Unformatted text preview: . If this occurs, then we have that J S = 1 and therefore I S = J S K S = 1 . Clearly, then, in this case X = S . If on the other hand, we have that & S =& > U we must have that J S = 0 and therefore I S = 0 . So, we need to search for min f n > S : K n = 1 g (which is the second success in the sequence of K n &s). To so we simply simulate another Geo ( & ) , independent of everything else and add it up to S , that is S ± S + Geo ( & ) and once again we need to verify if J S = 1 , which we do exactly as we did before. We then continue in this way until we obtain both J S = 1 and K S = 1 at which point we let X = S . 1...
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This note was uploaded on 04/19/2011 for the course IEOR 4404 taught by Professor C during the Spring '10 term at Columbia.
 Spring '10
 C

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