Solutions to Problem Set 5

Transcription

1 UC Berkeley, CS 74: Combinatorics and Discrete Probability (Fall 00 Solutions to Problem Set (MU 60 A family of subsets F of {,,, n} is called an antichain if there is no pair of sets A and B in F satisfying A B (a Give an example of F where F = n/ Choose every subset of size n/ (b Let be the number of sets in F with size k Show that k (Hint: Choose a random permutation of the numbers from to n, and let X k = if the first k numbers in your permutation yield a set in F If X = n X k, what can you say about X? Choose a random permutation (,, n, let X k = if the first k numbers yield a set in F, and let X = n X k Note that P[X k = ] = k Only for one value o is X k =, which means that E[X] So, we have E[X] = = E[X k ] k (c Argue that F n/ for any antichain F For a fixed n, the binomial coefficient is maximized at n/ So, we have n/ k Which implies that F = ( n n/ (MU 6 Consider the problem of whether graphs in G n,p have cliques of constant size k Suggest an appropriate threshold function for this property Generalize the argument used for cliques of size 4, using either the second moment method or the conditional expectation inequality, to prove that your threshold function is correct for cliques of size We suggest n /(k as the threshold function for whether subgraphs of G n,p have cliques of size k We generalize the argument found on page of the book Instead of the 4-cliques considered in the book, we now consider -cliques

2 Let p = f(n = o(n /4 For each set o vertices C i where i k, Xi is the indicator of whether C i is a -clique Let X = i= X i, then E[X] = p 0 As in Theorem 68, we get that E[X] = o(n o(n = o(, so E[X] < ɛ for large n Hence P[X ] E[X] < ɛ So, the probability that a subgraph of G n,p has a -clique is less than ɛ As in the Theorem 60, we find E[X X j = ] = i= E[X i X j = ] Consider the case where C i and C j share vertices Then there are ( ( ways we can pick the two shared vertices and n ways to pick the remaining vertices Then we are adding three unshared vertices, and we need to add nine new edges to get both X i = and X j = So, we multiply the above term with p 9 Using similar arguments for,,,and shared vertices, and using the conditional expectation inequality, we get P[X < 0] + p p 0 p p p 7 + (n p 4 This value approaches from below when n with p = f(n = ω(n / (MU 64 Consider a graph in G n,p, with p = /n Let X be the number of triangles in the graph, where a triangle is a clique with three edges Show that and that P[X ] /6 lim P[X ] /7 n Let C,, C ( n be an enumeration of all subsets of vertices in the graph For each i, X i indicates whether C i is a triangle Let X = i= X i For each i, we have P[X i = ] = E[X i ] = p and E[X] = p by linearity of expectation Now, applying Markov s inequality, we obtain the first bound: ( n P[X ] E[X] = (/n /6 Now, we will use the conditional expectation inequality First, we compute ( ( ( n n n E[X X i = ] = + p + p + p Now we can compute the bound P[X ] = P[X i = ] E[X X i = ] i= + p p + p + p Since p = /n, the expression above converges to /6 +/6+0+0 = /7 as n

3 4 (MU 7 Consider the two-state Markov chain with the following transition matrix [ ] p p p p Find a simple expression for P t 0,0 We can observe that P0,0 t+ = pp 0,0 t + ( pp 0, t and P 0, t = P 0,0 t From this, we can derive the recursion P0,0 t = (p P0,0 t + ( p, whose solution is t P0,0 t = (p t + ( p (p s = s=0 + (p t This can be verified by plugging the solution back into the recursion There is a second way to do this problem To be in state 0 at time t, either we never moved from state 0, or we took a number of trips to state and came back Hence, the number of steps of transition between the two states has to be even Note that no matter what state we are in, ( p is the probability of changing to the other state, and p is the probability of staying in the same state Hence, we need only the odd terms in (p + ( p t, (ie, all the terms where ( p is raised to an even power This allows us to derive the following equation: P t 0,0 = (t+/ B i+ (p, p, t where B k (a, b, t = ( t k a t k+ b k is the kth term in the binomial expansion of (a + b t This formula ca be verified by calculating the (0, 0-th element of the matrix P t (MU 7 Let X n be the sum of n independent rolls of a fair die Show that, for any k, lim n P[X n is divisible by k] = k Let Y t = X t (mod k, meaning that Y t is the remainder of X t divided by k Then Y 0, Y, is a Markov chain with k states, and Y 0 = 0 The transition probabilities are P i,j = I[i + a(mod k = j] (/6 a= X t is divisible by k if and only if Y n = 0 And we know that P[Y n = 0] = P0,0 n This is a finite ergodic Markov chain We will now show that the transition matrix is doubly stochastic and rely on the result of problem 7 (done in section to prove that the stationary distribution is

4 uniform For every j, we have P i,j = I[i + a(mod k = j] (/6 a= = (/6 = (/6 = (/6 = a= a= I[i + a(mod k = j] I[i = j a(mod k] This proves that the stationary distribution is uniform Therefore, lim n P[X n is divisible by k] = k 6 (MU 7 Consider a finite Markov chain on n states with stationary distribution π and transition probabilities P i,j Imagine starting the chain at time 0 and running it for m steps, obtaining the sequence of states X 0, X,, X m Consider the states in reverse order, X m, X m,, X 0 (a Argue that given X k+, the state X k is independent of X k+, X k+,, X m Thus the reverse sequence is Markovian We begin by simply writing out the definition of conditional expectation: P[X k X k+,, X m ] = P[X k, X k+,, X + m] P[X k+,, X m ] = P[X k]p[x k+ X k ]P[X k+,, X + m X k, X k+ ] P[X k+ ]P[X k+,, X + m X k ] = P[X k]p[x k+ X k ]P[X k+,, X + m X k ] P[X k+ ]P[X k+,, X + m X k ] = P[X k]p[x k+ X k ] P[X k+ ] Since this is a function only of X k and X k+, we have the desired Markovian dependency on only the previous state (b Argue that for the reverse sequence, the transition probabilities Q i,j are given by Q i,j = π jp j,i Using the result for part (a, we substitute the stationary distribution in for the marginals P[X k = j] = π j and P[X k+ = j] = π j : P[X k = j X k+ = i] = π jp[x k+ = j X k = i] = π jp j,i 4

5 (c Prove that if the original Markov chain is time reversible, so that P i,j = π j P j,i, then Q i,j = P i,j That is, the states follow the same transition probabilities whether viewed in forward order or reverse order This follows directly from part (c, where we obtain which can only be true if Q i,j = P i,j Q i,j = π j P j,i,