IMPORTANCE SAMPLING Importance Sampling Background: let x = (x 1,..., x n ), θ = E[h(X)] = h(x)f(x)dx 1 N. h(x i ) = Θ,

Transcription

1 1 IMPORTANCE SAMPLING Importance Sampling Background: let x = (x 1,..., x n ), θ = E[h(X)] = h(x)f(x)dx 1 N N h(x i ) = Θ, if X i F (X), and F (x) is cdf for f(x). For many problems, F (x) is difficult to sample from and/or V ar(h) is large. If a related, easily sampled pdf g(x) is available, could use θ = E g [ h(x)f(x h(x)f(x) ] = g(x)dx g(x) g(x) 1 N h(x i )f(x i ), N g(x i ) with X i G(X), for associated cdf G(X). Importance sampling: if V ar( h(x)f(x) g(x) ) is small, g(x) samples are concentrated where h(x)f(x) is important :

2 2 Importance Sampling Example: θ = 1 0 ex2 dx. Try g(x) = e x ; so θ = 1 0 ex2 x e x dx. To find G(x), note 1 0 ex dx = e 1, so G(x) = 1 x e t dt = (e x 1)/(e 1); e 1 0 then using X i = ln(1 + (e 1)U i ), 1 θ = (e 1) e x2 x ex (e 1) N dx e X2 i X i, e 1 N 0 N = 10000; U = rand(1,n); Y = exp(u.^2); disp( [mean(y) 2*std(Y)/sqrt(N)]) % simple MC e = exp(1); X = log(1+(e-1)*u); T = (e-1)*exp(x.*(x-1)); disp( [mean(t) 2*std(T)/sqrt(N)]) % importance Error reduction by 1/ exp(x), exp(x 2 ), and 1+x 2 +x 4 / x

3 3 Alternative: g(x) = 1 + x 2, G(x)? θ = with X i 3 4 X X3. 0 e x2 3(1 + x 2 ) 1 + x 2 4 dx 4 3N N e X2 i 1 + X 2 i N = 10000; U = rand(1,n); I = rand(1,n)<3/4; X = I.*U + (1-I).*U.^(1/3); T = 4*exp(X.^2)./(3*(1+X.^2)); disp( [mean(t) 2*std(T)/sqrt(N)]) % importance Better g(x) = 1 + x 2 + x 4 /2, G(x) = x x x5 ; θ = N 1 0 e x2 1 + x 2 + x 4 /2 N e X2 i 1 + X 2 i + X4 i /2, 30(1 + x 2 + x 4 /2) dx 43 with X i x x x5. N = ; U = rand(1,n); V = rand(1,n); I = V<30/43; J = V>40/43; X = I.*U + (1-I).*(1-J).*U.^(1/3)+ J.*U.^(1/5); T = 43*exp(X.^2)./(30*(1+X.^2+X.^4/2)); disp( [mean(t) 2*std(T)/sqrt(N)]) % importance ,

4 4 Importance Sampling Example: θ = 1 e z2 /2 z 3 e z dz. N = ; Z = randn(1,n); Y = Z.^3.*exp(Z); disp( [mean(y) 2*std(Y)/sqrt(N)]) % simple MC Try g(z) = 1 e (z 1)2 /2 = 1 e w2 /2, with w = z 1; so θ = e 1/2 z 3 g(z)dz = e1/2 (w + 1) 3 e w2 /2 dw. Simulation uses θ e1/2 N N (Z i+1) 3, with Z i Normal(0, 1). N = ; Z = randn(1,n); Y = exp(1/2)*(z+1).^3; disp( [mean(y) 2*std(Y)/sqrt(N)]) % importance Error reduction by 1/4. Note θ = e1/2 (3w 2 + 1)e w2 /2 dw = 4e 1/

5 5 Higher dimensional problems: often f(x) g(x) = g 1 (x 1 )g 2 (x 2 ) g n (x n ), so samples are from a sequence of 1-d samples. 2-d example: θ = e(x 1+x 2 ) 2 dx; if g(x) = e x 1e x 2; θ = e((x 1+x 2 ) 2 x 1 x 2 e x 1+x 2 dx. After scaling, with X ij = ln(1 + (e 1)U ij ), 1 1 θ = (e 1) 2 e ((x 1+x 2 ) 2 x 1 x 2 ) ex 1+x (e 1) 2dx (e 1)2 N e (X 1i+X 2i ) 2 X 1i X 2i. N N = 10000; U = rand(2,n); T = exp(sum(u).^2); disp( [mean(t) 2*std(T)/sqrt(N)]) % simple MC e = exp(1); X = log(1+(e-1)*u); T = (e-1)^2*exp(sum(x).^2-sum(x)); disp( [mean(t) 2*std(T)/sqrt(N)]) Better g(x) = e 2x 1e 2x 2, with g(1, 1) = f(1, 1)? Then G i (x) = e2x 1 e 2 1, X ij = ln(1 + (e 2 1)U ij )/2, and 1 1 θ = (e2 1) e = exp(1); X = log(1+(e^2-1)*u)/2; T = (e^2-1)^2*exp(sum(x).^2-2*sum(x))/4; disp( [mean(t) 2*std(T)/sqrt(N)]) Better g(x) = 1 + (x 1 + x 2 ) 2? e ((x 1+x 2 ) 2 2(x 1 +x 2 )) 4e2(x 1+x 2 ) (e 2 1) 2 dx,

