WEBVTT 00:00:00.000 --> 00:00:01.996 align:middle line:90% [SQUEAKING] 00:00:01.996 --> 00:00:03.493 align:middle line:90% [RUSTLING] 00:00:03.493 --> 00:00:05.988 align:middle line:90% [CLICKING] 00:00:05.988 --> 00:00:11.980 align:middle line:90% 00:00:11.980 --> 00:00:13.690 align:middle line:84% YUFEI ZHAO: Oftentimes, in applying 00:00:13.690 --> 00:00:15.700 align:middle line:84% the probabilistic method, we'll need 00:00:15.700 --> 00:00:19.810 align:middle line:84% to understand whether a certain random structure typically 00:00:19.810 --> 00:00:22.270 align:middle line:84% has a property that we're looking for. 00:00:22.270 --> 00:00:25.180 align:middle line:84% In this video, we'll look at one such property 00:00:25.180 --> 00:00:30.070 align:middle line:84% and understand how to analyze whether a certain random graph 00:00:30.070 --> 00:00:38.020 align:middle line:84% has the property of containing a triangle with high probability. 00:00:38.020 --> 00:00:40.270 align:middle line:84% The objects that we're going to be looking at today is 00:00:40.270 --> 00:00:45.820 align:middle line:84% the Erdos-Rényi random graph, commonly referred to as G(n,p). 00:00:45.820 --> 00:00:50.050 align:middle line:84% And this is the graph consisting of placing n vertices. 00:00:50.050 --> 00:00:55.240 align:middle line:84% And for each pair of vertices, we put down an edge between them 00:00:55.240 --> 00:01:00.400 align:middle line:84% with probability p independently for all pairs of vertices. 00:01:00.400 --> 00:01:03.520 align:middle line:84% In other words, we flip a biased coin 00:01:03.520 --> 00:01:06.640 align:middle line:84% that comes up heads with probability p independently 00:01:06.640 --> 00:01:12.200 align:middle line:84% for each possible edge and put down the edge randomly as such. 00:01:12.200 --> 00:01:14.150 align:middle line:90% This is a random graph. 00:01:14.150 --> 00:01:15.570 align:middle line:90% It's a random structure. 00:01:15.570 --> 00:01:16.490 align:middle line:90% It's a random object. 00:01:16.490 --> 00:01:19.820 align:middle line:84% And we would like to understand various properties 00:01:19.820 --> 00:01:21.520 align:middle line:90% of this random graph. 00:01:21.520 --> 00:01:24.050 align:middle line:84% And the property that we're going to look at in this video 00:01:24.050 --> 00:01:25.500 align:middle line:90% is the following. 00:01:25.500 --> 00:01:30.950 align:middle line:84% Does this random graph typically contain a triangle? 00:01:30.950 --> 00:01:34.820 align:middle line:84% In the rest of this video, we'll be looking at p 00:01:34.820 --> 00:01:37.090 align:middle line:90% as a function of n. 00:01:37.090 --> 00:01:41.510 align:middle line:84% So p is allowed to change, depending on n. 00:01:41.510 --> 00:01:45.030 align:middle line:84% Well, we can denote this dependence by a subscript, 00:01:45.030 --> 00:01:47.570 align:middle line:84% but out of convenience, we'll drop the subscript 00:01:47.570 --> 00:01:51.230 align:middle line:90% for the rest of the video. 00:01:51.230 --> 00:01:55.000 align:middle line:84% So G(n,p) Does it typically contain a triangle? 00:01:55.000 --> 00:01:59.180 align:middle line:84% Well, by "typically" I mean the following mathematically precise 00:01:59.180 --> 00:02:00.260 align:middle line:90% meaning. 00:02:00.260 --> 00:02:03.860 align:middle line:84% "Typically" means with probability that approaches 1 00:02:03.860 --> 00:02:06.970 align:middle line:90% as n goes to infinity. 00:02:06.970 --> 00:02:08.680 align:middle line:84% So the statements are going to be 00:02:08.680 --> 00:02:13.990 align:middle line:84% looking at are asymptotic as n becomes large. 00:02:13.990 --> 00:02:16.450 align:middle line:84% The main theorem that we'll prove in this video 00:02:16.450 --> 00:02:17.440 align:middle line:90% is the following. 00:02:17.440 --> 00:02:19.930 align:middle line:90% Well, there are two statements. 00:02:19.930 --> 00:02:25.810 align:middle line:84% I have conveniently labeled them theorem 0 and theorem 1. 00:02:25.810 --> 00:02:29.620 align:middle line:84% Theorem 0, sometimes referred to as the 0 statement, 00:02:29.620 --> 00:02:32.420 align:middle line:84% is the statement about not having any triangles, 00:02:32.420 --> 00:02:34.240 align:middle line:90% having zero triangles. 00:02:34.240 --> 00:02:37.480 align:middle line:84% And it says that if n times p-- remember, 00:02:37.480 --> 00:02:40.490 align:middle line:84% p is a function of n. p is allowed to depend on n. 00:02:40.490 --> 00:02:45.040 align:middle line:84% If n times p approaches 0 as n goes to infinity, 00:02:45.040 --> 00:02:51.130 align:middle line:84% then this random graph G(n,p) is triangle-free with probability 00:02:51.130 --> 00:02:55.000 align:middle line:84% approaching 1 as n goes to infinity. 