1 00:00:00,000 --> 00:00:01,100 2 00:00:01,100 --> 00:00:05,390 In this video, we'll look at an example in which we compute 3 00:00:05,390 --> 00:00:09,900 the expectation and cumulative density function of a mixed 4 00:00:09,900 --> 00:00:11,730 random variable. 5 00:00:11,730 --> 00:00:14,500 The problem is as follows. 6 00:00:14,500 --> 00:00:18,480 Al arrives at some bus stand or taxi stand at a given 7 00:00:18,480 --> 00:00:21,700 time-- let's say time t equals 0. 8 00:00:21,700 --> 00:00:25,500 He finds a taxi waiting for him with probability 2/3 in 9 00:00:25,500 --> 00:00:27,600 which he takes it. 10 00:00:27,600 --> 00:00:31,470 Otherwise, he takes the next arriving taxi or bus. 11 00:00:31,470 --> 00:00:34,080 12 00:00:34,080 --> 00:00:36,540 The time that the next taxi arrives between 0 and 10 13 00:00:36,540 --> 00:00:38,742 minutes, and it's uniformly distributed. 14 00:00:38,742 --> 00:00:42,111 The next bus leaves exactly in 5 minutes. 15 00:00:42,111 --> 00:00:46,920 So the question is, if X is Al's waiting time, what is the 16 00:00:46,920 --> 00:00:51,800 CDF and expectation of X? 17 00:00:51,800 --> 00:00:57,010 So one way to view this problem that's convenient is 18 00:00:57,010 --> 00:00:58,170 the tree structure. 19 00:00:58,170 --> 00:01:00,950 So I've drawn it for you here in which the events of 20 00:01:00,950 --> 00:01:05,920 interest are B1, B2, and B3, B1 being Al catches the 21 00:01:05,920 --> 00:01:10,370 waiting taxi, B2 being Al catches the next taxi, which 22 00:01:10,370 --> 00:01:15,460 arrives between 0 and 5 minutes, and B3 being Al 23 00:01:15,460 --> 00:01:18,290 catches the bus at the time t plus 5. 24 00:01:18,290 --> 00:01:20,850 Notice that these three events are disjoint. 25 00:01:20,850 --> 00:01:25,050 So Al catching the waiting taxi means he can't catch the 26 00:01:25,050 --> 00:01:28,840 bus or the next arriving taxi. 27 00:01:28,840 --> 00:01:32,920 And it also covers the entire set of outcomes. 28 00:01:32,920 --> 00:01:37,550 So, in fact, B1, B2, and B3 are a partition. 29 00:01:37,550 --> 00:01:40,845 So let's look at the relevant probabilities. 30 00:01:40,845 --> 00:01:43,740 31 00:01:43,740 --> 00:01:48,870 Whether or not B1 happens depends on whether or not the 32 00:01:48,870 --> 00:01:51,710 taxi's waiting for Al. 33 00:01:51,710 --> 00:01:56,680 So if the taxi is waiting for him, which happens with 2/3 34 00:01:56,680 --> 00:01:59,950 probability, B1 happens. 35 00:01:59,950 --> 00:02:03,450 Otherwise, with 1/3 probability, we see whether or 36 00:02:03,450 --> 00:02:07,130 not a taxi is going to arrive between 0 and 5 minutes. 37 00:02:07,130 --> 00:02:11,800 If it arrives, which is going to happen with what 38 00:02:11,800 --> 00:02:12,340 probability? 39 00:02:12,340 --> 00:02:15,160 Well, we know that the next taxi is going to arrive 40 00:02:15,160 --> 00:02:17,030 between 0 and 10 minutes uniform. 41 00:02:17,030 --> 00:02:18,610 It's a uniform distribution. 42 00:02:18,610 --> 00:02:21,300 And so half the mass is going to be between 0 and 5. 43 00:02:21,300 --> 00:02:23,930 And the other half is going to be between 5 and 10. 44 00:02:23,930 --> 00:02:29,050 And so this is going to be 1/2 and 1/2. 45 00:02:29,050 --> 00:02:33,190 And let's look at what X looks like. 46 00:02:33,190 --> 00:02:36,720 If B1 happens, Al isn't waiting at all, so x is going 47 00:02:36,720 --> 00:02:38,436 to be equal to 0. 48 00:02:38,436 --> 00:02:41,950 If B3 happens, which is the other easy case, Al's going to 49 00:02:41,950 --> 00:02:44,360 be waiting for 5 minutes exactly. 50 00:02:44,360 --> 00:02:48,985 And if B2 happens, well, it's going to be some value 51 00:02:48,985 --> 00:02:50,640 between 0 and 5. 52 00:02:50,640 --> 00:02:52,828 We can actually draw the density, so let's see if we 53 00:02:52,828 --> 00:02:55,070 can do that here. 54 00:02:55,070 --> 00:03:06,570 55 00:03:06,570 --> 00:03:12,895 The original next taxi was uniformly distributed 56 00:03:12,895 --> 00:03:14,204 between 0 and 10. 