Sampling Distributions Allen & Tildesley pgs and Numerical Recipes on random numbers

Transcription

1 Sampling Distributions Allen & Tildesley pgs and Numerical Recipes on random numbers Today I explain how to generate a non-uniform probability distributions. These are used in importance sampling to reduce the variance of a Monte Carlo evaluation or to simulate various physical processes. We start by assuming that there is software to generate udrn s (uniform RN) in the range (0,1). How do we sample an arbitrary p(x)dx? There are lots of tricks. 1

2 Discrete Distributions Any discrete distribution p k can be sampled by constructing the cumulant. 0 u 1 p a p b p c p d c k = k i=1 p i Sample 0<u<1. Find which region it is in, i.e. find k: c k-1 < u < c k Return label k. Make a Table c k = c k-1 + p k before simulation. The search can be done by bisection in log 2 (N) steps. For simple distributions, it might be even easier. There is a faster more complicated method (see notes on the class website). 2

3 Continuous Distributions Generalize this method to a continuous function. This is the mapping method. Construct the cumulant: Prob(x<y)=c(y) Sample u in (0,1) y c( y) = dxp(x) Find x=c -1 (u) Do inverse mapping from urnd to x. Then x is sampled from p(x) dx. Can always perform the inverse with a table lookup. Two analytic examples: P(x)=a e -ax 0 <x then x = -ln(u)/a P(x)=(a+1)x a 0 < x < 1 then x = u 1/(a+1) Problem: inverse mapping may be difficult to compute. 3

4 Sample x from q(x)dx Rejection technique Accept x with probability c(x), otherwise repeat What is distribution of accepted x s? p(x)dx= c(x)q(x)dx/normalization Hence choose c(x) = a p(x)/q(x) Where a is set so that c(x) 1 à a min[q(x)/p(x)] 1/a is the acceptance probability, the efficiency. Do not under sample, or else efficiency will be low. If inefficient, you will use lots of prns/step. 5

5 Composition method Combine several random numbers Add several udrn s. remember characteristic function limit for large k is a Gaussian Example: add two integers in (1,6) Example: add 12 udrn, i.e 12 ς = u i 6 u [0,1] i=1 Gaussian distributed on (-6,6). Take maximum of k udrns. x= max(u 1,.., u k ) Prove that P(x)= k x k-1 6

6 Normal distribution Inverse mapping is a little slow, also of infinite range. Trick: Generate 2 prns at a time: r=(x,y) Generate u 1 and u 2 Calculate y 1 = 2lnu 1 cos2πu 2 y 2 = 2lnu 1 sin2πu 2 Samples: p(x,y) dxdy = (2π) -1 exp( r 2 /2) = p(r) r dr dθ p(v) dv = 0.5*exp( v/2) with v= r 2 We can sample the angle using a rejection technique: Sample (x,y) in square Accept if x 2 +y 2 <1 Normalize to get the correct r. 7

7 Example: Box-Mueller Code to sample a Normal Distribution Normal distribution <x>=x 0 and <(x-x 0 ) 2 >=σ 2 While(r2 > 0.25) then {x=prng()-0.5, y=prng()-0.5 r2 = x*x+y*y} radius= sigma*sqrt(-2*ln(prng())/r2) xnormal = x0 + x*radius ynormal = y0 + y*radius No trig functions! Rejection mixes up regularity of random numbers Efficiency of angle generation is 4/π. Will get 2 ndrn s each time. 8

8 Multivariate normal distributions How to sample a correlated Gaussian? (say with D components) Suppose we want to sample <x i x j > =T ij & <x i >=0 1. Take the square root of T: SS T = T by making a Choleski decomposition (see Numerical Recipes or notes). We can assume S is a triangular matrix all entries above diagonal are zero. S= 2. Generate D normally distributed numbers y. 3. Transform to correlated random distribution x α = S αi y i where y i y j = δ ij i Proof: x α x β = S ai y i S β j y j = S ai S β j y i y j = S ai S β j δ ij = SS T = T 9

9 Reduction of Variance The statistical error goes as: error = sqrt(variance/computer time). DEFINE: Efficiency = ζ = 1/νT v = error 2 of mean and T = total CPU time How can you make simulation more efficient? Write a faster code, Get a faster computer Work on reducing the variance. Or all three We will talk about the third option: Importance sampling and correlated sampling 11