00:02:55.000 --> 00:02:57.700 align:middle line:84% In other words, with high probability, 00:02:57.700 --> 00:03:02.180 align:middle line:84% this random graph does not have any triangles. 00:03:02.180 --> 00:03:05.930 align:middle line:84% This makes sense in that this graph is pretty sparse 00:03:05.930 --> 00:03:07.520 align:middle line:90% when p is small. 00:03:07.520 --> 00:03:12.080 align:middle line:84% So this statement basically says that p is much less than 1 00:03:12.080 --> 00:03:14.430 align:middle line:84% over n, in terms of the growth rate. 00:03:14.430 --> 00:03:18.080 align:middle line:84% So when p is very small, you expect not so many edges. 00:03:18.080 --> 00:03:19.850 align:middle line:84% And it turns out that there's not 00:03:19.850 --> 00:03:24.620 align:middle line:84% going to be a triangle typically in this random graph. 00:03:24.620 --> 00:03:27.360 align:middle line:90% The second part is 0 and 1. 00:03:27.360 --> 00:03:29.270 align:middle line:90% The 1 statement. 00:03:29.270 --> 00:03:34.980 align:middle line:84% And it says that when (n,p) approaches infinity, again, 00:03:34.980 --> 00:03:40.370 align:middle line:84% p is a function of n, then this random graph G(n,p) typically 00:03:40.370 --> 00:03:41.960 align:middle line:90% contains a triangle. 00:03:41.960 --> 00:03:45.170 align:middle line:84% IE, it contains a triangle with probability 00:03:45.170 --> 00:03:50.920 align:middle line:84% that approaches 1 as n goes to infinity. 00:03:50.920 --> 00:03:54.430 align:middle line:84% So if you have a high edge probability, then typically, 00:03:54.430 --> 00:03:57.940 align:middle line:84% you would expect to see a triangle in a random 00:03:57.940 --> 00:04:03.070 align:middle line:84% instantiation of this random graph G(n,p). 00:04:03.070 --> 00:04:07.330 align:middle line:84% One way to summarize both statements is the following. 00:04:07.330 --> 00:04:10.420 align:middle line:84% One over n is a threshold for the property 00:04:10.420 --> 00:04:12.730 align:middle line:90% of containing a triangle. 00:04:12.730 --> 00:04:15.940 align:middle line:84% To interpret the sentence, imagine if p 00:04:15.940 --> 00:04:18.320 align:middle line:90% is much larger than one over n. 00:04:18.320 --> 00:04:20.970 align:middle line:84% Then we're in the theorem 1 setting. 00:04:20.970 --> 00:04:24.790 align:middle line:84% Then, the random graph typically contains a triangle. 00:04:24.790 --> 00:04:28.240 align:middle line:84% Whereas if p is much smaller than 1 00:04:28.240 --> 00:04:32.900 align:middle line:84% over n in the sense of being in the theorem 0 setting, 00:04:32.900 --> 00:04:37.240 align:middle line:84% then typically, the random graph does not contain any triangles. 00:04:37.240 --> 00:04:39.340 align:middle line:90% So 1 over n is a threshold. 00:04:39.340 --> 00:04:41.470 align:middle line:84% Depending on whether the edge probability 00:04:41.470 --> 00:04:44.330 align:middle line:84% p is much larger or much smaller than 1 over n. 00:04:44.330 --> 00:04:46.690 align:middle line:84% we expect very different behaviors 00:04:46.690 --> 00:04:50.620 align:middle line:84% from the perspective of whether the graph contains a triangle. 00:04:50.620 --> 00:04:54.670 align:middle line:84% These types of statements are widely 00:04:54.670 --> 00:04:59.230 align:middle line:84% studied in combinatorics and probability and understanding 00:04:59.230 --> 00:05:02.840 align:middle line:84% how to establish and prove these type of statements 00:05:02.840 --> 00:05:06.620 align:middle line:84% are some of the bread and butter of probabilistic combinatorics. 00:05:06.620 --> 00:05:12.320 align:middle line:84% So in this video, let us prove the statement in this box. 00:05:12.320 --> 00:05:15.290 align:middle line:84% Let's start with the easier direction, which, in this case, 00:05:15.290 --> 00:05:19.760 align:middle line:84% is the first statement, theorem 0. 00:05:19.760 --> 00:05:28.440 align:middle line:90% So let's prove theorem 0. 00:05:28.440 --> 00:05:31.030 align:middle line:90% 00:05:31.030 --> 00:05:32.750 align:middle line:90% We're looking at triangles. 00:05:32.750 --> 00:05:36.010 align:middle line:84% So let us define a random variable, x, 00:05:36.010 --> 00:05:47.920 align:middle line:84% to be the number of triangles in the random graph G(n,p). 00:05:47.920 --> 00:05:52.270 align:middle line:84% We can compute the expectation of x with not much difficulty, 00:05:52.270 --> 00:05:55.060 align:middle line:84% because by linearity of expectations, 00:05:55.060 --> 00:05:58.810 align:middle line:84% there are n choose three triples of vertices. 