6 6 Tilted Densities g(x) : given pdf f(x) let M(t) = e tx f(x)dx (the moment generating function). The tilted density for f(x) is f t (x) = etx f(x) M(t). Examples Exponential densities: if f(x) = λe λx, x [0, ), f t (x) = (λ t)e (λ t)x, t < λ. Bernoullli pmf s: f(x) = p x (1 p) 1 x, x = 0, 1. M(t) = E f [e tx ] = e t p + (1 p), so ( ) f t (x) = etx p x (1 p) 1 x e = t x ( p 1 x, e t p+(1 p) e t p+(1 p) e p+(1 p)) t a Bernoulli RV with p t = e t p e t p+(1 p). So f/f t = et p+(1 p) = e tx (e t p + (1 p)) e tx Generalization: if f(x) is a Binomial(n, p) pmf, f t (x) is Binomial(n, e t p + 1 p), with M(t) = (e t p + 1 p) n. Normal densities: if f(x) = e x2 /2, x (, ), e tx f(x) = ext e x2 /2 = e (x t)2 /2 e t2 /2 so f t (x) = e (x t)2 /2, Normal(t, 1), with M(t) = e t2 /2. Generalization: if f(x) is a Normal(µ, σ 2 ) pdf, then f t (x) is a Normal(µ + σ 2 t, σ 2 ) pdf. Choosing t: pick t with small V ar( h(x)f(x) f t (x) ). Text heuristic for exponentials and Bernoullis: if h = I{ X i > a}, choose t = t with E t [ X i ] a.

7 Examples: 1. Bernoulli RV Examples: if X i s are independent Bernoulli(p i ) RV s and θ = I{ n X i > a} = I{S > a}. n ˆθ = I{S > a}e ts (e t p i + (1 p i )), with E t [ n X i ] = n e t p i e t p i + (1 p i ). Example with n = 20, p i =.4, a = 16; choose t so that E t [S] = 20.4et.4e t +.6 = 16, with solution et = 6; then p t =.8, e t p + (1 p) = 3, and estimator is ˆθ = I{ X i > a}6 S 3 20 = 3 20 S I{ X i > a}/2 S. Matlab N = ; p =.4; n = 20; I = sum( rand(n,n) < p ) > 16; % Simple MC disp([mean(i) 2*std(I)/sqrt(N)]) 6e e-05 S = sum( rand(n,n) <.8 ); % importance I = 3.^(20-S).*( S > 16 )./2.^S; disp([mean(i) 2*std(I)/sqrt(N)]) e e-07 N = ; p =.4; n = 20; I = sum( rand(n,n) < p ) > 16; % Simple MC disp([mean(i) 2*std(I)/sqrt(N)]) 4.82e-05 Note: θ = e-06 ( 20 ) i (.4) i (.6) 20 i

8 8 2. Exponential RV Example: if X i Exp( 1 i+2 ), i = 1,..., 4, S(X) = 4 X i, find 4 x i θ = h(x) e i dx, 0 0 with h(x) = S(x)I{S(x) > 62}. Raw simulation uses X ij Exp( 1 i+2 ), to estimate θ 1 N h(x j ). N j=1 Matlab N = ; U = rand(4,n); X = -diag([3:6])*log(1-u); S = sum(x); h = S.*( S > 62 ); disp( [mean(h) 2*std(h)/sqrt(N)]) Note: to find E[S S > 62], divide by E[I(S > 62)] E = h/mean((s>62)); disp([mean(e) 2*std(E)/sqrt(N)])

9 For tilted density, use common tilt parameter t, so that X i Exp(1/(i + 2) t), θ = 4 C N [0, )4 h(x)e ts(x) i + 2 e 1 (i + 2)t 4 (i + 2) N 4 h(x j )e ts(xj), with C = j=1 4 x i ( 1 i+2 t) 4 i+2 1 (i+2)t) 1 1 (i + 2)t. Text estimates good t =.14, by approximately solving 4 E t [X i ] = 3 1 3t t t t = dx; But guess and check with Matlab finds better t.136. Matlab tests: t =.14; Cd = 1./(1-[3:6]*t); C = prod(cd); St = -([3:6].*Cd)*log(1-U); ht = C*St.*( St > 62 ).*exp(-t*st); disp([mean(ht) 2*std(ht)/sqrt(N)]) t =.136; Cd = 1./(1-[3:6]*t); C = prod(cd); St = -([3:6].*Cd)*log(1-U); ht = C*St.*( St > 62 ).*exp(-t*st); disp([mean(ht) 2*std(ht)/sqrt(N)]) E = ht/mean(c*(st > 62).*exp(-t*St)); disp([mean(e) 2*std(E)/sqrt(N)])% Expected Value Note smaller standard errors compared to raw sampling.

