WEBVTT 00:00:00.000 --> 00:00:12.750 align:middle line:84% [SQUEAKING] [RUSTLING] [CLICKING] 00:00:12.750 --> 00:00:14.400 align:middle line:84% YUFEI ZHAO: In this video, let us 00:00:14.400 --> 00:00:18.090 align:middle line:84% look at a basic yet important concept in probability known 00:00:18.090 --> 00:00:20.880 align:middle line:84% as linearity of expectations and use 00:00:20.880 --> 00:00:24.960 align:middle line:84% it to deduce some interesting consequences in combinatorics 00:00:24.960 --> 00:00:27.900 align:middle line:90% via the probabilistic method. 00:00:27.900 --> 00:00:31.290 align:middle line:84% Linearity of expectations says that, if you 00:00:31.290 --> 00:00:36.300 align:middle line:84% are given random variables, x1 to xn, and constants, 00:00:36.300 --> 00:00:42.030 align:middle line:84% C1 through Cn, then when we take a linear combination 00:00:42.030 --> 00:00:46.380 align:middle line:84% of these random variables, as such, C1 times x1-- 00:00:46.380 --> 00:00:49.650 align:middle line:84% so imagine these are real valued random variables and these are 00:00:49.650 --> 00:00:56.070 align:middle line:84% real constants, for instance-- then the sum of these C1 times 00:00:56.070 --> 00:01:01.740 align:middle line:84% x1 plus C2 times x2 and so on plus Cn times xn-- 00:01:01.740 --> 00:01:04.980 align:middle line:90% so this sum has expectation-- 00:01:04.980 --> 00:01:07.080 align:middle line:84% the following, which can be computed 00:01:07.080 --> 00:01:11.040 align:middle line:84% by distributing this expectation symbol across 00:01:11.040 --> 00:01:14.350 align:middle line:90% to the individual variables. 00:01:14.350 --> 00:01:17.030 align:middle line:84% So this is a basic and important property. 00:01:17.030 --> 00:01:21.190 align:middle line:84% And it's worth noting that a similar statement written 00:01:21.190 --> 00:01:23.270 align:middle line:90% for products is often not true. 00:01:23.270 --> 00:01:26.320 align:middle line:84% So it is not usually the case that the expectation 00:01:26.320 --> 00:01:28.990 align:middle line:84% of a product of two random variables 00:01:28.990 --> 00:01:32.120 align:middle line:84% is the product of their expectations, 00:01:32.120 --> 00:01:34.430 align:middle line:84% unless you're in some special circumstances, 00:01:34.430 --> 00:01:38.530 align:middle line:84% such as when x and y are independent or uncorrelated. 00:01:38.530 --> 00:01:42.130 align:middle line:84% Anyway, let us focus on the linearity of expectations 00:01:42.130 --> 00:01:46.360 align:middle line:84% and see some ways to use this in combinatorial applications. 00:01:46.360 --> 00:01:50.390 align:middle line:84% And the first example is the following question, 00:01:50.390 --> 00:01:53.980 align:middle line:84% which will be able to give a very quick and clean answer. 00:01:53.980 --> 00:01:56.530 align:middle line:84% The question is, what is the average number 00:01:56.530 --> 00:02:00.340 align:middle line:84% of fixed points of a permutation of 1 through n 00:02:00.340 --> 00:02:03.190 align:middle line:90% chosen uniformly at random? 00:02:03.190 --> 00:02:08.740 align:middle line:84% As a reminder, if we have numbers 1, 2, 3, 4-- here, 00:02:08.740 --> 00:02:09.860 align:middle line:90% n equals to 4-- 00:02:09.860 --> 00:02:14.300 align:middle line:84% then a permutation can be thought as some way 00:02:14.300 --> 00:02:17.990 align:middle line:84% to map these numbers, the sets back to themselves, 00:02:17.990 --> 00:02:20.210 align:middle line:90% in one to one correspondence. 00:02:20.210 --> 00:02:28.910 align:middle line:84% And a fixed point is some number that 00:02:28.910 --> 00:02:31.280 align:middle line:90% gets mapped back to themselves. 00:02:31.280 --> 00:02:36.700 align:middle line:84% So in this permutation, there are exactly two fixed points. 00:02:36.700 --> 00:02:39.370 align:middle line:90% OK, great. 00:02:39.370 --> 00:02:42.586 align:middle line:84% Let's answer this question using linearity of expectations. 