00:05:58.810 --> 00:06:02.870 align:middle line:84% And each triple forms a triangle with probability p cubed. 00:06:02.870 --> 00:06:05.900 align:middle line:90% 00:06:05.900 --> 00:06:08.690 align:middle line:84% We see that this quantity here is 00:06:08.690 --> 00:06:16.070 align:middle line:90% on the order of n cubed p cubed. 00:06:16.070 --> 00:06:22.100 align:middle line:84% We get to drop constant factors in this asymptotic estimate. 00:06:22.100 --> 00:06:28.530 align:middle line:84% Now when (n,p) goes to zero, as n goes to infinity, n cubed, 00:06:28.530 --> 00:06:30.710 align:middle line:90% p cubed also goes to zero. 00:06:30.710 --> 00:06:33.480 align:middle line:84% So this quantity is little o of 1. 00:06:33.480 --> 00:06:39.270 align:middle line:84% It's a quantity that goes to 0 as n goes to infinity. 00:06:39.270 --> 00:06:42.660 align:middle line:84% Let us now apply Markov's inequality. 00:06:42.660 --> 00:06:49.810 align:middle line:84% We explained and proved Markov's inequality in a different video. 00:06:49.810 --> 00:06:57.440 align:middle line:84% By Markov's inequality, we find that the probability 00:06:57.440 --> 00:07:03.230 align:middle line:84% that x is at least one is, at most, the expectation 00:07:03.230 --> 00:07:07.670 align:middle line:90% of x divided by 1. 00:07:07.670 --> 00:07:12.820 align:middle line:84% And from earlier, this quantity is little o of 1. 00:07:12.820 --> 00:07:13.320 align:middle line:90% All right. 00:07:13.320 --> 00:07:17.880 align:middle line:84% So the number of triangles is typically small in expectation. 00:07:17.880 --> 00:07:19.920 align:middle line:90% So it's small in expectation. 00:07:19.920 --> 00:07:22.730 align:middle line:84% And therefore, the probability that there 00:07:22.730 --> 00:07:28.850 align:middle line:84% is at least one triangle goes to 0 as n goes to infinity. 00:07:28.850 --> 00:07:35.460 align:middle line:84% Thus, the previous statement is equivalent to the statement 00:07:35.460 --> 00:07:42.660 align:middle line:84% that the probability that there are zero triangles approaches 1 00:07:42.660 --> 00:07:44.980 align:middle line:90% as n goes to infinity. 00:07:44.980 --> 00:07:48.930 align:middle line:90% 00:07:48.930 --> 00:07:53.040 align:middle line:84% This finishes the first part of the theorem. 00:07:53.040 --> 00:07:56.710 align:middle line:84% In the first part, the edge probability is very small, 00:07:56.710 --> 00:08:00.930 align:middle line:84% and so the expected number of triangles 00:08:00.930 --> 00:08:03.600 align:middle line:90% goes to 0 as n goes to infinity. 00:08:03.600 --> 00:08:06.810 align:middle line:84% And by Markov's inequality, we see that typically, there 00:08:06.810 --> 00:08:10.750 align:middle line:84% should not be any triangles in this random graph. 00:08:10.750 --> 00:08:13.720 align:middle line:84% Let us now move on to the second part. 00:08:13.720 --> 00:08:24.080 align:middle line:84% In the second part, namely theorem one, 00:08:24.080 --> 00:08:29.240 align:middle line:84% if we use the setup that we had earlier, 00:08:29.240 --> 00:08:33.400 align:middle line:84% we see that the number of triangles in expectation 00:08:33.400 --> 00:08:38.020 align:middle line:84% goes to infinity as n goes to infinity. 00:08:38.020 --> 00:08:41.470 align:middle line:84% However, this does not immediately 00:08:41.470 --> 00:08:46.840 align:middle line:84% imply that x is typically positive. 00:08:46.840 --> 00:08:50.830 align:middle line:84% Because it could be the case that x is still typically 00:08:50.830 --> 00:08:57.190 align:middle line:84% zero, but is very, very large a diminishing fraction 00:08:57.190 --> 00:08:59.470 align:middle line:90% of the time. 00:08:59.470 --> 00:09:01.510 align:middle line:84% And in that case, we would still be 00:09:01.510 --> 00:09:06.250 align:middle line:84% able to have the statement that the expectation of x 00:09:06.250 --> 00:09:09.160 align:middle line:84% goes to infinity while it would not be true 00:09:09.160 --> 00:09:11.170 align:middle line:90% that x is typically positive. 00:09:11.170 --> 00:09:14.500 align:middle line:84% So we'll need additional ideas to establish 00:09:14.500 --> 00:09:17.260 align:middle line:90% that x is typically positive. 00:09:17.260 --> 00:09:28.460 align:middle line:84% And what we will do is to use the second moment bound 00:09:28.460 --> 00:09:39.745 align:middle line:84% to show that x is typically close to its mean. 00:09:39.745 --> 00:09:44.300 align:middle line:90% 00:09:44.300 --> 00:09:48.740 align:middle line:84% So the x is very concentrated around its expectation. 