# CommentSolution4_18 - If this occurs then we have that J S...

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What is the algorithm in Q. 18 doing? In Q. 17 you have established that X = min f n ° 1 : I n = 1 g where the I n °s are independent (but NOT identically distributed) Bernoullies. In fact, P ( I n = 1) = ° n = 1 ± P ( I n = 0) . Now, we have that there exists ° 2 (0 ; 1) such that ° n ² 1 . So, we can think of I n having the following decomposi- tion. I n = J n K n , where f J n : n ° 1 g and f K n : n ° 1 g are two independent sequences of independent Bernoulli random variables. The J n °s are NOT iden- tically distributed, P ( J n = 1) = ° n = 1 ± P ( J n = 0) . The K n °s are i.i.d. with P ( K n = 1) = ° = 1 ± P ( K n = 0) . So, in order to simulate X we can do the following. First we simulate S = min f n ° 1 : K n = 1 g (note that S ² X ). Observe that S ³ Geo ( ° ) , we know then that K S = 1 to see if I S = 1 we must check if J S = 1 and to do so we simulate an independent uniform U ³ U [0 ; 1] and check if ° S
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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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