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### IEORE4404_001_2008 Problems_from_Ross

Course: IEOR E4404, Spring 2012
School: Columbia College
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Word Count: 922

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4404 Simulation Prof. IEOR Jose Blanchet Solutions to Chapter 8 Ross Book May 1, 2008 Page 1 of 4 Solutions to Chapter 8 Ross Book 8.3 (a) Dene the indicate function 1( 5=1 iXi 21.6), then E [1( 5=1 iXi 21.6)] = i i P ( 5=1 iXi 21.6). So, 1( 5=1 iXi 21.6) is an estimator. Because logU gives a i i an exponential random variable with mean 1, denote h(U) = 1( 5=1 ilogUi 21.6), i where Ui is Uniform r.v for...

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4404 Simulation Prof. IEOR Jose Blanchet Solutions to Chapter 8 Ross Book May 1, 2008 Page 1 of 4 Solutions to Chapter 8 Ross Book 8.3 (a) Dene the indicate function 1( 5=1 iXi 21.6), then E [1( 5=1 iXi 21.6)] = i i P ( 5=1 iXi 21.6). So, 1( 5=1 iXi 21.6) is an estimator. Because logU gives a i i an exponential random variable with mean 1, denote h(U) = 1( 5=1 ilogUi 21.6), i where Ui is Uniform r.v for all i = 1, ..., 5. (b) (b) Let Y1 = h(U) and Y2 = h(1 U), so Y1 +Y2 is an antithetic estimator. Because 2 h(U) is decreasing in each Ui , Y1 and Y2 are negatively correlated. (c) Yes, it is ecient, because the new estimator has smaller variance. 8.5 (a) Consider the control variate Z , so the estimator is Z 3 +eZ +c Z , where c = Cov (Z 3 + 1 eZ , Z ) and can be approximated by c = n1 n (Xi X )(Zi Z ), where Xi = i=1 3 + eZi . Let X + c Y be the controlled estimator with variance 2 = V ar (X + c Y ) = Zi 1 n (V ar (X ) Cov (X, Z )) (b) The 95% condence interval is (X + c Y 1.96 length is within 0.1 require that 2 1.96 2 M,X + c Y + 1.96 2 M ). That the 2 0.1 M 1 96 i.e., M ( 20..1 )2 2 . 2 can be approximated to 11.7611 by running 104 iterations. Therefore, we need to at least M = 2.0 104 replications. By running simulation, we get the 95% condence interval (1.5976, 1.6889). The following is the Matlab code: clear all; N=2*10^4; X=zeros(N,1); Z=zeros(N,1); for i=1:N Z(i)=randn; X(i)=Z(i)^3+exp(Z(i)); end covXZ=cov(X,Z); cov_XZ=covXZ(1,2);% convariance(X,Z) c_star=-cov_XZ/var(Z); % c^* theta_bar=mean(X)+c_star*mean(Z) % sample mean sigma_square=(var(X)-cov_XZ^2) Confidence_interval=[theta_bar-1.96*sqrt(sigma_square/N), theta_bar+1.96*sqrt(sigma_square/N)] 8.10 (a) We compute c as follows: c = cov (I, X ) var(X ) 2 IEOR 4404, Solutions to Chapter 8 Ross Book Now, cov (I, X ) = E [IX ] E [I ]E [X ]. And, E [IX ] = E [IX |X a]P [X a] = 1 E [X |X a]P [X a] = a2 /2, while E [I ]E [X ] = a/2. Therefore, cov (I, X ) = 2 a(a 1 1) 0. Recall V ar(X ) = 12 and V ar(I ) = a a2 . So, the percentage of variance reduction is [ 1 a(a 1)]2 [Cov (I, X )]2 = 21 = 3a(1 a) V ar(I )V ar(X ) 12 a(1 a) (b) If X were exponential with mean 1, E [IX ] = E [X |X a]P [X a] = 1 (1 + a)ea , E [X ] = 1, V ar(X ) = 1. So, Cov (I, X ) = (1 a) (1 + a)ea . So, the percentage of variance reduction is [(1 a) (1 + a)ea ]2 [Cov (I, X )]2 = V ar(I )V ar(X ) a(1 a) (c) Here, note that as X increases, I has a tendency to decrease which justies the negative correlation. 8.17 (a) The raw estimator would be to generate and X Y and average the number of times we get that X + Y t. (b) Using indicator function, P (X + Y t) = E [1(X + Y t)]. We can condition on either X or Y . (c) Once again, you can use either X or Y here as control variables. For example, E [1(X + Y t)|X ] + c(Y y ) is new estimator. 8.28 & 8.29 When using importance sampling, we rst need to nd the appropriate tilting that would give E [X ] 5. To do this recall that, under the tilted distribution, component i has a Bernoulli distribution with parameter p,i = pi e pi e + 1 pi where pi = 0.5 + i/50. Therefore, under the tilted distribution, X has mean 20 f () = E [X ] = 20 p,i = i=1 i=1 pi e pi e + 1 pi After plotting the function f () it can be seen that f (2.165) = 4.9990, so we choose = 2.165. The moment generating function of X is the product of the moment generating functions of the components, that is, MX () = 20 (pi e + 1 pi ) i=1 The importance sampling estimator for = P (X 5) is then Z = I (X 5)e X MX ( ) where the components dening X are generated as Bernoulli random variables with parameters p,i , i = 1, 2, ..., 20. For 106 simulations the following values were obtained Z (n) = 6.7621e 10 S 2 (n) = 3.7427e 14 where Z (n) is the sample mean, and S 2 (n) is the sample variance. Similarly, dene Z 1 = I (X == 5)e X MX ( ) IEOR 4404, Solutions to Chapter 8 Ross Book . By running simulation, we get sample mean and sample variance Z (n) = 6.6980e 10 S 2 (n) = 3.7386e 14 P (X 5) = 6.7621e 10 0 P (X = 5) P (X = 5|X 5) = 0.9905 P (X 5) % function f() function y=f(x) p=zeros(20,1); pp=zeros(20,1); for i=1:20 p(i)=.5+i/50; pp(i)=p(i)*exp(x)/(p(i)*exp(x)+1-p(i)); end y=sum(pp); Then, by plotting to nd % prob 3 clear all; for i=1:10/0.001 x(i)=-5+.001*i; y(i)=f(x(i)); if i>1 && (y(i)-5)*(y(i-1)-5)<=0 y(i-1) x(i-1) plot(x,y) break; end end The code for the importance sampling % inportance sampling clear all; theta=-2.165; p=zeros(20,1); p_M=zeros(20,1); for i=1:20 p(i)=.5+i/50; p_M(i)=p(i)*exp(theta)+1-p(i); end x_Moment=prod(p_M); n=10^6; % number of simulations z=zeros(n,1); z1=zeros(n,1); x=zeros(20,1); for j=1:n 3 4 IEOR 4404, Solutions to Chapter 8 Ross Book for k=1:20 if rand<=p(k) x(k)=1; else x(k)=0; end end if sum(x)<=5 z(j)=exp(-theta*sum(x))*x_Moment; else z(j)=0; end if sum(x)==5 z1(j)=exp(-theta*sum(x))*x_Moment; else z1(j)=0; end end Sample_mean=mean(z) % p(X<=5) Sample_variance=var(z) Sample_mean1=mean(z1) %p(x==5) Sample_variance1=var(z1) Conditional_prob=Sample_mean1/Sample_mean
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