S6880 #13. Variance Reduction Methods

Transcription

1 S6880 #13 Variance Reduction Methods

2 1 Variance Reduction Methods Variance Reduction Methods 2 Importance Sampling Importance Sampling 3 Control Variates Control Variates Cauchy Example Revisited 4 Antithetic Variates Antithetic Variates 5 Conditioning Conditioning Outline (WMU) S6880 #13 S6880, Class Notes #13 2 / 29

3 General Techniques Reduction of simulation cost variance reduction. General Techniques 1 Importance Sampling 2 Control and antithetic variates 3 Conditioning 4 Use of special features of a problem (WMU) S6880 #13 S6880, Class Notes #13 3 / 29

4 A note about hit-or-miss Monte Carlo integretation = less efficient than original Monte Carlo φ(x) c 0 a θ = b a φ(x)dx φ(x) c over [a, b] b draw uniform sample from U([a, b] [0, c]) ˆθ = c(b a) {proportion of sample items falling in shaded area} Var(hit-or-miss Monte-Carlo) Var(original Monte-Carlo) (WMU) S6880 #13 S6880, Class Notes #13 4 / 29

5 1 Variance Reduction Methods Variance Reduction Methods 2 Importance Sampling Importance Sampling 3 Control Variates Control Variates Cauchy Example Revisited 4 Antithetic Variates Antithetic Variates 5 Conditioning Conditioning Outline (WMU) S6880 #13 S6880, Class Notes #13 5 / 29

6 What is Importance Sampling Want: Estimate θ = φ(x)f (x)dx = E f φ(x), X p.d.f. f estimated by ˆθ f = 1 n n i=1 φ(x i.i.d. i), X 1,..., X n f. Method: Instead of sampling from f, we sample from another p.d.f. i.i.d. g. i.e., X 1,..., X n g. ˆθ g = 1 n n i=1 ψ(x i), where ψ = φ f g. Note that [ θ = φ(x)f (x)dx = φ(x) f (x) ] g(x)dx g(x) = ψ(x)g(x)dx = Eψ(X), X g. (WMU) S6880 #13 S6880, Class Notes #13 6 / 29

7 Unbiasedness of Importance Sampling Estimator Note that 1 ˆθ g is unbiased (since E ˆθ g = Eψ(X) = θ). 2 Var(ˆθ g ) = Var g (ψ(x))/n = 1 (ψ(x) θ) 2 g(x)dx n = 1 ( φ(x) f (x) ) 2 n g(x) θ g(x)dx. which can be made small if we choose g such that ψ = φ f g is closed to a constant. = Choose g such that { g c φf g is easier to sample than f (WMU) S6880 #13 S6880, Class Notes #13 7 / 29

8 Cauchy Example, continued θ = P(Cauchy > 2), f Cauchy, φ(x) = I(x > 2). 1 So, want g I(x > 2). π(1+x 2 ) Try g(x) 2 2 I(x > 2) p.d.f. of x 2 U(0,1) p.d.f. of 1 U(0,1/2). Then ψ(x) = φ(x) f (x) g(x) = 1 x 2 2π I(x > 2) which is a (monotone) 1+x 2 increasing function over (2, ) with values bounded by 4 5 and 1. = constant. Can show Var g (ψ(x)) and hence Var(ˆθ g ) n (WMU) S6880 #13 S6880, Class Notes #13 8 / 29

9 θ = = = = = Note about Cauchy Example /2 1/2 0 1/2 0 ψ(x)g(x)dx = 1 2π 1 2π 2 1 g(x)dx 1 + x 2 1 x 2 2π 1 + x 2 g(x)dx y 2 g(y 1 )( y 2 )dy (let y = x 1 ) 1 [1 π(1 + y 2 ) 2 g(y 1 )y 2] dy 1 π(1 + y 2 ) dy (Note : g(y 1 ) = 2y 2 ) = Monte Carlo integration 1/2 0 f (y)dy see class notes 12. (WMU) S6880 #13 S6880, Class Notes #13 9 / 29

10 1 Variance Reduction Methods Variance Reduction Methods 2 Importance Sampling Importance Sampling 3 Control Variates Control Variates Cauchy Example Revisited 4 Antithetic Variates Antithetic Variates 5 Conditioning Conditioning Outline (WMU) S6880 #13 S6880, Class Notes #13 10 / 29

11 What are Control Variates Goal: Estimate θ = Eφ(X). Method: Find ψ( ) such that µ ψ = Eψ(X) = R ψ(x)f (x)dx is known and easy-to-calculate. Idea is to choose ψ( ) so that ψ φ, ψ(x) is a control variate. θ = Eφ(X) = E[φ(X) ψ(x)] + Eψ(X) = E[φ(X) ψ(x)] + µ ψ. Then ˆθ = µ ψ + 1 n n i=1 [φ(x i) ψ(x i )] where i.i.d. X 1,..., X n f. Var(ˆθ) = 1 n Var[φ(X) ψ(x)] 1 Varφ(X) if ψ φ. n (WMU) S6880 #13 S6880, Class Notes #13 11 / 29