00:09:48.740 --> 00:09:52.640 align:middle line:84% And in that case, having expectation 00:09:52.640 --> 00:09:55.910 align:middle line:84% that goes to infinity would then imply 00:09:55.910 --> 00:09:58.490 align:middle line:90% that x is typically positive. 00:09:58.490 --> 00:10:02.390 align:middle line:90% So let's carry out this idea. 00:10:02.390 --> 00:10:08.960 align:middle line:84% Let's consider the following probability. 00:10:08.960 --> 00:10:23.410 align:middle line:84% We see that the probability that x is equal to 0. 00:10:23.410 --> 00:10:26.890 align:middle line:84% Well, if x is equal to 0, and here x 00:10:26.890 --> 00:10:31.030 align:middle line:84% is some non-negative quantity, then this statement 00:10:31.030 --> 00:10:39.570 align:middle line:84% is equivalent to the statement that x deviates 00:10:39.570 --> 00:10:45.980 align:middle line:84% from this mean in the downward direction 00:10:45.980 --> 00:10:52.740 align:middle line:90% by minus the expectation of x. 00:10:52.740 --> 00:10:54.130 align:middle line:90% So this is the same as x. 00:10:54.130 --> 00:10:56.910 align:middle line:84% Less than equal to 0, but x is always non-negative. 00:10:56.910 --> 00:11:00.870 align:middle line:84% So these two probabilities are equal to each other. 00:11:00.870 --> 00:11:17.710 align:middle line:84% We can then relax the event inside as such. 00:11:17.710 --> 00:11:20.600 align:middle line:90% 00:11:20.600 --> 00:11:23.270 align:middle line:84% Again, noting that the expectation of x 00:11:23.270 --> 00:11:26.210 align:middle line:90% is always non-negative. 00:11:26.210 --> 00:11:30.350 align:middle line:84% Now we apply a Chebyshev's inequality. 00:11:30.350 --> 00:11:33.160 align:middle line:84% So in this step, we can apply Chebyshev's inequality. 00:11:33.160 --> 00:11:35.770 align:middle line:84% And you'll find an explanation and proof 00:11:35.770 --> 00:11:40.600 align:middle line:84% of Chebyshev's inequality in a different video, which 00:11:40.600 --> 00:11:43.990 align:middle line:84% allows us to conclude that this probability is, 00:11:43.990 --> 00:11:50.473 align:middle line:84% at most, the variance of x divided by the expectation of x 00:11:50.473 --> 00:11:50.973 align:middle line:90% squared. 00:11:50.973 --> 00:11:57.240 align:middle line:90% 00:11:57.240 --> 00:11:58.050 align:middle line:90% All right. 00:11:58.050 --> 00:12:02.280 align:middle line:84% So what have we learned from this calculation? 00:12:02.280 --> 00:12:04.890 align:middle line:90% Let me state a corollary. 00:12:04.890 --> 00:12:22.760 align:middle line:84% If the expectation of x is much smaller, so if the variance of x 00:12:22.760 --> 00:12:28.940 align:middle line:84% is much smaller than the square of expectation, 00:12:28.940 --> 00:12:32.510 align:middle line:84% then the right-hand side goes to 0 00:12:32.510 --> 00:12:38.630 align:middle line:84% as n goes to infinity, which then implies 00:12:38.630 --> 00:12:44.720 align:middle line:84% that x is positive with high probability, 00:12:44.720 --> 00:12:47.900 align:middle line:84% meaning that this probability that x is positive 00:12:47.900 --> 00:12:51.120 align:middle line:90% goes to 1 as n goes to infinity. 00:12:51.120 --> 00:12:54.337 align:middle line:90% 00:12:54.337 --> 00:12:55.920 align:middle line:84% So this is the conclusion that we will 00:12:55.920 --> 00:12:58.955 align:middle line:90% use to establish theorem 1. 00:12:58.955 --> 00:13:02.080 align:middle line:90% 00:13:02.080 --> 00:13:03.970 align:middle line:84% So the number of triangles, we'll 00:13:03.970 --> 00:13:07.720 align:middle line:84% be able to write in the following way. 00:13:07.720 --> 00:13:16.030 align:middle line:84% x is equal to the sum over all triples of vertices 00:13:16.030 --> 00:13:23.320 align:middle line:84% of the following quantity, xij, xik, xjk, where we define 00:13:23.320 --> 00:13:25.725 align:middle line:90% these x sub ij's as follows. 00:13:25.725 --> 00:13:29.150 align:middle line:90% 00:13:29.150 --> 00:13:41.140 align:middle line:84% x sub ij is equal to 1 if ij is an edge and zero 00:13:41.140 --> 00:13:43.575 align:middle line:90% if ij is not an edge. 00:13:43.575 --> 00:13:52.520 align:middle line:90% 00:13:52.520 --> 00:14:07.860 align:middle line:84% The vertices are labeled by numbers 1 through n. 00:14:07.860 --> 00:14:12.060 align:middle line:84% So you see this sum, which is indexed over triples 00:14:12.060 --> 00:14:21.890 align:middle line:84% of vertices, each term is 1 if these vertices i, j, and k is 00:14:21.890 --> 00:14:24.710 align:middle line:90% a triangle and is 0 otherwise. 00:14:24.710 --> 00:14:28.000 align:middle line:84% So this is the number of triangles in the graph. 