1 00:00:02,240 --> 00:00:05,620 We now look at an example similar to the previous one, 2 00:00:05,620 --> 00:00:08,750 in which we have again two scenarios, but in which we 3 00:00:08,750 --> 00:00:11,160 have both discrete and continuous 4 00:00:11,160 --> 00:00:13,600 random variables involved. 5 00:00:13,600 --> 00:00:15,620 You have $1 and the opportunity 6 00:00:15,620 --> 00:00:17,200 to play in the lottery. 7 00:00:17,200 --> 00:00:21,730 With probability 1/2, you do nothing and you're left with 8 00:00:21,730 --> 00:00:24,426 the dollar that you started with. 9 00:00:24,426 --> 00:00:28,330 With probability 1/2, you decide to play the lottery. 10 00:00:28,330 --> 00:00:31,780 And in that case, you get back an amount of money which is 11 00:00:31,780 --> 00:00:34,120 random and uniformly distributed 12 00:00:34,120 --> 00:00:37,100 between zero and two. 13 00:00:37,100 --> 00:00:40,170 Is the random variable, X, discrete? 14 00:00:40,170 --> 00:00:44,120 The answer is no, because it takes values on 15 00:00:44,120 --> 00:00:46,830 a continuous range. 16 00:00:46,830 --> 00:00:50,670 Is the random variable, X, continuous? 17 00:00:50,670 --> 00:00:56,790 The answer is no, because the probability that X takes the 18 00:00:56,790 --> 00:01:00,920 value of exactly one is equal to 1/2. 19 00:01:04,209 --> 00:01:07,900 Even though X takes values in a continuous range, this is 20 00:01:07,900 --> 00:01:10,530 not enough to make it a continuous random variable. 21 00:01:10,530 --> 00:01:13,460 We defined continuous random variables to be those that can 22 00:01:13,460 --> 00:01:15,460 be described by a PDF. 23 00:01:15,460 --> 00:01:18,930 And you have seen it in such a case, any individual point 24 00:01:18,930 --> 00:01:20,880 should have zero probability. 25 00:01:20,880 --> 00:01:25,060 But this is not the case here, and so X is not continuous. 26 00:01:25,060 --> 00:01:28,840 We call X a mixed random variable. 27 00:01:28,840 --> 00:01:32,770 More generally, we can have a situation where the random 28 00:01:32,770 --> 00:01:37,300 variable X with some probability is the same as a 29 00:01:37,300 --> 00:01:41,320 particular discrete random variable, and with some other 30 00:01:41,320 --> 00:01:44,140 probability it is equal to some other 31 00:01:44,140 --> 00:01:46,420 continuous random variable. 32 00:01:46,420 --> 00:01:50,550 Such a random variable, X, does not have a PMF because it 33 00:01:50,550 --> 00:01:52,240 is not discrete. 34 00:01:52,240 --> 00:01:56,880 Also, it does not have a PDF because it is not continuous. 35 00:01:56,880 --> 00:01:59,479 How do we describe such a random variable? 36 00:01:59,479 --> 00:02:02,450 Well, we can describe it in terms of a cumulative 37 00:02:02,450 --> 00:02:04,070 distribution function. 38 00:02:04,070 --> 00:02:07,760 CDFs are always well defined for all 39 00:02:07,760 --> 00:02:10,000 kinds of random variables. 40 00:02:10,000 --> 00:02:13,290 We have two scenarios, and so we can use the Total 41 00:02:13,290 --> 00:02:18,890 Probability Theorem and write that the CDF is equal to the 42 00:02:18,890 --> 00:02:23,600 probability of the first scenario, which is p, times 43 00:02:23,600 --> 00:02:27,240 the probability that the random variable Y is less than 44 00:02:27,240 --> 00:02:28,820 or equal to x. 45 00:02:28,820 --> 00:02:31,900 This is a conditional model under the first scenario. 46 00:02:31,900 --> 00:02:34,870 And with some probability, we have the second scenario. 