57 00:03:14,204 --> 00:03:17,100 58 00:03:17,100 --> 00:03:19,030 But now, we're told two pieces of information. 59 00:03:19,030 --> 00:03:21,690 We're told that B2 happens, which means that there's no 60 00:03:21,690 --> 00:03:24,190 taxi waiting, and the next taxi arrives 61 00:03:24,190 --> 00:03:25,300 between 0 and 5 minutes. 62 00:03:25,300 --> 00:03:28,300 Well, the fact that there was no taxi waiting has no bearing 63 00:03:28,300 --> 00:03:30,090 on that density. 64 00:03:30,090 --> 00:03:33,450 But the fact that the next taxi arrives between 0 and 5 65 00:03:33,450 --> 00:03:36,300 does make a difference, because the density then is 66 00:03:36,300 --> 00:03:42,925 going to be definitely 0 in any region outside 0 and 5. 67 00:03:42,925 --> 00:03:45,070 Now, the question is, how is it going to look 68 00:03:45,070 --> 00:03:46,930 between 0 and 5? 69 00:03:46,930 --> 00:03:48,440 Well, it's not going to look crazy. 70 00:03:48,440 --> 00:03:49,800 It's not going to look like something different. 71 00:03:49,800 --> 00:03:54,100 It's simply going to be a scale version of the original 72 00:03:54,100 --> 00:03:56,970 density between 0 and 5. 73 00:03:56,970 --> 00:04:03,185 You can verify this by looking at the actual formula for when 74 00:04:03,185 --> 00:04:07,550 you condition events on a random variable. 75 00:04:07,550 --> 00:04:10,280 Here, it's going to be 1/5 in order for this to 76 00:04:10,280 --> 00:04:11,530 integrate out to 1. 77 00:04:11,530 --> 00:04:14,340 78 00:04:14,340 --> 00:04:18,610 And now we can jump right into figuring out the expectation. 79 00:04:18,610 --> 00:04:21,260 Now, notice that X is actually a mixed random variable? 80 00:04:21,260 --> 00:04:22,060 What does that mean? 81 00:04:22,060 --> 00:04:26,820 Well, X either takes on values according to either a discrete 82 00:04:26,820 --> 00:04:29,040 probability law or a continuous one. 83 00:04:29,040 --> 00:04:32,607 So if B1 happens, for example, X is going to be exactly equal 84 00:04:32,607 --> 00:04:35,660 to 0 with probability 1, which is a 85 00:04:35,660 --> 00:04:38,200 discrete probability problem. 86 00:04:38,200 --> 00:04:43,480 On the other hand, if B2 happens, then the value of X 87 00:04:43,480 --> 00:04:47,525 depends on the density, which is going to be continuous. 88 00:04:47,525 --> 00:04:48,960 So X is going to be a continuous 89 00:04:48,960 --> 00:04:51,730 random variable here. 90 00:04:51,730 --> 00:04:56,540 So how do you define an expectation in this case? 91 00:04:56,540 --> 00:05:04,680 Well, you can do it so that it satisfies the total 92 00:05:04,680 --> 00:05:11,350 expectation theorem, which means that the expectation of 93 00:05:11,350 --> 00:05:18,320 X is the probability of B1 times the expectation given B1 94 00:05:18,320 --> 00:05:27,220 plus the probability of B2 times the expectation given B2 95 00:05:27,220 --> 00:05:31,121 plus the probability of B3 times the 96 00:05:31,121 --> 00:05:33,220 expectation given B3. 97 00:05:33,220 --> 00:05:39,340 98 00:05:39,340 --> 00:05:41,530 So this will satisfy the total expectation theorem. 99 00:05:41,530 --> 00:05:45,180 100 00:05:45,180 --> 00:05:50,436 So the probability of B1 is going to be exactly 2/3. 101 00:05:50,436 --> 00:05:53,390 102 00:05:53,390 --> 00:05:59,320 It's simply the probability of a taxi waiting for Al. 103 00:05:59,320 --> 00:06:03,442 The expected value of X-- well, when B1 happens, X is 104 00:06:03,442 --> 00:06:04,970 going to be exactly equal to 0. 105 00:06:04,970 --> 00:06:08,740 So the expected value is also going to be 0. 106 00:06:08,740 --> 00:06:11,600 The probability of B2 happening is the probability 107 00:06:11,600 --> 00:06:15,050 of a taxi not being there times the probability of a 108 00:06:15,050 --> 00:06:17,520 taxi arriving between 0 and 5. 