10 Given the integral: Importance Sampling I = dxf (x) = f (x) How should we sample x to maximize the efficiency? Transform the integral: I = f (x) dx p(x) = f (x) p(x) p(x) p Estimator variance is: 2 f( x) f( x) v σ = I dxp( x) I px ( ) = px ( ) 2 2 p 2 Optimal sampling: δv δ p(x) = 0 with dx p(x) = 1 The mean value is independent of p(x), but variance v is not! Assume CPU-time/sample is independent of p(x), and vary p(x) to minimize v. 12

11 Finding Optimal p*(x) for Sampling Trick to parameterize as a positive definite PDF, define q(x): Solution: p *(x) = dx f (x) f (x) Estimator: p(x) = f (x) p*(x) [ q(x) ] 2 dx [ q(x) ] 2 sign( f (x)) = dx f (x) If f(x) is entirely positive or negative, estimator is constant. zero variance We can t generally sample p * (x), because, if we could, then we would have solved problem analytically! But the form of p*(x) is a guide to lowering the variance. Importance sampling is a general technique: it works in any dimension. 13

12 Example of importance sampling Suppose f(x) is given by: Try to sample from a Gaussian: 2 f (x) = e x /2 1+ x 2 Value is independent of a. p(x) = e x2 /2a (2πa) 1/2 CPU time is not 14

13 v σ 2 = dx p(x) f (x) p(x) I 2 = dx e x 2 (1 1/ 2a) I 2 (1+ x 2 ) 2 Importance sampling functions Variance integrand 15

14 What are the allowed values of a? Clearly for p(x) to exist: 0<a a For finite estimator 1<a For finite variance 0.5<a Obvious value a=1 Optimal value a=0.6 v = dx f (x) 2 p(x) = e x2 (1 1/ 2a) 1+ x 2 dx (2πa)1/2 16

15 What does infinite variance look like? Spikes Long tails on the distributions Near optimal 17

16 General Approach to Importance Sampling Basic idea of importance sampling is to sample more in regions where function is large. Find a convenient approximation to f(x). Do not under-sample -- could cause infinite variance. Over-sampling -- loss in efficiency but not infinite variance. Always derive conditions for finite variance analytically. Always test that estimated value is independent of important sampling. Sign problem: zero variance is not possible for oscillatory integral. Monte Carlo can add but not subtract. 18

17 Correlated Sampling Suppose we want to compute a function G(F 1,F 2 ) where F s are given by integrals: F 2 = dx f 2 x Suppose we use the same p(x) and the same random numbers to do both integrals. F 1 = dxf 1 ( x) ( ) What is the optimal p(x)? p *(x) f 1 (x) dg df 1 + f 2 (x) dg df 2 It is a weighted average of the distributions for F 1 and F 2. 19

18 Sampling the Boltzmann distribution Suppose we want to calculate a whole set of averages: O k = dr O k (R)e V ( R)/kT dr e V ( R)/kT Optimal sampling is: p k * (x) e V (R)/kT (O k (R) < O k >) We need to sample the first factor because we want several different properties. Avoid undersampling. But the Boltzmann distribution is very highly peaked. 20

19 Independent Sampling for exp(-v/kt)? What is the variance of the free energy and how does it depend on the number of particles? Var(Z) = Z(T)= exp(-v/kt). Sample R uniformly in (0,L): P(R) = Ω -N =1 dr e βv ( R) Z(β) = Z(2β) Z(β) 2 var(β F) = var(z) / Z 2 = Z(2β) Z(β) 2 1= e2β[ F (β ) F (2β )] 1 β[ F (β ) F (2β )] error(β F) = e Variance grows exponentially versus N since F is extensive! The number of sample points will grow exponentially in N, just like a grid based method. We haven t reduced the complexity! 2 21

20 Intuitive explanation Throw N points in a box, area A. Say probability of no overlap is q. Throw 2N points in a box, area 2A. Probability of no overlap is q 2. Throw mn points in a box with area ma Probability of no overlap is p=q m. Probability of getting a success is p=exp(m ln(q)). Success defined as a reasonable sample of a configuration. This general argument shows independent sample has exponential scaling. We need to sample only near the peak of the distribution: to do so use random walks. 22