1 00:00:00,000 --> 00:00:01,245 2 00:00:01,245 --> 00:00:04,350 Hi, In this problem, we'll be looking at the PDF the 3 00:00:04,350 --> 00:00:06,660 absolute value of x. 4 00:00:06,660 --> 00:00:10,330 So if we know a random variable, x, and we know it's 5 00:00:10,330 --> 00:00:14,180 PDF, how can we use that information to help us find 6 00:00:14,180 --> 00:00:16,280 the PDF of another random variable-- the 7 00:00:16,280 --> 00:00:18,090 absolute value of x? 8 00:00:18,090 --> 00:00:21,210 And so throughout this problem, we'll define a new 9 00:00:21,210 --> 00:00:22,450 random variable called y. 10 00:00:22,450 --> 00:00:26,100 And we'll define that y to be equal to the absolute value of 11 00:00:26,100 --> 00:00:28,950 x, just to make things simpler. 12 00:00:28,950 --> 00:00:32,439 So we'll do a couple of concrete examples, and then 13 00:00:32,439 --> 00:00:34,320 we'll try to generalize at the end. 14 00:00:34,320 --> 00:00:36,830 The first example that we'll deal with in part A is 15 00:00:36,830 --> 00:00:39,040 this PDF for x. 16 00:00:39,040 --> 00:00:43,000 So we're told that the PDF of x is 1/3 between negative 2 17 00:00:43,000 --> 00:00:45,650 and 1, and 0 otherwise. 18 00:00:45,650 --> 00:00:47,420 And here's a picture of what it looks like. 19 00:00:47,420 --> 00:00:52,110 It's just a rectangle from negative 2 to 1. 20 00:00:52,110 --> 00:00:55,220 So now we want to find out what is the PDF of the 21 00:00:55,220 --> 00:00:58,140 absolute value of x, which we've called y? 22 00:00:58,140 --> 00:01:02,210 And at this point, it may be helpful to step back and think 23 00:01:02,210 --> 00:01:05,940 about this problem from the discrete point of view again. 24 00:01:05,940 --> 00:01:09,980 So if x were a discrete random variable, the problem would 25 00:01:09,980 --> 00:01:15,310 be, what is the probability that the absolute value of x 26 00:01:15,310 --> 00:01:18,480 is equal to, say, 1/2? 27 00:01:18,480 --> 00:01:20,840 Well, the probability that the absolute value of 28 00:01:20,840 --> 00:01:22,480 x is equal to 1/2-- 29 00:01:22,480 --> 00:01:25,090 that can occur in two different ways. 30 00:01:25,090 --> 00:01:28,610 One is that x itself is 1/2. 31 00:01:28,610 --> 00:01:32,810 Or x could be negative 1/2, in which case, the absolute value 32 00:01:32,810 --> 00:01:35,030 of x would still be 1/2. 33 00:01:35,030 --> 00:01:37,740 So those two events are mutually exclusive. 34 00:01:37,740 --> 00:01:41,460 And so the probability of either one of them happening 35 00:01:41,460 --> 00:01:43,010 is you can just add them up. 36 00:01:43,010 --> 00:01:46,190 And so the probability of the absolute value of x being 1/2 37 00:01:46,190 --> 00:01:50,440 would have two contributions, one from x being 1/2, and one 38 00:01:50,440 --> 00:01:52,030 from x being negative 1/2. 39 00:01:52,030 --> 00:01:54,650 40 00:01:54,650 --> 00:01:57,280 The analogous idea carries over to the continuous case, 41 00:01:57,280 --> 00:01:58,600 when you have a PDF. 42 00:01:58,600 --> 00:02:02,260 So now let's say that we're interested in the case where 43 00:02:02,260 --> 00:02:06,620 we want to know the PDF of y at 1/2. 44 00:02:06,620 --> 00:02:11,680 Well, that again, is going to have two contributions, one 45 00:02:11,680 --> 00:02:18,410 from where x is 1/2, and one from where x is minus 1/2. 46 00:02:18,410 --> 00:02:23,980 And so you can just imagine that each one of these values 47 00:02:23,980 --> 00:02:26,990 for y-- and remember, y has to be non-negative, because it's 48 00:02:26,990 --> 00:02:28,830 an absolute value-- 49 00:02:28,830 --> 00:02:34,470 has two contributions, one from the right side of 0, and 50 00:02:34,470 --> 00:02:38,490 one from the left, or negative, side of 0. 