1 00:00:00,000 --> 00:00:00,490 2 00:00:00,490 --> 00:00:04,410 In this video, we're going to do an example in which we 3 00:00:04,410 --> 00:00:09,090 derive the probability density function of the sum of two 4 00:00:09,090 --> 00:00:10,630 random variables. 5 00:00:10,630 --> 00:00:12,550 The problem tells us the following. 6 00:00:12,550 --> 00:00:16,870 We're given that X and Y are independent random variables. 7 00:00:16,870 --> 00:00:20,035 X is a discrete random variable with PMF Px. 8 00:00:20,035 --> 00:00:23,620 Y is continuous with PDF Fy. 9 00:00:23,620 --> 00:00:26,760 And we'd like to compute the PDF of Z which is equal to X 10 00:00:26,760 --> 00:00:30,750 plus Y. We're going to use the standard approach here-- 11 00:00:30,750 --> 00:00:33,920 compute the CDF of Z and then take the 12 00:00:33,920 --> 00:00:37,060 derivative to get the PDF. 13 00:00:37,060 --> 00:00:49,250 So in this case, the CDF, which is Fz, by definition is 14 00:00:49,250 --> 00:00:51,880 the random variable Z being less than little z. 15 00:00:51,880 --> 00:01:00,860 But Z is just X plus Y. So now, we'd actually like to, 16 00:01:00,860 --> 00:01:03,890 instead of having to deal with two random variables, X and Y, 17 00:01:03,890 --> 00:01:06,130 we'd like to deal with one at a time. 18 00:01:06,130 --> 00:01:12,700 And the total probability theorem allows us to do this 19 00:01:12,700 --> 00:01:16,070 by conditioning on one of the two random variables. 20 00:01:16,070 --> 00:01:18,720 Conditioning on Y here is a bit tricky, because Y is 21 00:01:18,720 --> 00:01:22,160 continuous, and you have to be careful with your definitions. 22 00:01:22,160 --> 00:01:25,290 So conditioning on X seems like the way to go. 23 00:01:25,290 --> 00:01:26,540 So let's do that. 24 00:01:26,540 --> 00:01:38,280 25 00:01:38,280 --> 00:01:41,500 This is just the probability that X equals little x, which 26 00:01:41,500 --> 00:01:46,270 is exactly equal to the PMF of X evaluated at x. 27 00:01:46,270 --> 00:01:50,770 28 00:01:50,770 --> 00:01:57,570 Now we're given we're fixing X equal to little x. 29 00:01:57,570 --> 00:02:00,320 So we can actually replace every instance of the random 30 00:02:00,320 --> 00:02:01,570 variable with little x. 31 00:02:01,570 --> 00:02:13,860 32 00:02:13,860 --> 00:02:16,630 And now I'm going to just rearrange this so that it 33 00:02:16,630 --> 00:02:17,610 looks a little nicer. 34 00:02:17,610 --> 00:02:21,770 So I'm going to have Y on the left and say Y is less than z 35 00:02:21,770 --> 00:02:29,420 minus x, where z minus x is just a constant. 36 00:02:29,420 --> 00:02:32,820 Now, remember that X and Y are independent. 37 00:02:32,820 --> 00:02:37,120 So telling us something about X shouldn't change our beliefs 38 00:02:37,120 --> 00:02:39,920 about Y. So in this case, we can actually drop the 39 00:02:39,920 --> 00:02:41,170 conditioning. 40 00:02:41,170 --> 00:02:51,140 41 00:02:51,140 --> 00:02:55,293 And this is exactly the CDF of Y evaluated at z minus x. 42 00:02:55,293 --> 00:02:59,990 43 00:02:59,990 --> 00:03:05,710 So now we've simplified as far as we could. 44 00:03:05,710 --> 00:03:08,330 So let's take the derivative and see where that takes us. 45 00:03:08,330 --> 00:03:13,230 46 00:03:13,230 --> 00:03:19,250 So the PDF of Z is, by definition, the derivative of 47 00:03:19,250 --> 00:03:27,850 the CDF, which we just computed here. 48 00:03:27,850 --> 00:03:36,723 This is sum over x Fy z minus x Px. 49 00:03:36,723 --> 00:03:38,940 What next? 50 00:03:38,940 --> 00:03:41,375 Interchange the derivative and the summation. 51 00:03:41,375 --> 00:03:51,920 52 00:03:51,920 --> 00:03:54,690 And a note of caution here. 53 00:03:54,690 --> 00:03:57,960 54 00:03:57,960 --> 00:04:02,200 So if x took on a finite number of values, you'd have a 55 00:04:02,200 --> 00:04:04,260 finite number of terms here. 56 00:04:04,260 --> 00:04:07,030 And this would be completely valid. 57 00:04:07,030 --> 00:04:10,050 You can just do this. 58 00:04:10,050 --> 00:04:13,440 But if x took on, for example, a countably 59 00:04:13,440 --> 00:04:15,000 infinite number of values-- 60 00:04:15,000 --> 00:04:18,200 a geometric random variable, for example-- 61 00:04:18,200 --> 00:04:22,059 this would actually require some formal justification. 62 00:04:22,059 --> 00:04:25,760 But I'm not going to get into that. 63 00:04:25,760 --> 00:04:31,420 So here, the derivative with respect to z-- this is 64 00:04:31,420 --> 00:04:33,030 actually z-- 65 00:04:33,030 --> 00:04:35,520 is you use chain rule here. 66 00:04:35,520 --> 00:04:41,450 Px doesn't matter, because it's not a function of z. 67 00:04:41,450 --> 00:04:48,046 So we have Fy evaluated at z minus x according to the chain 68 00:04:48,046 --> 00:04:50,480 rule, and then the derivative of the inner quantity, z minus 69 00:04:50,480 --> 00:04:52,180 x, which is just 1. 70 00:04:52,180 --> 00:04:55,170 So we don't need to put anything there. 71 00:04:55,170 --> 00:04:59,490 And we get Px of x. 72 00:04:59,490 --> 00:05:00,490 So there we go. 73 00:05:00,490 --> 00:05:04,130 We've derived the PDF of z. 74 00:05:04,130 --> 00:05:07,650 Notice that this looks quite similar to the convolution 75 00:05:07,650 --> 00:05:12,040 formula when you assume that both X and Y are either 76 00:05:12,040 --> 00:05:13,690 continuous or discrete. 77 00:05:13,690 --> 00:05:18,710 And so that tells us that this looks right. 78 00:05:18,710 --> 00:05:25,390 So in summary, we've basically computed the PDF of X plus Y 79 00:05:25,390 --> 00:05:29,300 where X is discrete and Y is continuous. 80 00:05:29,300 --> 00:05:32,120 And we've used the standard two-step approach-- 81 00:05:32,120 --> 00:05:35,950 compute the CDF and then take the derivative to get the PDF. 82 00:05:35,950 --> 00:05:37,200