00:02:42.586 --> 00:02:45.210 align:middle line:90% 00:02:45.210 --> 00:02:46.800 align:middle line:84% One way to think about the problem 00:02:46.800 --> 00:02:50.550 align:middle line:84% is that there are n factorial permutations. 00:02:50.550 --> 00:02:54.073 align:middle line:84% And maybe we want to try to go about counting how many of them 00:02:54.073 --> 00:02:55.740 align:middle line:84% have zero fixed points, how many of them 00:02:55.740 --> 00:02:57.990 align:middle line:84% have one fixed points, how many have two fixed points. 00:02:57.990 --> 00:03:01.270 align:middle line:84% But that method can get pretty cumbersome and pretty difficult, 00:03:01.270 --> 00:03:03.270 align:middle line:90% quite quickly. 00:03:03.270 --> 00:03:05.010 align:middle line:84% However, if we look at this problem 00:03:05.010 --> 00:03:08.080 align:middle line:84% through the lens of linearity of expectations, 00:03:08.080 --> 00:03:11.910 align:middle line:84% there turns out to be a very quick solution. 00:03:11.910 --> 00:03:16.210 align:middle line:84% And the method is to introduce some random variables. 00:03:16.210 --> 00:03:20.160 align:middle line:84% Let xi be a random variable that equals 00:03:20.160 --> 00:03:30.910 align:middle line:84% to 1 if i is a fixed point, meaning that in this permutation 00:03:30.910 --> 00:03:35.270 align:middle line:84% i gets mapped to itself and 0 otherwise. 00:03:35.270 --> 00:03:38.470 align:middle line:90% 00:03:38.470 --> 00:03:42.820 align:middle line:84% So x sub i is the indicator random variable 00:03:42.820 --> 00:03:45.130 align:middle line:84% for the element i being a fixed point 00:03:45.130 --> 00:03:46.970 align:middle line:90% of this random permutation. 00:03:46.970 --> 00:03:51.350 align:middle line:90% 00:03:51.350 --> 00:03:54.390 align:middle line:84% What's the expectation of xi Well, 00:03:54.390 --> 00:03:59.215 align:middle line:84% this is the probability that i is a fixed point. 00:03:59.215 --> 00:04:05.290 align:middle line:90% 00:04:05.290 --> 00:04:08.410 align:middle line:84% Well, this being a uniform permutation chosen uniformly 00:04:08.410 --> 00:04:12.070 align:middle line:84% at random, i is sent to each of the elements 1 through n 00:04:12.070 --> 00:04:13.670 align:middle line:90% with equal probabilities. 00:04:13.670 --> 00:04:15.580 align:middle line:84% So in particular, it is sent back to itself 00:04:15.580 --> 00:04:18.310 align:middle line:90% with probability 1 over n. 00:04:18.310 --> 00:04:28.100 align:middle line:84% And then the number of fixed points equals to the sum 00:04:28.100 --> 00:04:32.830 align:middle line:90% of the xi's-- 00:04:32.830 --> 00:04:36.160 align:middle line:90% so x1 plus x2, and so on, to xn. 00:04:36.160 --> 00:04:42.790 align:middle line:84% And now, we can take expectation on both sides 00:04:42.790 --> 00:04:51.600 align:middle line:84% and apply linearity of expectations 00:04:51.600 --> 00:04:57.230 align:middle line:84% and see that, well, each individual term is 1 over n 00:04:57.230 --> 00:04:58.870 align:middle line:90% and there are n such terms. 00:04:58.870 --> 00:05:00.300 align:middle line:90% So the answer is 1. 00:05:00.300 --> 00:05:03.870 align:middle line:84% And that is the answer to this question. 00:05:03.870 --> 00:05:05.610 align:middle line:84% So the average number of fixed points 00:05:05.610 --> 00:05:08.220 align:middle line:84% of a permutation chosen uniformly at random 00:05:08.220 --> 00:05:09.838 align:middle line:90% is exactly 1. 00:05:09.838 --> 00:05:11.880 align:middle line:84% And you see that there's a very quick calculation 00:05:11.880 --> 00:05:14.430 align:middle line:84% once you get the hang of the idea of linearity 00:05:14.430 --> 00:05:17.130 align:middle line:90% of expectations. 00:05:17.130 --> 00:05:20.640 align:middle line:84% Let us look at a slightly more interesting example. 00:05:20.640 --> 00:05:23.520 align:middle line:84% And for this example, we'll consider 00:05:23.520 --> 00:05:25.180 align:middle line:90% the concept of a tournament. 00:05:25.180 --> 00:05:31.070 align:middle line:90% 00:05:31.070 --> 00:05:35.180 align:middle line:84% So a tournament is a concept in graph theory referring 00:05:35.180 --> 00:05:38.750 align:middle line:90% to the following. 