47 00:02:34,870 --> 00:02:38,370 And under that scenario, X will take a value less than 48 00:02:38,370 --> 00:02:42,800 little x, if and only if our random variable Z will take a 49 00:02:42,800 --> 00:02:45,280 value less than little x. 50 00:02:45,280 --> 00:02:51,510 Or in CDF notation, this is p times the CDF of the random 51 00:02:51,510 --> 00:02:57,620 variable Y evaluated at this particular x plus another 52 00:02:57,620 --> 00:03:07,520 weighted term involving the CDF of the random variable Z. 53 00:03:07,520 --> 00:03:11,820 We can also define the expected value of X in a way 54 00:03:11,820 --> 00:03:15,940 that is consistent with the Total Expectation Theorem, 55 00:03:15,940 --> 00:03:20,290 namely define the expected value of X to be the 56 00:03:20,290 --> 00:03:23,850 probability of the first scenario, in which case X is 57 00:03:23,850 --> 00:03:26,870 discrete times the expected value of the associated 58 00:03:26,870 --> 00:03:30,120 discrete random variable, plus the probability of the second 59 00:03:30,120 --> 00:03:34,740 scenario, under which X is continuous, times the expected 60 00:03:34,740 --> 00:03:40,200 value of the associated continuous random variable. 61 00:03:40,200 --> 00:03:42,670 Going back to our original example, we have two 62 00:03:42,670 --> 00:03:49,242 scenarios, the scenarios that we can call A1 and A2. 63 00:03:49,242 --> 00:03:54,110 Under the first scenario, we have a uniform PDF, and the 64 00:03:54,110 --> 00:03:58,010 corresponding CDF is as follows. 65 00:03:58,010 --> 00:04:01,880 It's flat until zero, then it rises linearly. 66 00:04:01,880 --> 00:04:04,880 And then it stays flat, and the value 67 00:04:04,880 --> 00:04:06,900 here is equal to one. 68 00:04:06,900 --> 00:04:11,750 So the slope here is 1/2. 69 00:04:11,750 --> 00:04:15,050 So the slope is equal to the corresponding PDF. 70 00:04:15,050 --> 00:04:18,029 Under the second scenario, we have a discrete, actually a 71 00:04:18,029 --> 00:04:19,529 constant random variable. 72 00:04:19,529 --> 00:04:25,020 And so the CDF is flat at zero until this value, and at that 73 00:04:25,020 --> 00:04:29,800 value we have a jump equal to one. 74 00:04:29,800 --> 00:04:32,690 We then use the Total Probability Theorem, which 75 00:04:32,690 --> 00:04:36,700 tells us that the CDF of the mixed random variable will be 76 00:04:36,700 --> 00:04:41,450 1/2 times the CDF under the first scenario plus 1/2 times 77 00:04:41,450 --> 00:04:43,850 the CDF under the second scenario. 78 00:04:43,850 --> 00:04:48,590 So we take 1/2 of this plot and 1/2 of that plot 79 00:04:48,590 --> 00:04:49,990 and add them up. 80 00:04:49,990 --> 00:04:56,905 What we get is a function that rises now at the slope of 1/4. 81 00:05:01,700 --> 00:05:05,970 Then we have a jump, and the size of that to jump is going 82 00:05:05,970 --> 00:05:09,970 to be equal to 1/2. 83 00:05:09,970 --> 00:05:16,770 And then it continues at a slope of 1/4 until it reaches 84 00:05:16,770 --> 00:05:17,470 this value. 85 00:05:17,470 --> 00:05:19,915 And after that time, it remains flat. 86 00:05:22,540 --> 00:05:25,090 So this is a simple illustration that for mixed 87 00:05:25,090 --> 00:05:27,720 random variables it's not too hard to obtain the 88 00:05:27,720 --> 00:05:31,590 corresponding CDF even though this random variable does not 89 00:05:31,590 --> 00:05:34,370 have a PDF or a PMF of its own.