109 00:06:17,520 --> 00:06:22,650 It's going to be 1/3 times 1/2. 110 00:06:22,650 --> 00:06:29,420 And the expected value of X given B2 is going to be the 111 00:06:29,420 --> 00:06:31,780 expected value of this density. 112 00:06:31,780 --> 00:06:34,173 The expected value of this density is the midpoint 113 00:06:34,173 --> 00:06:35,770 between 0 and 5. 114 00:06:35,770 --> 00:06:37,670 And so it's going to be 5/2. 115 00:06:37,670 --> 00:06:40,470 116 00:06:40,470 --> 00:06:44,170 And the probability of B3 is going to be 1/3 times 1/2. 117 00:06:44,170 --> 00:06:48,090 118 00:06:48,090 --> 00:06:51,260 Finally, the expected value of X given B3. 119 00:06:51,260 --> 00:06:54,400 Well, when B3 happens, X is going to be 120 00:06:54,400 --> 00:06:57,250 exactly equal to 5. 121 00:06:57,250 --> 00:07:01,810 So the expected value is also going to be 5. 122 00:07:01,810 --> 00:07:08,820 Now we're left with 5/12 plus 5/6, which 123 00:07:08,820 --> 00:07:11,770 is going to be 15/12. 124 00:07:11,770 --> 00:07:14,526 125 00:07:14,526 --> 00:07:19,290 And we can actually fill that in here so that we can clear 126 00:07:19,290 --> 00:07:23,650 up the board to do the other part. 127 00:07:23,650 --> 00:07:28,980 128 00:07:28,980 --> 00:07:36,520 Now we want to compute the CDF of X. Well, what is the CDF? 129 00:07:36,520 --> 00:07:41,120 Well, the CDF of X is going to be equal to the probability 130 00:07:41,120 --> 00:07:43,460 that the random variable X is less than or equal 131 00:07:43,460 --> 00:07:45,235 to some little x. 132 00:07:45,235 --> 00:07:48,650 It's a constant [INAUDIBLE]. 133 00:07:48,650 --> 00:07:55,120 Before we jump right in, let's try to understand what's the 134 00:07:55,120 --> 00:07:57,226 form of the CDF. 135 00:07:57,226 --> 00:08:00,220 And let's consider some interesting cases. 136 00:08:00,220 --> 00:08:02,790 You know that the random variable X, the waiting time, 137 00:08:02,790 --> 00:08:06,400 is going to be somewhere between 0 and 5, right? 138 00:08:06,400 --> 00:08:12,230 So let's consider what happens if little x is going to be 139 00:08:12,230 --> 00:08:13,690 less than 0. 140 00:08:13,690 --> 00:08:17,410 That's basically saying, what's the probability of the 141 00:08:17,410 --> 00:08:20,160 random variable X being less than some number 142 00:08:20,160 --> 00:08:23,010 that's less than 0? 143 00:08:23,010 --> 00:08:27,070 Waiting time can't be negative, so the probablility 144 00:08:27,070 --> 00:08:31,030 of this is going to be 0. 145 00:08:31,030 --> 00:08:37,400 Now, what if X is between equaling 0 and 146 00:08:37,400 --> 00:08:38,750 strictly less than 5? 147 00:08:38,750 --> 00:08:41,740 148 00:08:41,740 --> 00:08:47,640 In that case, either X can fall between 0 and 5 according 149 00:08:47,640 --> 00:08:52,000 to this case, in the case of B2, or X can be 150 00:08:52,000 --> 00:08:52,760 exactly equal to 0. 151 00:08:52,760 --> 00:08:54,800 It's not clear. 152 00:08:54,800 --> 00:08:57,461 So let's do that later. 153 00:08:57,461 --> 00:08:58,980 Let's fill that in later. 154 00:08:58,980 --> 00:09:05,702 What about if x is greater than or equal to 5? 155 00:09:05,702 --> 00:09:07,130 Little x, right? 156 00:09:07,130 --> 00:09:10,600 That's the probability that the random variable X is less 157 00:09:10,600 --> 00:09:14,060 than some number that's bigger than or equal to 5. 158 00:09:14,060 --> 00:09:17,733 The waiting time X, the random variable, is definitely going 159 00:09:17,733 --> 00:09:21,812 to be less than or equal to 5, so the probability of this is 160 00:09:21,812 --> 00:09:23,062 going to be 1. 161 00:09:23,062 --> 00:09:27,130 162 00:09:27,130 --> 00:09:30,160 So now this case. 163 00:09:30,160 --> 00:09:31,720 How do we do it? 164 00:09:31,720 --> 00:09:34,940 Well, let's try to use a similar kind of approach that 165 00:09:34,940 --> 00:09:39,940 we did for the expected value and use the total probability 166 00:09:39,940 --> 00:09:41,290 theorem in this case. 167 00:09:41,290 --> 00:09:44,350 So let's try to review this. 168 00:09:44,350 --> 00:09:51,960 First of all, let's assume that this is true, that little 169 00:09:51,960 --> 00:09:55,500 x is between 0 and 5, including 0. 