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Unformatted text preview: Columbia University IEOR 4404: Simulation Fall 2009 Solutions to Assignment 5 1. (Exercise 5.14) (a) Consider X with distribution G . Since the denominator of the frac- tion is equal to G ( b )- G ( a ), we can easily guess that the infor- mation we have to use in order to write F as a conditional dis- tribution is a ≤ X ≤ b . Let us prove it. Assume therefore that F ( x ) = P ( X ≤ x | a ≤ X ≤ b ) and prove that F ( x ) = G ( x ) − G ( a ) G ( b ) − G ( a ) : F ( x ) = P ( X ≤ x | a ≤ X ≤ b ) = P ( X ≤ x | a ≤ X ≤ b ) P ( a ≤ X ≤ b ) P ( a ≤ X ≤ b ) = P ( X ≤ x, a ≤ X ≤ b ) P ( a ≤ X ≤ b ) using Bayes rule = P ( a ≤ X ≤ x ) P ( a ≤ X ≤ b ) since x ∈ [ a, b ] = G ( x )- G ( a ) G ( b )- G ( a ) Therefore, we proved that F can be written as F ( x ) = P ( X ≤ x | a ≤ X ≤ b ). (b) It is important to notice that the distribution F is defined only on the interval [ a, b ]. Therefore, by differentiating F , we obtain the following density: f ( x ) = F ′ ( x ) = g ( x ) G ( b )- G ( a ) 1 [ a,b ] ( x ) . Now, we have to find an upper bound c for the ratio f ( y ) g ( y ) for all y ∈ [ a, b ]. Given the definition of f above, it is easy to see that c = 1 G ( b ) − G ( a ) works. Therefore, we are able to apply the usual ac- ceptance/rejection algorithm to generate Y with distribution F : STEP 1. Generate a random variable X with distribution G . STEP 2. Generate U ∼ Uniform[0 , 1] STEP 3. If U ≤ f ( X ) c g ( X ) , then set Y=X. Otherwise, go back to STEP 1. By running the algorithm above, we know that Y will have the de- sired distribution F . Nonetheless, the question asks us to show that the following modified version of the usual acceptance/rejection al-...
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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