Exercise 5 Release: Due:

Transcription

1 Stochastic Modeling and Simulation Winter 28 Prof. Dr. I. F. Sbalzarini, Dr. Christoph Zechner (MPI-CBG/CSBD TU Dresden, 87 Dresden, Germany Exercise 5 Release: Due: Question : Variance of MC Integration vs Importance Sampling a umerically compute the mean and variance of (x2 dx using simple MC integration with a standard uniform PDF proposal p(x and importance sampling with the proposal PDF q(x = 2x, ( x. Use trials with = samples for each method. Compare the mean and variances with the analytical solution (see lecture notes. For simple Monte Carlo integration, the estimate is: x 2 dx Θ = x 2 i Analytically, the solution is: x 2 dx = x = The variance of Θ is: { V ar{θ = V ar x 2 i = V ar{x2 = ( E[x 4 ] E[x 2 ] 2 ( = ( x 4 p(xdx x 2 p(xdx = ( x5 x 2 5 ( = ( 5 = ( 4 45

2 import numpy a s np 2 import s c i p y. s t a t s as s s import m a t p l o t l i b. pyplot as p l t 4 from s c i p y. s p e c i a l import e r f 5 from s c i p y. s p e c i a l import e r f i n v 6 7 # simple mc i n t e g r a t i o n 8 def mci ( a, b, n : 9 x = np. random. uniform ( a, b, n f x = x 2 i n t e g r a l = (b a /n np. sum( f x 2 return i n t e g r a l 4 r e s u l t s = np. array ( [ ] 5 f o r i in range ( : 6 r e s u l t s = np. append ( r e s u l t s, mci (,, 7 r e s u l t s. std ( 2 # v a r i a n c e 8 r e s u l t s. mean ( # mean Using simple Monte Carlo integration, the mean and variance are.697 and.8655, respectively. The variance roughly matches the analytical solution Θ = ( 4 45 = for =. For importance sampling with a proposal distribution of q(x = 2x: x 2 dx Θ 2 = x 2 i q(x i The variance of Θ 2 is: { { V ar{θ 2 x 2 i = V ar = V ar 2x i = { x V ar = 2 4 V ar{x = ( E[x 2 ] E[x] 2 4 = 4 = 4 = 4 = 4 ( ( x 2 q(xdx ( ( x 2 (2xdx 2x4 4 ( 2 ( 2x 2 ( 2 = x i 2 xq(xdx x(2xdx ( 72 2

3 # importance sampling 2 def impsam ( a, b, n : x = np. random. uniform (,, n 4 y = np. s q r t ( x 5 f y = ( y 2 /(2 y 6 i n t e g r a l = /n np. sum( f y 7 return i n t e g r a l 8 9 r e s u l t s = np. array ( [ ] f o r i in range ( : r e s u l t s = np. append ( r e s u l t s, impsam (,, 2 r e s u l t s. std ( 2 # v a r i a n c e r e s u l t s. mean ( # mean With importance sampling, the mean and variance are.2826 and , respectively. The variance from importance sampling is lower than from simple Monte Carlo integration. b Compute the variance from using simple MC integration with a uniform proposal distribution from X U(, and importance sampling with a normal proposal distribution from X (5, 4 to estimate (x x2 dx. Solve this problem analytically and numerically using trials with = samples for each method. You may use numerical solvers to evaluate definite integrals in the analytical solution. The analytical solution of (x x2 dx is: ( x (x x 2 2 dx = 2 x = = 5 2 For simple MC integration, the variance is: { V ar{θ = V ar (x i x 2 i = ( E[(x x 2 ] (E[(x x 2 ] = = = ( (x x 2 p(xdx ( (x x 2 dx ( ( 5 = V ar{x x2 ( ( = 5 9 (x x 2 p(xdx x x 2 dx

4 # simple mc i n t e g r a t i o n 2 def mci ( a, b, n : x = np. random. uniform ( a, b, n 4 f x = x x 2 5 i n t e g r a l = (b a /n np. sum( f x 6 return i n t e g r a l 7 8 r e s u l t s = np. array ( [ ] 9 f o r i in range ( : r e s u l t s = np. append ( r e s u l t s, mci (,, r e s u l t s. std ( 2 # v a r i a n c e 2 r e s u l t s. mean ( # mean Using simple Monte Carlo integration with = samples, the mean and variance are and , respectively. The analytical solution of variance is 5/9 = For importance sampling, the variance is: { V ar{θ 2 (x i x 2 i = V ar q(x i { (x i x 2 i = V ar q(x i = { (x x 2 V ar q(x ( [ ((x = ] x 2 2 [ ] (x x 2 2 E E q(x q(x ( = ( (x x 2 2 ( (x x 2 q(xdx q(x q(x ( = ( (x x 2 dx (x x 2 dx q(x = ( (x x 2 dx (x x 2 dx (x µ2 e 2σ 2 2πσ 2 ( = ( = 2.52 q(xdx 4

