:03:11,670 --> 00:03:18,420 continuous if there is a legitimate joint PDF that can 53 00:03:18,420 --> 00:03:21,890 be used to calculate the associated probabilities 54 00:03:21,890 --> 00:03:24,710 through this particular formula. 55 00:03:24,710 --> 00:03:27,770 So we have really an indirect definition. 56 00:03:27,770 --> 00:03:32,290 Instead of defining the joint PDF as a probability, we 57 00:03:32,290 --> 00:03:37,610 actually define it indirectly by saying what it does, how it 58 00:03:37,610 --> 00:03:42,220 will be used to calculate probabilities. 59 00:03:42,220 --> 00:03:44,780 A picture will be helpful here. 60 00:03:44,780 --> 00:03:48,200 Here's a plot of a possible joint PDF. 61 00:03:48,200 --> 00:03:53,000 These are the x and y-axes. 62 00:03:53,000 --> 00:03:58,990 And the function being plotted is the joint PDF of these two 63 00:03:58,990 --> 00:04:01,580 random variables. 64 00:04:01,580 --> 00:04:05,730 This joint PDF is higher at some places and lower at 65 00:04:05,730 --> 00:04:08,800 others, indicating that certain regions of the x,y 66 00:04:08,800 --> 00:04:11,480 plane are more likely than others. 67 00:04:11,480 --> 00:04:15,630 The joint PDF determines the probability of a set B by 68 00:04:15,630 --> 00:04:21,470 integrating over that set B. Let's say it's this set. 69 00:04:21,470 --> 00:04:24,940 Integrating the PDF over that set. 70 00:04:24,940 --> 00:04:30,190 Pictorially, what this means is that we look at the volume 71 00:04:30,190 --> 00:04:36,790 that sits on top of that set, but below the PDF, below the 72 00:04:36,790 --> 00:04:41,780 joint PDF, and so we obtain some three-dimensional object 73 00:04:41,780 --> 00:04:43,100 of this kind. 74 00:04:43,100 --> 00:04:47,940 And this integral corresponds to actually finding this 75 00:04:47,940 --> 00:04:52,740 volume here, the volume that sits on top of the set B but 76 00:04:52,740 --> 00:04:55,060 which is below the joint PDF. 77 00:04:58,080 --> 00:05:01,410 Let us now develop some additional understanding of 78 00:05:01,410 --> 00:05:03,680 joint PDFs. 79 00:05:03,680 --> 00:05:12,430 As we just discussed, for any given set B, we can integrate 80 00:05:12,430 --> 00:05:15,890 the joint PDF over that set. 81 00:05:15,890 --> 00:05:18,770 And this will give us the probability of 82 00:05:18,770 --> 00:05:21,120 that particular set. 83 00:05:21,120 --> 00:05:24,460 Of particular interest is the case where we're dealing with 84 00:05:24,460 --> 00:05:29,420 a set which is a rectangle, in which case the situation is a 85 00:05:29,420 --> 00:05:30,750 little simpler. 86 00:05:30,750 --> 00:05:33,380 So suppose that we have a rectangle where the 87 00:05:33,380 --> 00:05:38,460 x-coordinate ranges from A to B and the y-coordinate ranges 88 00:05:38,460 --> 00:05:44,420 from some C to some D. Then, the double integral over this 89 00:05:44,420 --> 00:05:48,520 particular rectangle can be written in a form where we 90 00:05:48,520 --> 00:05:52,620 first integrate with respect to one of the variables that 91 00:05:52,620 --> 00:05:57,950 ranges from A to B. And then, we integrate over all possible 92 00:05:57,950 --> 00:06:01,480 values of y as they range from C to D. 93 00:06:01,480 --> 00:06:05,270 Of particular interest is the special case where we're 94 00:06:05,270 --> 00:06:10,290 dealing with a small rectangle such as this one. 95 00:06:10,290 --> 00:06:15,990 A rectangle with sizes equal to some delta where delta is a 96 00:06:15,990 --> 00:06:19,000 small number. 97 00:06:19,000 --> 00:06:22,920 In that case, the double integral, which is the volume 98 00:06:22,920 --> 00:06:28,810 on top of that rectangle, is simpler to evaluate. 