10 Tilting for Normal Densities: if f(x) = 1 e (x µ)2 /2, tilted density f t (x) = f(x)ext M(t) is a shifted normal. Example: θ = 1 e z2 /2 z 3 e z dz. Pick t to match point (mode) where integrand a(z) = Ke (z 1)2 /2 z 3 is max. 10 exp(x x 2 /2) x 3 /sqrt(2 π) x

11 11 Or, pick t to match mean at t = 2.5. For either case: θ = 1 e (z t)2 /2 z 3 e z zt+t2 /2 dz = 1 e w2 /2 (w+t) 3 e (w+t)(1 t)+t2 /2), using z = w+t. Matlab tests: N = ; W = randn(1,n); t = (1+sqrt(13)/2; Y = (W+t).^3.*exp((W+t)*(1-t)+t^2/2); disp( [mean(y) 2*std(Y)/sqrt(N)]) % mode tilt t = 2.5; Y = (W+t).^3.*exp((W+t)*(1-t)+t^2/2); disp( [mean(y) 2*std(Y)/sqrt(N)]) % mean tilt Compare: t = 1; Y = (W+t).^3.*exp((W+t)*(1-t)+t^2/2); disp( [mean(y) 2*std(Y)/sqrt(N)]) % tilt = t=0; Y = (W+t).^3.*exp((W+t)*(1-t)+t^2/2); disp( [mean(y) 2*std(Y)/sqrt(N) ]) % t=0, raw MC

12 Tilting for Multidimensional Normal Density Problems: use vector t. Choice of t? Try to make V ar( h(x)f(x) f(x t) ) small: a) choose point t where h(x)f(x) is maximum (mode), or b) choose t = E[xh(x)]/E[h(x)] (mean). Asian Option example: this has S m = S m 1 e (r σ2 2 )δ+σ δz m, with δ = T/M, Z m Normal(0, 1) and expected profit θ = E[e rt max( 1 M = e rt = M S i (Z) K, 0)] max( 1 M M M zi 2/2 h(z) e ( ) dz, m with h(z) = e rt max( 1 M M S i(z) K, 0). 12 M S i (z) K, 0) e zi 2/2 ( ) dz m For method a), find t to maximize h(z)e M zi 2/2. For method b), t can be estimated from data. Given t, use ˆθ = = h(z) e M z 2 i /2 e M (z i t i ) 2 /2 M (y i +t i ) 2 /2 e h(y + t) e e M y 2 i /2 e M (z i t i ) 2 /2 ( dz ) m M yi 2/2 ( ) dy. m

13 Example with M = 16, S 0 = K = 50, T = 1, r =.05, σ =.1. Matlab test using method b): M = 16; S0 = 50; K = 50; T = 1; dlt = T/M; r = 0.05; s = 0.1; rd = ( r - s^2/2 )*dlt; N = 10000; z = randn(m,n); % Simple MC S = S0*exp(cumsum(rd + s*sqrt(dlt)*z)); h = exp(-r*t)*max( mean(s)-k, 0 ); disp([mean(h) var(h) 2*std(h)/sqrt(N)]) t = z*h /sum(h); % Approx. Mean Tilt Vector y = z; z = y + t*ones(1,n); S = S0*exp(cumsum(rd + s*sqrt(dlt)*z)); h = exp(-r*t)*max( mean(s)-k, 0 ); ht = h.*exp(sum(y.*y-z.*z)/2); % Importance disp([mean(ht) var(ht) 2*std(ht)/sqrt(N)]) Notice variance reduction from tilted sampling. Using method a) with Matlab fminsearch to find t: Sf hf *z/2)*max(mean(sf(z))-k,0); t = fminsearch(@(z)-hf(z), ones(m,1) ); % Tilt t y = z; z = y + t*ones(1,n); S = S0*exp(cumsum(rd + s*sqrt(dlt)*z)); h = exp(-r*t)*max( mean(s)-k, 0 ); ht = h.*exp(sum(y.*y-z.*z)/2); % Importance disp([mean(ht) var(ht) 2*std(ht)/sqrt(N)])