00:14:28.000 --> 00:14:30.980 align:middle line:84% And let me, just because it will make 00:14:30.980 --> 00:14:37.130 align:middle line:84% the notation a little easier, rewrite this quantity as such. 00:14:37.130 --> 00:14:45.980 align:middle line:84% Here, t ranges over all triples of vertices. 00:14:45.980 --> 00:14:50.920 align:middle line:84% Onward the triples of vertices, and x sub t 00:14:50.920 --> 00:14:54.970 align:middle line:84% is this term so we can rewrite this term as just 00:14:54.970 --> 00:14:59.115 align:middle line:90% a single variable x sub ijk. 00:14:59.115 --> 00:15:03.760 align:middle line:90% 00:15:03.760 --> 00:15:06.040 align:middle line:84% To apply the corollary, we will need 00:15:06.040 --> 00:15:09.760 align:middle line:84% to get some estimate from the variance of x. 00:15:09.760 --> 00:15:12.610 align:middle line:84% And we'll do so through covariance. 00:15:12.610 --> 00:15:23.960 align:middle line:84% So as a reminder, the covariance of two random variables, say 00:15:23.960 --> 00:15:33.120 align:middle line:84% y and z, is defined to be the expectation of y times 00:15:33.120 --> 00:15:44.070 align:middle line:84% z minus the expectation of y times the expectation of z. 00:15:44.070 --> 00:15:48.160 align:middle line:84% In particular, if y and z are independent random variables, 00:15:48.160 --> 00:15:51.180 align:middle line:90% then their covariance is 0. 00:15:51.180 --> 00:15:56.580 align:middle line:84% And the variance of a random variable x 00:15:56.580 --> 00:15:59.420 align:middle line:84% is equal to the covariance of x with itself. 00:15:59.420 --> 00:16:02.450 align:middle line:90% 00:16:02.450 --> 00:16:09.270 align:middle line:84% So let us compute the variance of x, x 00:16:09.270 --> 00:16:12.600 align:middle line:90% being the number of triangles. 00:16:12.600 --> 00:16:17.320 align:middle line:84% We can rewrite it as a covariance of x with itself. 00:16:17.320 --> 00:16:20.450 align:middle line:90% 00:16:20.450 --> 00:16:25.370 align:middle line:84% Here, x, we can write it as a sum of various terms indexed 00:16:25.370 --> 00:16:27.250 align:middle line:90% by triples of vertices. 00:16:27.250 --> 00:16:38.760 align:middle line:90% 00:16:38.760 --> 00:16:41.640 align:middle line:84% I'll use t for the first index in the first sum 00:16:41.640 --> 00:16:45.720 align:middle line:84% and t prime for the indices in the second sum. 00:16:45.720 --> 00:16:48.030 align:middle line:84% The nice thing about covariance is 00:16:48.030 --> 00:16:50.790 align:middle line:90% that it is a bilinear function. 00:16:50.790 --> 00:16:55.390 align:middle line:84% If you split y in the covariance of y and z, 00:16:55.390 --> 00:17:00.100 align:middle line:84% if you split y into a linear combination of various sums, 00:17:00.100 --> 00:17:03.450 align:middle line:84% then you'll see that the dependence in this formula 00:17:03.450 --> 00:17:06.630 align:middle line:84% is linear on splitting y, so you will 00:17:06.630 --> 00:17:12.380 align:middle line:84% be able to write the covariance correspondingly as a sum. 00:17:12.380 --> 00:17:16.300 align:middle line:84% So given this covariance of these two sums, 00:17:16.300 --> 00:17:25.420 align:middle line:84% we can distribute this quantity and write it 00:17:25.420 --> 00:17:31.390 align:middle line:84% as the covariance over all pairs of triples T and T 00:17:31.390 --> 00:17:39.430 align:middle line:84% prime, the covariance of x sub T against x sub T prime. 00:17:39.430 --> 00:17:44.130 align:middle line:90% 00:17:44.130 --> 00:17:47.820 align:middle line:84% Now we need to understand this term, the covariance of x sub 00:17:47.820 --> 00:17:52.410 align:middle line:84% T against the covariance of x sub T prime. 00:17:52.410 --> 00:17:59.530 align:middle line:84% Towards this end, let us think about what happens to this term. 00:17:59.530 --> 00:18:02.830 align:middle line:90% So let me do a side calculation. 00:18:02.830 --> 00:18:06.110 align:middle line:84% From the definition of covariance, 00:18:06.110 --> 00:18:13.600 align:middle line:84% we see that this quantity equals to the expectation of x sub T, x 00:18:13.600 --> 00:18:19.690 align:middle line:84% sub T prime minus the expectation of x sub 00:18:19.690 --> 00:18:23.070 align:middle line:84% T times the expectation of x sub T prime. 00:18:23.070 --> 00:18:27.610 align:middle line:90% 00:18:27.610 --> 00:18:30.225 align:middle line:84% There are a few cases that we need to consider. 00:18:30.225 --> 00:18:33.700 align:middle line:90% 00:18:33.700 --> 00:18:41.830 align:middle line:84% First, T and T prime are two triples of vertices 00:18:41.830 --> 00:18:44.120 align:middle line:84% and they could be in various positions. 00:18:44.120 --> 00:18:48.250 align:middle line:84% The corresponding full triangles could be disjoint, 00:18:48.250 --> 00:18:49.880 align:middle line:90% or they could overlap. 