99 00:06:28,810 --> 00:06:33,010 It is equal to the value of the function that we're 100 00:06:33,010 --> 00:06:36,640 integrating at some point in the rectangle --- let's take 101 00:06:36,640 --> 00:06:37,940 that corner --- 102 00:06:37,940 --> 00:06:42,640 times the area of that little rectangle, which is equal to 103 00:06:42,640 --> 00:06:44,390 delta square. 104 00:06:44,390 --> 00:06:47,820 So we have an interpretation of the joint PDF in terms of 105 00:06:47,820 --> 00:06:50,820 probabilities of small rectangles. 106 00:06:50,820 --> 00:06:54,140 Joint PDFs are not probabilities. 107 00:06:54,140 --> 00:06:57,560 But rather, they are probability densities. 108 00:06:57,560 --> 00:07:03,020 They tell us the probability per unit area. 109 00:07:03,020 --> 00:07:05,190 And one more important comment. 110 00:07:05,190 --> 00:07:08,450 For the case of a single continuous random variable, we 111 00:07:08,450 --> 00:07:12,750 know that any single point has 0 probability. 112 00:07:12,750 --> 00:07:16,620 This is again, true for the case of two jointly continuous 113 00:07:16,620 --> 00:07:17,440 random variables. 114 00:07:17,440 --> 00:07:19,460 But more is true. 115 00:07:19,460 --> 00:07:24,350 If you take a set B that has 0 area. 116 00:07:24,350 --> 00:07:26,890 For example, a certain curve. 117 00:07:26,890 --> 00:07:34,890 Suppose that this curve is the entire set B. Then, the volume 118 00:07:34,890 --> 00:07:40,320 under the joint PDF that's sitting on top of that curve 119 00:07:40,320 --> 00:07:43,570 is going to be equal to 0. 120 00:07:43,570 --> 00:07:48,460 So 0 area sets have 0 probability. 121 00:07:48,460 --> 00:07:50,930 And this is one of the characteristic features of 122 00:07:50,930 --> 00:07:54,590 jointly continuous random variables. 123 00:07:54,590 --> 00:07:57,920 Now, let's think of a particular situation. 124 00:07:57,920 --> 00:08:04,160 Suppose that X is a continuous random variable, and let Y be 125 00:08:04,160 --> 00:08:08,160 another random variable, which is identically equal to X. 126 00:08:08,160 --> 00:08:11,730 Since X is a continuous random variable, Y is also a 127 00:08:11,730 --> 00:08:13,690 continuous random variable. 128 00:08:13,690 --> 00:08:17,480 However, in this situation, we are certain that the outcome 129 00:08:17,480 --> 00:08:20,890 of the experiment is going to fall on the line 130 00:08:20,890 --> 00:08:23,080 where x equals y. 131 00:08:23,080 --> 00:08:26,910 All the probability lies on top of a line, and 132 00:08:26,910 --> 00:08:30,140 a line has 0 area. 133 00:08:30,140 --> 00:08:33,840 So we have positive probability on the set of 0 134 00:08:33,840 --> 00:08:37,159 area, which contradicts what we discussed before. 135 00:08:37,159 --> 00:08:41,100 Well, this simply means that X and Y are not jointly 136 00:08:41,100 --> 00:08:42,230 continuous. 137 00:08:42,230 --> 00:08:44,420 Each one of them is continuous, but together 138 00:08:44,420 --> 00:08:47,840 they're not jointly continuous. 139 00:08:47,840 --> 00:08:52,230 Essentially, joint continuity is something more than 140 00:08:52,230 --> 00:08:56,370 requiring each random variable to be continuous by itself. 141 00:08:56,370 --> 00:09:01,010 For joint continuity, we want the probability to be really 142 00:09:01,010 --> 00:09:03,820 spread over two dimensions. 143 00:09:03,820 --> 00:09:07,630 Probability is not allowed to be concentrated on a 144 00:09:07,630 --> 00:09:09,140 one-dimensional set. 145 00:09:09,140 --> 00:09:11,850 On the other hand, in this example, the probability is 146 00:09:11,850 --> 00:09:14,450 concentrated on a one-dimensional set. 147 00:09:14,450 --> 00:09:16,220 And we do not have joint continuity.