00:05:38.750 --> 00:05:40.220 align:middle line:90% There are n vertices. 00:05:40.220 --> 00:05:42.420 align:middle line:84% Think of n players in some tournament. 00:05:42.420 --> 00:05:44.780 align:middle line:84% And between every pair of vertices, 00:05:44.780 --> 00:05:47.870 align:middle line:84% we have some directed edge pointing 00:05:47.870 --> 00:05:50.180 align:middle line:90% in one of these two directions. 00:05:50.180 --> 00:05:53.660 align:middle line:84% So every pair of vertices, there is some edge pointing 00:05:53.660 --> 00:05:55.486 align:middle line:90% to one of the two directions. 00:05:55.486 --> 00:06:04.510 align:middle line:90% 00:06:04.510 --> 00:06:08.860 align:middle line:84% OK, so that's an example of a tournament. 00:06:08.860 --> 00:06:11.440 align:middle line:84% I'll need to introduce another concept, which 00:06:11.440 --> 00:06:12.675 align:middle line:90% is that of a Hamilton path. 00:06:12.675 --> 00:06:23.870 align:middle line:90% 00:06:23.870 --> 00:06:28.700 align:middle line:84% So a Hamilton path is a directed path, 00:06:28.700 --> 00:06:32.510 align:middle line:84% meaning we travel along the directions of these edges 00:06:32.510 --> 00:06:34.090 align:middle line:90% according to their directions. 00:06:34.090 --> 00:06:46.240 align:middle line:84% So directed path that passes through each vertex, so 00:06:46.240 --> 00:06:51.045 align:middle line:84% every vertex of the graph exactly once, no more, no less. 00:06:51.045 --> 00:06:54.000 align:middle line:90% 00:06:54.000 --> 00:07:00.160 align:middle line:84% So let's see if we can find any Hamilton paths in this example. 00:07:00.160 --> 00:07:01.320 align:middle line:90% Well, I see one. 00:07:01.320 --> 00:07:03.450 align:middle line:84% So if we start with a middle vertex 00:07:03.450 --> 00:07:07.950 align:middle line:84% and go along this one, this edge, then this edge, and then 00:07:07.950 --> 00:07:10.560 align:middle line:84% this edge, so that's a Hamilton path 00:07:10.560 --> 00:07:14.820 align:middle line:84% that goes through all four vertices, each vertex passing 00:07:14.820 --> 00:07:16.210 align:middle line:90% through it exactly once. 00:07:16.210 --> 00:07:18.815 align:middle line:84% And it always traverses along the direction of the edges. 00:07:18.815 --> 00:07:23.830 align:middle line:90% 00:07:23.830 --> 00:07:25.410 align:middle line:84% Let us prove the following theorem. 00:07:25.410 --> 00:07:28.640 align:middle line:90% 00:07:28.640 --> 00:07:48.186 align:middle line:84% For every n, there exists a tournament 00:07:48.186 --> 00:08:03.550 align:middle line:84% of n vertices with at least n factorial times 2 00:08:03.550 --> 00:08:10.935 align:middle line:84% to the minus n plus 1 Hamilton paths. 00:08:10.935 --> 00:08:20.160 align:middle line:90% 00:08:20.160 --> 00:08:24.320 align:middle line:84% So in other words, for every n, there 00:08:24.320 --> 00:08:26.930 align:middle line:84% is some way to orient the edges of the complete graph 00:08:26.930 --> 00:08:29.960 align:middle line:84% on n vertices so that it has lots 00:08:29.960 --> 00:08:34.669 align:middle line:84% and lots of Hamilton paths, specifically at least this many. 00:08:34.669 --> 00:08:36.790 align:middle line:84% So that's a theorem that we're aiming to prove. 00:08:36.790 --> 00:08:39.640 align:middle line:90% 00:08:39.640 --> 00:08:42.159 align:middle line:84% We will not prove this theorem by explicitly 00:08:42.159 --> 00:08:43.880 align:middle line:90% constructing such a tournament. 00:08:43.880 --> 00:08:46.480 align:middle line:84% Instead, we'll invoke the probabilistic method 00:08:46.480 --> 00:08:50.620 align:middle line:84% and show that a random tournament has an expectation 00:08:50.620 --> 00:08:53.800 align:middle line:90% of this property. 00:08:53.800 --> 00:08:56.470 align:middle line:90% So here's the proof. 00:08:56.470 --> 00:09:09.020 align:middle line:84% Let's consider a random tournament 00:09:09.020 --> 00:09:16.905 align:middle line:84% on n vertices chosen uniformly at random. 