Then one would hope to show that that function is

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: iven nite set of nodes, in which each pair of nodes is joined by an edge independently with probability p. We could simulate, or construct," such a random graph as follows: for each pair of nodes i j , generate a random number Uij  U 0; 1 , and join nodes i and j with an edge if Uij  p. Here is a problem: show that the probability of the resulting graph being connected is nondecreasing in p. That is, for p1 p2 , we want to show that Pp1 fgraph connectedg  Pp2 fgraph connectedg: I would say that this is intuitively obvious, but we want to give an actual proof. Again, the example is just meant to illustrate the idea of coupling, not to give an example that can be solved only with coupling! One way that one might approach this problem is to try to nd an explicit expression for the probability of being connected as a function of p. Then one would hope to show that that function is increasing, perhaps by di erentiating with respect to p and showing that the derivative is nonnegative. That is conceptually a straightforward approach, but you may become discouraged at the rst step|I don't think there is an obvious way of writing down the probability the graph is connected. Anyway, doesn't it seem somehow very ine cient, or at least overkill," to have to give a precise expression for the desired probability if all one desires is to show the inituitively obvious monotonicity property? Wouldn't you hope to give an argument that somehow simply formalizes the intuition that we all have? Stochastic Processes J. Chang, March 30, 1999 1. MARKOV CHAINS Page 1-24 One nice way to show that probabilities are ordered is to show that the corresponding events are ordered: if A B then PA  PB . So let's make two events by making two random graphs G1 and G2 , with each edge of G1 having probability p1 and each edge of G2 having probability p2 . We could do that by using two sets of U 0; 1 random variables: fUij g for G1 and fVij g for G2. OK, so now we ask: is it true that 1.49 fG1 connectedg fG2 connectedg? The answer is no; indeed, the random graphs G1 and G2 are independent, so that clearly P fG1 connected; G2 not connectedg = PfG1 connectedgPfG2 not connectedg 0: The problem is that we have used di erent, independent random numbers in constructing the graphs G1 and G2 , so that, for example, it is perfectly possible to have simultaneously Uij  p1 and Vij p2 for all i j , in which the graph G1 would be completely connected and the graph G2 would be completely disconnected. Here is a simple way to x the argument: use the same random numbers in de ning the two graphs. That is, draw the edge i; j  in graph G1 if Uij  p1 and the edge i; j  in graph G2 if Uij  p2 . Now notice how the picture has changed: with the modi ed de nitions it is obvious that, if an edge i; j  is in the graph G1 , then that edge is also in G2 . From this, it is equally obvious that 1.49 now holds. This establishes the desired monotonicity of the probability of being connected. Perfectly obvious, isn't it? So, what characterizes a c...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online