00:18:49.880 --> 00:18:53.440 align:middle line:84% And we need to consider all of those possibilities. 00:18:53.440 --> 00:18:58.780 align:middle line:84% Now before doing those cases, let's observe that this quantity 00:18:58.780 --> 00:19:04.330 align:middle line:84% here, if you have, for example, t and t prime 00:19:04.330 --> 00:19:10.560 align:middle line:84% being in any number of configurations. 00:19:10.560 --> 00:19:14.360 align:middle line:84% So these are the vertices of T and this is the vertices 00:19:14.360 --> 00:19:16.370 align:middle line:90% of T prime, for instance. 00:19:16.370 --> 00:19:18.860 align:middle line:84% So they could share some edges or they 00:19:18.860 --> 00:19:20.425 align:middle line:90% may overlap in some edges. 00:19:20.425 --> 00:19:25.960 align:middle line:90% 00:19:25.960 --> 00:19:30.550 align:middle line:84% But no matter what the configurations are, 00:19:30.550 --> 00:19:33.640 align:middle line:84% the first term always equals to p 00:19:33.640 --> 00:19:37.960 align:middle line:84% raised to the number of edges contained 00:19:37.960 --> 00:19:43.150 align:middle line:90% in the union of T and T prime. 00:19:43.150 --> 00:19:45.850 align:middle line:90% So in the complete graph. 00:19:45.850 --> 00:19:50.500 align:middle line:84% So what I mean here is the following. 00:19:50.500 --> 00:19:55.340 align:middle line:90% 00:19:55.340 --> 00:20:01.100 align:middle line:84% So here is p raised to the number of-- 00:20:01.100 --> 00:20:02.810 align:middle line:90% so p raised to-- 00:20:02.810 --> 00:20:04.850 align:middle line:84% let me still write it this way and then we'll 00:20:04.850 --> 00:20:07.820 align:middle line:90% see some specific examples. 00:20:07.820 --> 00:20:12.230 align:middle line:84% Minus p raised to the number of edges in T and p 00:20:12.230 --> 00:20:14.820 align:middle line:84% raised to the number of edges in t prime. 00:20:14.820 --> 00:20:17.870 align:middle line:84% So here, the exponents over correspond 00:20:17.870 --> 00:20:20.030 align:middle line:84% to the number of edges in the complete graph 00:20:20.030 --> 00:20:22.940 align:middle line:84% spanned by these vertices, T and T prime. 00:20:22.940 --> 00:20:26.870 align:middle line:84% So let's look at a few different possibilities. 00:20:26.870 --> 00:20:33.080 align:middle line:84% The first is if the two triangles, T and T prime, 00:20:33.080 --> 00:20:36.570 align:middle line:90% do not intersect in any edges. 00:20:36.570 --> 00:20:39.810 align:middle line:84% So the number of vertices they intersect is at most one. 00:20:39.810 --> 00:20:43.070 align:middle line:84% So the two scenarios are the one that I just drew 00:20:43.070 --> 00:20:48.830 align:middle line:84% and if these two triangles are even disjoint in their vertex 00:20:48.830 --> 00:20:49.370 align:middle line:90% set. 00:20:49.370 --> 00:20:55.490 align:middle line:84% So then there's no edge overlap, in which case the edges-- 00:20:55.490 --> 00:20:57.380 align:middle line:84% so the random variables that correspond 00:20:57.380 --> 00:20:58.920 align:middle line:84% to the edges of these two triangles, 00:20:58.920 --> 00:21:00.840 align:middle line:84% they are independent of each other. 00:21:00.840 --> 00:21:02.850 align:middle line:84% And so the covariance should be 0. 00:21:02.850 --> 00:21:07.400 align:middle line:84% And indeed, the first term here, you have p raised to the power 6 00:21:07.400 --> 00:21:11.790 align:middle line:84% and then minus p cubed and p cubed. 00:21:11.790 --> 00:21:15.830 align:middle line:84% And so in this case, this covariance is 0. 00:21:15.830 --> 00:21:24.910 align:middle line:84% In the second case, when these two triples T and T 00:21:24.910 --> 00:21:28.870 align:middle line:84% prime overlapping exactly two vertices. 00:21:28.870 --> 00:21:33.515 align:middle line:90% Well, here they are-- 00:21:33.515 --> 00:21:36.400 align:middle line:90% 00:21:36.400 --> 00:21:38.830 align:middle line:84% so the union of these two triangles-- 00:21:38.830 --> 00:21:40.790 align:middle line:90% so I think I misspoke earlier. 00:21:40.790 --> 00:21:43.030 align:middle line:84% So by this notation, I really mean 00:21:43.030 --> 00:21:46.630 align:middle line:84% viewing T as a triangle and T prime as a triangle, 00:21:46.630 --> 00:21:48.920 align:middle line:84% and looking at the union of these two triangles. 00:21:48.920 --> 00:21:52.430 align:middle line:84% So the union of these two triangles have five edges. 00:21:52.430 --> 00:21:55.930 align:middle line:84% So that's the number of terms here, 00:21:55.930 --> 00:22:00.270 align:middle line:84% dismissing repetition in the first term, 00:22:00.270 --> 00:22:03.810 align:middle line:90% the number of factors. 