170 00:09:55,500 --> 00:09:58,230 171 00:09:58,230 --> 00:10:04,900 And let's use the total probability theorem, and use 172 00:10:04,900 --> 00:10:08,502 the partitions B1, B2, and B3. 173 00:10:08,502 --> 00:10:34,080 174 00:10:34,080 --> 00:10:37,770 So what's the probability of B1? 175 00:10:37,770 --> 00:10:42,300 It's the probability that Al catches waiting taxi, which 176 00:10:42,300 --> 00:10:43,890 happens with probability 2/3. 177 00:10:43,890 --> 00:10:46,600 178 00:10:46,600 --> 00:10:50,460 What's the probability that the random variable X, which 179 00:10:50,460 --> 00:10:55,560 is less than or equal to little x under this condition, 180 00:10:55,560 --> 00:10:56,220 when B1 happens? 181 00:10:56,220 --> 00:11:00,010 Well, if B1 happens, then random variable X is going to 182 00:11:00,010 --> 00:11:02,680 be exactly equal to 0, right? 183 00:11:02,680 --> 00:11:06,286 So in that case, it's definitely going to be less 184 00:11:06,286 --> 00:11:11,630 than or equal to any value of x, including 0. 185 00:11:11,630 --> 00:11:16,010 So the probability will be 1. 186 00:11:16,010 --> 00:11:21,280 What's the probability that B2 happens now? 187 00:11:21,280 --> 00:11:24,912 The probability that B2 happens is 1/3 times 1/2, as 188 00:11:24,912 --> 00:11:26,162 we did before. 189 00:11:26,162 --> 00:11:29,500 190 00:11:29,500 --> 00:11:33,220 And the probability that the random variable X is less than 191 00:11:33,220 --> 00:11:36,970 or equal to little x when B2 happens. 192 00:11:36,970 --> 00:11:39,800 Well, if B2 happens, this is your density. 193 00:11:39,800 --> 00:11:41,430 And this is our condition. 194 00:11:41,430 --> 00:11:42,980 And so x is going to be somewhere in 195 00:11:42,980 --> 00:11:45,330 between these spots. 196 00:11:45,330 --> 00:11:47,920 And we'd like to compute what's the probably that 197 00:11:47,920 --> 00:11:51,180 random variable X is less than or equal to little x. 198 00:11:51,180 --> 00:11:53,470 So we want this area. 199 00:11:53,470 --> 00:11:59,980 And that area is going to have height of 1/5 and width of x. 200 00:11:59,980 --> 00:12:04,580 And so the area's going to be 1/5 times x. 201 00:12:04,580 --> 00:12:08,818 And finally, the probability that B3 happens is going to be 202 00:12:08,818 --> 00:12:15,840 1/3 times 1/2 again times the probability that the random 203 00:12:15,840 --> 00:12:18,500 variable X is less than or equal to little x given B3. 204 00:12:18,500 --> 00:12:23,216 Well, when B3 happens, X is going to be exactly 5 as a 205 00:12:23,216 --> 00:12:24,650 random variable. 206 00:12:24,650 --> 00:12:28,440 But little x, you know-- we're assuming in this condition-- 207 00:12:28,440 --> 00:12:32,920 is going to be between 0 and 5, but strictly less than 5. 208 00:12:32,920 --> 00:12:38,220 So there's no way that if the random variable X is 5 and 209 00:12:38,220 --> 00:12:40,410 this is strictly less than 5, this is going to be true. 210 00:12:40,410 --> 00:12:43,380 And so that probability will be 0. 211 00:12:43,380 --> 00:12:48,455 So we're now left with 2/3 plus 1/30. 212 00:12:48,455 --> 00:12:53,090 213 00:12:53,090 --> 00:12:56,470 And now we can fill this in. 214 00:12:56,470 --> 00:12:59,730 2/3 plus 1/30 x. 215 00:12:59,730 --> 00:13:02,260 216 00:13:02,260 --> 00:13:07,910 And this is our CDF. 217 00:13:07,910 --> 00:13:13,370 So now we've finished the problem, computed the expected 218 00:13:13,370 --> 00:13:18,880 value here and then the CDF here, and this was a great 219 00:13:18,880 --> 00:13:22,040 illustration of how you would do so for a 220 00:13:22,040 --> 00:13:23,290 mixed random variable. 221 00:13:23,290 --> 00:13:24,190