12 How to Find Control Variates i.i.d. Example: X 1,..., X n F with mean µ (known). Want to estimate θ = Eφ(X) = Eφ(X 1,, X n ) = E(median) = EX (m+1) if n = 2m + 1. Choose ψ(x) = X = the sample mean, then Eψ(X) = µ(= µ ψ ). Then estimate θ by ˆθ = µ + 1 N N [φ(x (i) ) ψ(x (i) )], where i=1 X (i) = i th sample of size n = 2m + 1 = (X (i) i,, X (i) n ) i.i.d. F for i = 1,, N. ˆθ often has smaller variance than 1 N N i=1 φ(x(i) ). (WMU) S6880 #13 S6880, Class Notes #13 12 / 29

13 General Result In general, ψ(x) can be taken as an approximation of φ(x) based on some basis {b k (x)}. That is, ψ(x) = K k=1 a kb k (x), where Eb k (X) = µ k and is known. If a k s are unknown, then estimate a k s by least square estimates based on initial sample and obtain â k s. Use ˆψ(x) = K k=1 âkb k (x) and estimate θ by ˆθ = K â k µ k + 1 n k=1 n [φ(x (i) ) ˆψ(X (i) )]. i=1 (WMU) S6880 #13 S6880, Class Notes #13 13 / 29

14 Cauchy Example, revisited 1 The first few terms of the Taylor expansion of 1 1+x 2 suggest that we can consider x 2 and x 4. θ = P(X > 2) = 1 2 P(0 X 2) = π 1 + x 2 dx = f (x)g(x)dx, f (x) = 1 1 = p.d.f. of Cauchy π 1 + x 2 = π Eφ(X) g(x) = 1 2 I [0,2](x) = p.d.f. of U(0,2) where φ(x) = 1 1+x 2 with X U(0, 2). If we approximate φ(x) by ψ(x) = a 0 + a 2 x 2 + a 4 x 4 : sample x 1,, x 5 U(0, 2) ( initial sample) yielded: (WMU) S6880 #13 S6880, Class Notes #13 14 / 29

15 Cauchy Example, revisited 1, continued obs. x φ(x) regress y = φ(x) on x 2, x 4 = â 0 =.943, â 2 =.513, â 4 = That is, ˆψ(x) = x x 4. Then ˆθ = π [( x x 4 ) + 1 n = π [ n n [φ(x i ) ˆψ(x i )]] i=1 n [φ(x i ) ˆψ(x i )]] i=1 here EX k = 2 0 x k 1 2k 2dx = k+1 so EX 2 = 4 3, EX 4 = (WMU) S6880 #13 S6880, Class Notes #13 15 / 29

16 Cauchy Example, revisited 1, continued Var(ˆθ) = 4 π 2 n Var(φ(X) ˆψ(X)) = 4 π 2 n ( ) /n Plotting φ(x) and ˆψ(x) suggests that we would do better by including terms in x and x 3. Regress y on x, x 2, x 3, x 4 â 0 = 1.012, â 1 =.069, â 2 = 1.076, â 3 =.813, â 4 =.181 with variance /n φ(x) ψ^(x) (WMU) S6880 #13 S6880, Class Notes #13 16 / 29

17 Cauchy Example, revisited 1, continued Note: can increase the precision by increasing the initial sample size. For instance, an initial sample of size 10 yielded 0.036, 1.642, 1.829, 1.420, 0.532, 1.974, 1.245, 1.615, 0.303, Fit φ(x) on polynomial of degree 4 on x yielded ˆψ(x) = x.962x x 3.141x 4 with Var(ˆθ) /n. (WMU) S6880 #13 S6880, Class Notes #13 17 / 29

18 Cauchy Example, revisited 2 θ = P(X > 2) = 1 2 Ef (Y ), Y U(0, 1 2 ) = 1 1 Eφ(Y ) where φ(y) = 2π 1 + y 2, Y U(0, 1 2 ) Approximate φ(y) by ψ(y) = a 0 + a 1 y + a 2 y 2 + a 3 y 3 + a 4 y 4 : obs. y z = φ(y) where y s were sampled from U(0, 1 2 ). Regress z on y, y 2, y 3, y 4 : â 0 =.997, â 1 =.056, â 2 = 1.376, â 3 = 1.078, â 4 =.246. That is, ˆψ(y) = y 1.376y y 3.246y 4. (WMU) S6880 #13 S6880, Class Notes #13 18 / 29

19 Cauchy Example revisited 2, continued EY k = 1/2 0 y k 2dy = 2 k k+1, so ˆθ = 1 [ π n ] n (φ(y) ˆψ(y)). i=1 Var(ˆθ) = 1 4π 2 n Var(φ(Y ) ˆψ(Y )), where Y U(0, 1 2 ) /n = constant. Taking n = 1 (sample of size 1!) gives an accurate enough approximation to θ. (WMU) S6880 #13 S6880, Class Notes #13 19 / 29

20 1 Variance Reduction Methods Variance Reduction Methods 2 Importance Sampling Importance Sampling 3 Control Variates Control Variates Cauchy Example Revisited 4 Antithetic Variates Antithetic Variates 5 Conditioning Conditioning Outline (WMU) S6880 #13 S6880, Class Notes #13 20 / 29