00:22:03.810 --> 00:22:11.035 align:middle line:84% And minus the later two terms are both p cubed and p cubed. 00:22:11.035 --> 00:22:16.200 align:middle line:90% 00:22:16.200 --> 00:22:34.030 align:middle line:84% Finally, if T and T prime are identical, 00:22:34.030 --> 00:22:39.850 align:middle line:84% then the number of edges in the first term 00:22:39.850 --> 00:22:44.355 align:middle line:90% is 3 minus p cubed p cubed. 00:22:44.355 --> 00:22:47.890 align:middle line:90% 00:22:47.890 --> 00:22:54.580 align:middle line:84% So recall that p is some quantity that goes to 0 as-- 00:22:54.580 --> 00:22:58.900 align:middle line:84% well, it doesn't have to go to 0, but we-- 00:22:58.900 --> 00:23:03.520 align:middle line:84% p is some quantity that for now you can think of as going to 0 00:23:03.520 --> 00:23:06.970 align:middle line:84% so that the first term is always dominant. 00:23:06.970 --> 00:23:10.280 align:middle line:84% Even if p is constant up to a constant factor, 00:23:10.280 --> 00:23:11.865 align:middle line:84% still, the first term is dominant. 00:23:11.865 --> 00:23:15.510 align:middle line:90% 00:23:15.510 --> 00:23:18.220 align:middle line:84% So let us now continue the earlier conversation-- 00:23:18.220 --> 00:23:20.020 align:middle line:90% the earlier computation. 00:23:20.020 --> 00:23:26.010 align:middle line:84% And we see that this sum is equal to the following. 00:23:26.010 --> 00:23:30.480 align:middle line:84% There are-- we need to think about how each of these cases 00:23:30.480 --> 00:23:31.740 align:middle line:90% can arise. 00:23:31.740 --> 00:23:34.270 align:middle line:84% The first case contributes zero to the sum, 00:23:34.270 --> 00:23:36.360 align:middle line:84% so we don't have to worry about it. 00:23:36.360 --> 00:23:39.030 align:middle line:84% The second term, the second possibility 00:23:39.030 --> 00:23:40.450 align:middle line:90% contributes the following. 00:23:40.450 --> 00:23:43.580 align:middle line:90% It contributes n choose 2. 00:23:43.580 --> 00:23:46.080 align:middle line:84% Well, we need to count how many different configurations can 00:23:46.080 --> 00:23:47.880 align:middle line:84% we have like this, where you have 00:23:47.880 --> 00:23:52.440 align:middle line:84% these two, the red and the blue triangles overlapping 00:23:52.440 --> 00:23:53.290 align:middle line:90% by an edge. 00:23:53.290 --> 00:23:59.035 align:middle line:84% So this n choose 2 for the two overlapping vertices and n times 00:23:59.035 --> 00:23:59.535 align:middle line:90% n minus-- 00:23:59.535 --> 00:24:02.910 align:middle line:90% 00:24:02.910 --> 00:24:10.183 align:middle line:84% times n minus 2 Times N minus 3 for the remaining two vertices. 00:24:10.183 --> 00:24:13.360 align:middle line:90% 00:24:13.360 --> 00:24:20.760 align:middle line:84% And in this case, we have p to the 5 minus p to the 6. 00:24:20.760 --> 00:24:23.610 align:middle line:90% 00:24:23.610 --> 00:24:28.050 align:middle line:84% And then the third possibility, n 00:24:28.050 --> 00:24:36.655 align:middle line:84% choose 3 times p cubed minus p to the sixth power. 00:24:36.655 --> 00:24:44.590 align:middle line:90% 00:24:44.590 --> 00:24:46.390 align:middle line:84% This is some expression, and we only 00:24:46.390 --> 00:24:49.270 align:middle line:84% need to get some bounds so we can play around with asymptotics 00:24:49.270 --> 00:24:51.680 align:middle line:84% and not think about constant factors. 00:24:51.680 --> 00:24:54.760 align:middle line:84% So we'll basically ignore the constant factors 00:24:54.760 --> 00:24:56.120 align:middle line:90% and do an upper bound. 00:24:56.120 --> 00:24:57.740 align:middle line:90% So up to a constant factor. 00:24:57.740 --> 00:24:58.540 align:middle line:90% So big O. 00:24:58.540 --> 00:25:05.170 align:middle line:84% The first term here involves n to the fourth p fifth power. 00:25:05.170 --> 00:25:08.300 align:middle line:84% So we can ignore the negative contributions. 00:25:08.300 --> 00:25:10.900 align:middle line:84% They don't matter for this calculation. 00:25:10.900 --> 00:25:19.330 align:middle line:84% And then the second term here is n cubed p cubed. 00:25:19.330 --> 00:25:21.580 align:middle line:84% At most, on the order of n cubed p cubed. 00:25:21.580 --> 00:25:27.450 align:middle line:90% 00:25:27.450 --> 00:25:35.650 align:middle line:84% So under the hypothesis that np goes to infinity, 00:25:35.650 --> 00:25:40.790 align:middle line:84% Both these terms are little o of n to the six. 00:25:40.790 --> 00:25:42.620 align:middle line:90% p to the sixth. 