51 00:02:38,490 --> 00:02:41,800 So you can come up and write an algebraic expression for 52 00:02:41,800 --> 00:02:45,670 this, and we'll do that in Part C. But you can also look 53 00:02:45,670 --> 00:02:48,500 at this from a visual point of view. 54 00:02:48,500 --> 00:02:53,570 And you can take the PDF diagram itself and imagine 55 00:02:53,570 --> 00:02:56,410 transforming it to find out what the PDF of the absolute 56 00:02:56,410 --> 00:02:58,220 value of x would look like. 57 00:02:58,220 --> 00:03:03,110 So the way to do it would be you take what's on the 58 00:03:03,110 --> 00:03:04,690 negative side. 59 00:03:04,690 --> 00:03:08,290 You flip it over and take the mirror image, and then you 60 00:03:08,290 --> 00:03:11,350 stack it on top of what you have on the right-hand side, 61 00:03:11,350 --> 00:03:13,700 or the positive side. 62 00:03:13,700 --> 00:03:15,750 So take this, flip it over, and stack it on top. 63 00:03:15,750 --> 00:03:18,310 You can imagine just taking this block, flipping it over. 64 00:03:18,310 --> 00:03:22,230 And just think of it as like a Tetris block that's falling 65 00:03:22,230 --> 00:03:23,470 down from above. 66 00:03:23,470 --> 00:03:26,050 And it stacks on top of wherever it lands. 67 00:03:26,050 --> 00:03:28,520 So it'll turn it into something 68 00:03:28,520 --> 00:03:29,770 that looks like this. 69 00:03:29,770 --> 00:03:38,490 70 00:03:38,490 --> 00:03:46,490 So there's already a block of height 1/3 from 0 to 1. 71 00:03:46,490 --> 00:03:47,460 That's from the original 1. 72 00:03:47,460 --> 00:03:50,530 And now we take this, and flip it over, and drop it on top. 73 00:03:50,530 --> 00:03:56,110 Well, this part is going to fall on top of the segment 74 00:03:56,110 --> 00:03:57,320 from 0 to 1. 75 00:03:57,320 --> 00:04:00,090 And then this part gets flipped over and 76 00:04:00,090 --> 00:04:01,070 dropped over here. 77 00:04:01,070 --> 00:04:03,290 And it falls down here. 78 00:04:03,290 --> 00:04:09,610 And so the final PDF actually looks like this kind of 79 00:04:09,610 --> 00:04:15,400 staircase, where this is 2/3 now, because this has two 80 00:04:15,400 --> 00:04:17,640 contributions of 1/3 each, and this is 1/3. 81 00:04:17,640 --> 00:04:20,839 82 00:04:20,839 --> 00:04:26,090 So that is the graphical way of approaching this. 83 00:04:26,090 --> 00:04:37,750 And the PDF for completeness, the PDF of y would be 2/3 for 84 00:04:37,750 --> 00:04:51,440 y between 0 and 1, 1/3 for y from 1 to 2, and 0 otherwise. 85 00:04:51,440 --> 00:04:56,650 86 00:04:56,650 --> 00:04:59,615 All right, so let's move on to part B, and 87 00:04:59,615 --> 00:05:00,540 get some more practice. 88 00:05:00,540 --> 00:05:07,210 Part B, we're given that this PDF of x now is 2 times e to 89 00:05:07,210 --> 00:05:10,810 the negative 2x for x positive, and 0 otherwise. 90 00:05:10,810 --> 00:05:14,030 Now you may just recognize this as an exponential random 91 00:05:14,030 --> 00:05:17,100 variable with a parameter of 2. 92 00:05:17,100 --> 00:05:19,480 And again, we can graph this and see what it looks like. 93 00:05:19,480 --> 00:05:22,640 94 00:05:22,640 --> 00:05:29,240 And it turns out that it's going to start out at 2 and 95 00:05:29,240 --> 00:05:30,490 fall off exponentially. 96 00:05:30,490 --> 00:05:32,520 97 00:05:32,520 --> 00:05:34,990 So in this case, this is actually quite simple. 98 00:05:34,990 --> 00:05:40,280 Because if you look at it, x is already positive. 99 00:05:40,280 --> 00:05:41,930 It doesn't have any negative parts. 100 00:05:41,930 --> 00:05:46,240 So in fact, the absolute value of x is the same as x itself, 101 00:05:46,240 --> 00:05:49,070 because x is never negative. 102 00:05:49,070 --> 00:05:54,620 And so y is just the same thing as x. 103 00:05:54,620 --> 00:06:00,360 And so in this case, actually, the PDF of y is exactly the 104 00:06:00,360 --> 00:06:01,860 same as the PDF of x. 105 00:06:01,860 --> 00:06:06,960 It's just 2e to the minus 2y, for y 106 00:06:06,960 --> 00:06:11,030 positive and zero otherwise. 