21 Antithetic Variates Method Goal: Estimate θ = Eφ(X). Method: Find ψ such that ψ(x) has the same distributions as φ(x) such that cor(φ, ψ) < 0. Then θ = 1 2E[φ(X) + ψ(x)] and estimate θ by ˆθ = 1 n 2n i=1 [φ(x i) + ψ(x i )]. Var(ˆθ) = 1 Var[φ(X) + ψ(x)] 4n = 1 [Varφ(X) + Varψ(X) + 2cov(φ(X), ψ(x))] 4n = 1 [Varφ(X) + cov(φ(x), ψ(x))] 2n = Varφ(X) [1 + cor(φ(x), ψ(x))]. 2n Consequently, there are gains (in reduction of variance) since cor(φ(x), ψ(x)) < 0. (WMU) S6880 #13 S6880, Class Notes #13 21 / 29

22 How to Choose Antithetic Variates Theorem (Ripley) If U U(0, 1) then { Eφ(U) = Eφ(1 U) If, in addition, φ is monotone, then Varφ(U) = Varφ(1 U) cor(φ(u), φ(1 U)) < 0. (WMU) S6880 #13 S6880, Class Notes #13 22 / 29

23 Remarks on the Theorem 1 Reasons for monotone: θ = Eφ(X) = EZ, Z = φ(x) F Z (since F Z is monotone, so is F 1 1 Z ). θ = EZ = EFZ (U). So we can choose ψ(x) = F 1 1 Z (1 U) = FZ (1 F Z (Z )). 2 Theorem (Ripley, about generalization): If X is symmetric about c, then can choose ψ(x) = φ(2c X). 3 Use other method if F 1 Z is hard to calculate or φ is not monotone. (WMU) S6880 #13 S6880, Class Notes #13 23 / 29

24 Example φ is symmetric about 1 2, cor(φ(u), φ(1 U)) 1. Better use g(x) = { x, x < x, x > 1 2 ψ(x) = φ(g(x)). Then Eψ(X) = Eφ(X), for X U(0, 1) (WMU) S6880 #13 S6880, Class Notes #13 24 / 29

25 Cauchy Example, revisited 2 θ = π(1 + x 2 ) dx = 1 2 Eh(X), h(x) = 2 π(1 + x 2, X U(0, 2) ) = 1 2 Eh(2U) = 1 2 Eφ(U) by letting X = 2U, U U(0, 1), φ(u) = h(2u) = 2 π(1+4u 2 ). So, we can estimate θ by since φ is monotone. ˆθ = n n [φ(u i ) + φ(1 U i )] i=1 Var(ˆθ) = 1 4n Var[φ(U) + φ(1 U)] ( )/n. (WMU) S6880 #13 S6880, Class Notes #13 25 / 29

26 1 Variance Reduction Methods Variance Reduction Methods 2 Importance Sampling Importance Sampling 3 Control Variates Control Variates Cauchy Example Revisited 4 Antithetic Variates Antithetic Variates 5 Conditioning Conditioning Outline (WMU) S6880 #13 S6880, Class Notes #13 26 / 29

27 About Conditioning Recall: 1 E[X Y = y] = h(y), a function of y. 2 EX = E[E[X Y ]] 3 Var(X) = Var(E[X Y ]) + E[Var(X Y )] so that Var(E[X Y ]) Var(X) with the equality iff X = g(y ) (in which case Var(X Y = y) = Var(g(y) y) = 0!) Goal: Estimate θ = Eφ(X). Method: If there exists Y such that ψ(y) = E[φ(X) Y = y] can be easily calculated, then estimate θ by ˆθ = 1 n n i=1 ψ(y i) with variance Var(ˆθ) = 1 n Var(ψ(Y )) = 1 Var(E[φ(X) Y ]) n 1 n Var(φ(X)). (WMU) S6880 #13 S6880, Class Notes #13 27 / 29

28 Example Suppose X = Z W, Z N(0, 1), W > 0, and W and Z are independent (recall swindle ). The goal is to estimate θ = P(X n > c), where X n is the sample mean of the sample X 1,..., X n. No conditioning (naive!): where X (j) n 1 j N. ˆθ = 1 (j) #{X n > c, 1 j N}, N is the sample mean of the j th sample {X (j) 1,, X (j) n }, (WMU) S6880 #13 S6880, Class Notes #13 28 / 29

29 Example, continued Conditioning: Denote W = (W 1,..., W n ) t, w = (w 1,..., w n ). Given W = w, n X n = 1 n i=1 where σ 2 (w) = ( Z i N 0, 1 n W i n 2 n i=1 1 w 2 i i=1 ( ) c So, P(X n > c W = w) = 1 Φ σ(w) ) 1 wi 2 N(0, σ 2 (w)), ψ(w), where Φ( ) is the standard normal c.d.f. Therefore, a better estimate is ˆθ = 1 N N i=1 ψ(w(j) ), where W (j) denotes the j th sample {W (1) 1,, W (j) n } from W, 1 j N. (WMU) S6880 #13 S6880, Class Notes #13 29 / 29