5 # importance sampling 2 def impsam ( a, b, n : x = np. random. normal ( 5, 2, n 4 # x only within bounds 5 x = x [ np. where ( ( x >= & ( x <= ] 6 # q ( x i s normalized to account f o r bounds 7 q = s s. norm. pdf ( x, l o c =5, s c a l e =2/ e r f (5/(2 np. s q r t ( 2 8 f x = ( x x 2 /q 9 i n t e g r a l = /( l e n ( x np. sum( f x return i n t e g r a l 2 r e s u l t s = np. array ( [ ] f o r i in range ( : 4 r e s u l t s = np. append ( r e s u l t s, impsam (,, 5 r e s u l t s. std ( 2 # v a r i a n c e 6 r e s u l t s. mean ( # mean Using importance sampling with = samples, the mean and variance are and.525, respectively. The analytical solution of variance is 2.52/ =.25, which is smaller than the variance from simple Monte Carlo integration. Question 2: Optimum Proposal Distribution Find the optimum σ 2 for the proposal distribution q(x (.5, σ 2 to estimate the CDF of X Beta(2, 2 by importance sampling. Solve this problem numerically by scanning through σ 2 values from. to. in increments of (../2. Use trials with = samples to obtain the variances. The normal distribution (.5, σ 2 that most closely matches the Beta(2, 2 distribution will be the best proposal distribution to use. # p l o t s normal d i s t r i b u t i o n s and a beta d i s t r i b u t i o n 2 x = np. l i n s p a c e (,, 2 f o r i in norm vars : 4 y = s s. norm. pdf ( x, l o c =. 5, s c a l e = np. s q r t ( i 5 p l t. p l o t ( x, y 6 y = s s. beta. pdf ( x, a = 2, b = 2 7 p l t. p l o t ( x, y 8 p l t. show ( 5

6 # importance sampling 2 def importance sampling (n, var : x = np. random. normal ( l o c =. 5, s c a l e = np. s q r t ( var, s i z e = n 4 # x only within bounds 5 x = x [ np. where ( ( x >= & ( x <= ] 6 # n o r m a l i z a t i o n f a c t o r k 7 k =. 5 ( e r f ((.5 /( np. s q r t (2 var +e r f (. 5 / np. s q r t (2 var 8 # normalized q ( x 9 q = s s. norm. pdf ( x, l o c =. 5, s c a l e = np. s q r t ( var /k f x = s s. beta. pdf ( x, a = 2, b = 2 /q i n t e g r a l = / l e n ( x np. sum( f x 2 return i n t e g r a l 4 # run importance sampling at d i f f e r e n t v a r i a n c e s 5 norm vars = np. l i n s p a c e (.,., 2 6 mc vars = np. array ( [ ] 7 f o r i in norm vars : 8 r e s u l t s = np. array ( [ ] 9 f o r j in range ( : 2 r e s u l t s = np. append ( r e s u l t s, importance sampling (, i 2 mc vars = np. append ( mc vars, r e s u l t s. std ( p l t. p l o t ( norm vars, mc vars 24 p l t. show ( The importance sampling estimate variance was lowest using a proposal distribution of (.5,

7 Question : Variance Reduction by Antithetic Variates a Estimate (/(x+dx using MC integration with 2 = samples from a uniform proposal distribution X U(, and with = 5 samples from X together with its antithetic variates X. Compare the mean and variance between the two approaches from trials. # simple MC i n t e g r a t i o n 2 def mci ( a, b, n : x = np. random. uniform ( a, b, n 4 f x = /( x+ 5 i n t e g r a l = (b a /n np. sum( f x 6 return i n t e g r a l 7 8 r e s u l t s = np. array ( [ ] 9 f o r i in range ( : r e s u l t s = np. append ( r e s u l t s, mci (,, p r i n t ( r e s u l t s. mean ( 2 p r i n t ( r e s u l t s. std ( 2 Mean and variance from simple Monte Carlo integration are and.989, respectively. # using a n t i t h e t i c v a r i a t e s 2 def m c i a n t i ( a, b, n : x = np. random. uniform ( a, b, i n t ( n/2 4 a n t i x = x # a n t i t h e t i c v a r i a t e s 5 a l l x = np. append ( x, a n t i x # combine a l l x 6 f x = /( a l l x + 7 i n t e g r a l = (b a /n np. sum( f x 8 return i n t e g r a l 9 r e s u l t s = np. array ( [ ] f o r i in range ( : 2 r e s u l t s = np. append ( r e s u l t s, m c i a n t i (,, p r i n t ( r e s u l t s. mean ( 4 p r i n t ( r e s u l t s. std ( 2 Mean and variance from simple Monte Carlo integration using antithetic variates are and.88, respectively. Variance is lower with antithetic variates. 7

8 b Estimate (x x2 dx by importance sampling with a proposal distribution p(x (5, 4 using antithetic variates from a uniform distribution. Use trials and a total of 2 = samples to obtain the importance sampling variance. How different is it from the variance obtained in question b? Hint: Generate random numbers and their antithetic variates from a uniform distribution and do a transformation to a normal distribution for importance sampling OR generate antithetic variates of random numbers from a normal distribution by getting their mirror image values from the mean. # importance sampling with a n t i t h e t i c v a r i a t e s 2 def impsam anti ( a, b, n : x = np. random. uniform (,, i n t ( n/2 4 a n t i x = x 5 a l l x = np. append ( x, a n t i x 6 y = 2 np. s q r t ( 2 e r f i n v (2 a l l x +5 7 y = y [ np. where ( ( y >= & ( y <= ] 8 q = s s. norm. pdf ( y, l o c =5, s c a l e =2/ e r f (5/(2 np. s q r t ( 2 9 f y = ( y y 2 /q i n t e g r a l = /( l e n ( y np. sum( f y return i n t e g r a l 2 r e s u l t s = np. array ( [ ] 4 f o r i in range ( : 5 r e s u l t s = np. append ( r e s u l t s, impsam anti (,, 6 p r i n t ( r e s u l t s. mean ( 7 p r i n t ( r e s u l t s. std ( 2 Mean and variance from importance sampling using antithetic variates are and 59.97, respectively. The variance is higher as compared to the importance sampling in b. Antithetic variates don t reduce the variance of Monte Carlo integration estimates when integrating non-monotonic functions. 8