107 00:06:11,030 --> 00:06:16,690 Now you can also see this graphically also, because to 108 00:06:16,690 --> 00:06:20,630 the left of 0, the negative part, there is no PDF. 109 00:06:20,630 --> 00:06:21,470 The PDF is 0. 110 00:06:21,470 --> 00:06:24,460 And so if you were to take this, flip it over, and drop 111 00:06:24,460 --> 00:06:27,090 it on top, you wouldn't get anything, because there's 112 00:06:27,090 --> 00:06:27,590 nothing there. 113 00:06:27,590 --> 00:06:32,610 And so the entire PDF, even after you take the absolute 114 00:06:32,610 --> 00:06:35,660 value, is just the original one. 115 00:06:35,660 --> 00:06:42,180 So to generalize, what I said at the beginning was that, 116 00:06:42,180 --> 00:06:50,910 remember, the probability in the discrete case, if you 117 00:06:50,910 --> 00:06:53,520 wanted the probability that the absolute value of a random 118 00:06:53,520 --> 00:06:59,630 variable equals something, that would just be the 119 00:06:59,630 --> 00:07:04,400 probability that the random variable equals that value of 120 00:07:04,400 --> 00:07:10,160 little x, or the random variable equals 121 00:07:10,160 --> 00:07:15,450 negative little x. 122 00:07:15,450 --> 00:07:17,870 In either of those two cases, the absolute 123 00:07:17,870 --> 00:07:19,400 value would equal x. 124 00:07:19,400 --> 00:07:20,980 So you get those two contributions. 125 00:07:20,980 --> 00:07:24,300 And so to generalize in the continuous case with PDFs, you 126 00:07:24,300 --> 00:07:26,450 get something that looks very similar. 127 00:07:26,450 --> 00:07:39,790 So in this case, the PDF or y is just the PDF of x at y. 128 00:07:39,790 --> 00:07:48,426 So this is the case where x is just equal to y, plus the PDF 129 00:07:48,426 --> 00:07:50,650 of x evaluated negative y. 130 00:07:50,650 --> 00:07:54,700 So you, again, have both of these two contributions. 131 00:07:54,700 --> 00:08:00,030 And we can rewrite this top one to make 132 00:08:00,030 --> 00:08:03,020 it look more similar. 133 00:08:03,020 --> 00:08:08,840 So the PMF of some discrete [? number ?] y, where this is 134 00:08:08,840 --> 00:08:10,580 a discrete random variable that's equal to the absolute 135 00:08:10,580 --> 00:08:15,760 value of x, would be the PMF of x evaluated at y, plus the 136 00:08:15,760 --> 00:08:19,120 PMF of x evaluated at negative y. 137 00:08:19,120 --> 00:08:21,790 So in both the discrete and continuous cases, you have the 138 00:08:21,790 --> 00:08:24,040 same thing. 139 00:08:24,040 --> 00:08:29,380 So the overall summary of this problem is that, when you take 140 00:08:29,380 --> 00:08:30,070 a transformation-- 141 00:08:30,070 --> 00:08:31,540 in this case, an absolute value-- 142 00:08:31,540 --> 00:08:34,659 you can reason about it and figure out how to decompose 143 00:08:34,659 --> 00:08:37,720 that into arguments about the original random variable, just 144 00:08:37,720 --> 00:08:38,760 plain old x. 145 00:08:38,760 --> 00:08:43,159 And for the specific case of the absolute value, it just 146 00:08:43,159 --> 00:08:47,320 becomes taking a mirror image and popping it on top of what 147 00:08:47,320 --> 00:08:48,280 you originally had. 148 00:08:48,280 --> 00:08:50,780 So remember, you always have these two contributions. 149 00:08:50,780 --> 00:08:53,810 And so if you ever have a random variable that you need 150 00:08:53,810 --> 00:08:55,880 to take an absolute value of, you don't have to be scared. 151 00:08:55,880 --> 00:08:58,790 All you have to do is consider both of these contributions 152 00:08:58,790 --> 00:09:02,982 and add them up, and you have the PDF that you want. 153 00:09:02,982 --> 00:09:04,232 So I'll see you next time. 154 00:09:04,232 --> 00:09:05,350