00:25:42.620 --> 00:25:44.210 align:middle line:84% So that's something that one can check 00:25:44.210 --> 00:25:46.560 align:middle line:90% from the from the hypothesis. 00:25:46.560 --> 00:25:49.360 align:middle line:90% 00:25:49.360 --> 00:25:55.670 align:middle line:84% Thus, the variance of x is little o 00:25:55.670 --> 00:26:00.740 align:middle line:90% of the expectation of x squared. 00:26:00.740 --> 00:26:07.070 align:middle line:84% And so by the corollary of Chebyshev earlier, 00:26:07.070 --> 00:26:16.780 align:middle line:84% we see that with probability approaching 1, x is positive. 00:26:16.780 --> 00:26:21.190 align:middle line:84% So that there's-- typically, this random graph contains 00:26:21.190 --> 00:26:22.570 align:middle line:90% a triangle. 00:26:22.570 --> 00:26:27.190 align:middle line:84% And that finishes the proof of the second part, this theorem 1. 00:26:27.190 --> 00:26:33.520 align:middle line:84% So to recap, we showed that 1 over n is a threshold 00:26:33.520 --> 00:26:37.240 align:middle line:84% for the property of the random graph G(n,p) to contain 00:26:37.240 --> 00:26:38.000 align:middle line:90% a triangle. 00:26:38.000 --> 00:26:39.990 align:middle line:90% So the statement has two parts. 00:26:39.990 --> 00:26:43.550 align:middle line:84% The first part is that when p is quite small, 00:26:43.550 --> 00:26:45.340 align:middle line:84% it's much smaller than 1 over n, then 00:26:45.340 --> 00:26:48.440 align:middle line:84% typically, this random graph does not contain a triangle. 00:26:48.440 --> 00:26:50.470 align:middle line:84% And the way we establish this fact 00:26:50.470 --> 00:26:53.980 align:middle line:84% is by showing that the number of triangles, x, 00:26:53.980 --> 00:26:59.560 align:middle line:84% the number of triangles, has expected value going to 0. 00:26:59.560 --> 00:27:02.200 align:middle line:84% And thus, by Markov's inequality, 00:27:02.200 --> 00:27:07.960 align:middle line:84% this number of triangles is 0 with high probability. 00:27:07.960 --> 00:27:14.110 align:middle line:84% The second part, where p now is much larger than 1 over n, 00:27:14.110 --> 00:27:16.760 align:middle line:84% here, the expected number of triangles 00:27:16.760 --> 00:27:20.180 align:middle line:84% goes to infinity as n goes to infinity. 00:27:20.180 --> 00:27:23.570 align:middle line:84% But that alone does not allow us to conclude 00:27:23.570 --> 00:27:28.220 align:middle line:84% that the number of triangles is positive typically. 00:27:28.220 --> 00:27:31.100 align:middle line:84% Instead, we'll need to apply a second moment argument 00:27:31.100 --> 00:27:33.440 align:middle line:84% to show that the number of triangles 00:27:33.440 --> 00:27:36.230 align:middle line:84% is typically concentrated around its mean. 00:27:36.230 --> 00:27:39.260 align:middle line:84% And that was a calculation that we did. 00:27:39.260 --> 00:27:42.710 align:middle line:84% This was a second moment calculation, namely 00:27:42.710 --> 00:27:46.910 align:middle line:84% understanding and computing the variance of this random variable 00:27:46.910 --> 00:27:50.930 align:middle line:84% and showing that the variance is much smaller 00:27:50.930 --> 00:27:53.420 align:middle line:84% than the square of the expectation 00:27:53.420 --> 00:27:56.270 align:middle line:84% from which then we can use Chebyshev's inequality 00:27:56.270 --> 00:28:01.730 align:middle line:84% to deduce that the number of triangles is typically positive. 00:28:01.730 --> 00:28:03.540 align:middle line:84% And combining these two proofs together, 00:28:03.540 --> 00:28:06.230 align:middle line:84% we obtain this threshold statement. 00:28:06.230 --> 00:28:10.940 align:middle line:84% Now what we just showed is an important, but fairly basic 00:28:10.940 --> 00:28:15.080 align:middle line:84% application of using probabilistic methods 00:28:15.080 --> 00:28:18.440 align:middle line:84% to understand typicality statements. 00:28:18.440 --> 00:28:22.840 align:middle line:84% And these types of statements and proof techniques 00:28:22.840 --> 00:28:27.160 align:middle line:84% are quite ubiquitous in studies involving 00:28:27.160 --> 00:28:29.300 align:middle line:84% the probabilistic method and beyond. 00:28:29.300 --> 00:28:33.640 align:middle line:84% And there are much more difficult results 00:28:33.640 --> 00:28:36.130 align:middle line:84% of similar flavor, but that involve 00:28:36.130 --> 00:28:37.880 align:middle line:90% much more advanced techniques. 00:28:37.880 --> 00:28:40.870 align:middle line:84% And what you saw here is simply a taste 00:28:40.870 --> 00:28:45.450 align:middle line:84% of what such a statement and its technique could look like. 00:28:45.450 --> 00:28:52.000 align:middle line:90%