00:09:16.905 --> 00:09:25.830 align:middle line:90% 00:09:25.830 --> 00:09:29.770 align:middle line:84% One way to do this is to take a complete graph on n vertices 00:09:29.770 --> 00:09:32.340 align:middle line:84% and for every edge flip a fair coin 00:09:32.340 --> 00:09:35.550 align:middle line:84% and use that coin to decide which way the edge orients 00:09:35.550 --> 00:09:37.360 align:middle line:90% of the two different directions. 00:09:37.360 --> 00:09:40.370 align:middle line:90% 00:09:40.370 --> 00:09:43.250 align:middle line:84% Now, let's think about the number of Hamilton paths. 00:09:43.250 --> 00:09:50.960 align:middle line:90% So each path of the-- 00:09:50.960 --> 00:10:05.430 align:middle line:84% so each-- so each of the n factorial permutations 00:10:05.430 --> 00:10:18.245 align:middle line:84% of vertices forms a directed path. 00:10:18.245 --> 00:10:22.660 align:middle line:90% 00:10:22.660 --> 00:10:26.710 align:middle line:84% OK, so first, consider a permutation of vertices-- 00:10:26.710 --> 00:10:30.470 align:middle line:90% for example, 2, 1, 3, 4. 00:10:30.470 --> 00:10:33.700 align:middle line:84% And think about what is the probability that the edges that 00:10:33.700 --> 00:10:38.450 align:middle line:84% go in the direction of this permutation-- 00:10:38.450 --> 00:10:43.860 align:middle line:84% that the edges are oriented according to the permutation. 00:10:43.860 --> 00:10:46.555 align:middle line:84% Well, we have to flip n minus 1 coins. 00:10:46.555 --> 00:10:48.430 align:middle line:84% And all of them have to come up in such a way 00:10:48.430 --> 00:10:51.660 align:middle line:84% so that the edges are pointing in the direction 00:10:51.660 --> 00:10:54.520 align:middle line:84% of this permutation, in the order of this permutation. 00:10:54.520 --> 00:11:00.930 align:middle line:84% So the probability that this is a directed path for each 00:11:00.930 --> 00:11:03.210 align:middle line:84% of these permutations is precisely 00:11:03.210 --> 00:11:08.030 align:middle line:84% 2 to the minus parentheses n minus 1. 00:11:08.030 --> 00:11:11.090 align:middle line:90% 00:11:11.090 --> 00:11:14.960 align:middle line:84% And now, we invoke the linearity of expectations 00:11:14.960 --> 00:11:24.110 align:middle line:84% to claim that the expected number of Hamilton paths 00:11:24.110 --> 00:11:26.600 align:middle line:84% must then be-- well, each of the n 00:11:26.600 --> 00:11:31.520 align:middle line:84% factorial permutations has probability of 2 00:11:31.520 --> 00:11:38.090 align:middle line:84% to the minus n minus 1 of being a directed path. 00:11:38.090 --> 00:11:42.020 align:middle line:84% So this is a calculation that is analogous to the calculation 00:11:42.020 --> 00:11:45.850 align:middle line:84% that we did in the earlier part of this video. 00:11:45.850 --> 00:11:50.500 align:middle line:84% Well, this is what happens in expectation, on average. 00:11:50.500 --> 00:11:55.150 align:middle line:84% And thus, there must be some instance 00:11:55.150 --> 00:11:58.810 align:middle line:84% where we can beat this average or at least be at least as 00:11:58.810 --> 00:12:00.200 align:middle line:90% large as this average. 00:12:00.200 --> 00:12:09.040 align:middle line:84% So thus, there exists a tournament 00:12:09.040 --> 00:12:19.685 align:middle line:84% with at least this many Hamilton paths. 00:12:19.685 --> 00:12:24.020 align:middle line:90% 00:12:24.020 --> 00:12:27.470 align:middle line:84% And that concludes the proof of this theorem 00:12:27.470 --> 00:12:30.590 align:middle line:90% that we laid out earlier. 00:12:30.590 --> 00:12:34.520 align:middle line:84% So this is an example of applying linearity 00:12:34.520 --> 00:12:38.510 align:middle line:84% of expectations as a step in the probabilistic method 00:12:38.510 --> 00:12:42.020 align:middle line:84% to prove this nice and simple result, 00:12:42.020 --> 00:12:47.590 align:middle line:84% that there exists tournaments with lots of Hamilton paths. 00:12:47.590 --> 00:12:55.000 align:middle line:90%