In a simple graph, G=(V, E), for each pair of distinct vertices u,v, edge(u,v) appears with probability 1/2. A triangle is a cycle of length three. X is a random variable denoting the number of triangles in a random graph G. Given the expected value of X,
E(X) = (n choose 3)*(1/8)
for n = # of nodes, and variance,
Var(X) = (n choose 3) * (n +5) / 64
give an upper bound using Chernoff bounds on the probability that X exceeds its expected value by 10%. Give general expression in terms of n. (Keep in mind that X is the sum of DEPENDENT random variables, as each triangle's probability of being present is affect by those around it. i.e. if triangle (u, v, w) is present, triangle (u, v, x) has a high probability of also being present as one of its edges is already known to exist. If this assumption is wrong, please explain why.)
2007-01-21
12:48:05
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1 answers
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Anonymous
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Science & Mathematics
➔ Mathematics