Monte Carlo Methods So far we have discussed Monte Carlo methods based on a uniform distribution of random numbers on the interval [0,1] p(x) = 1 0  x  1 p(x)= 0 otherwise the hit and miss algorithm generates pairs of points (x,y) and either accepts or rejects the point the sample mean method samples points uniformly on the interval [a,b]

Choose p(x) so that f(x)/p(x) is a fairly constant function

Nonuniform probability distributions Consider a probability density p(x) such that p(x)dx is the probability that event x is in the interval between x and x+dx +  We have  p(x) dx = 1 -  x Calculate P(x) =  p(x’) dx’ = r =  dr -  p(x’)dx’ = dr ==> p(x’)= dr/dx’ r is a uniform random number on the interval [0,1] Invert this and solve for x in terms of r

Eg.1 suppose we want p(x)= 1/(b-a), a  x  b = 0 otherwise x Calculate P(x) =  p(x’) dx’ a r = (x-a)/(b-a) Solving for x, x= a + (b-a) r obvious!

Eg.2 p(x) = (1/ ) exp(-x/ ), [0,  ] = 0 x < 0 Hence r = P(x) = 1 - exp(-x/ ) And x = - ln(1-r) = - ln(r) This technique is only feasible if the inversion process can be carried out.

Eg.3 p(x) = (1/ 2  2 ) 1/2 exp(-x 2 /2  2 ) P(x) = ? However the two-dimensional distribution p(x,y)dxdy = (1/ 2  2 ) exp(-(x 2 +y 2 )/2  2 )dxdy can be integrated by a change of variable  = r 2 /2= (x 2 +y 2 )/2  2, tan  =y/x p( ,  )d  d  = (1/2  ) exp(-  )d  d  Generate  uniformly on the interval [0,2  ] i.e.  = 2  r Generate  according to the exponential distribution with = 1 i.e.  = -ln r x= (2  2 ) 1/2 cos  and y=(2  2 ) 1/2 sin  are Gaussian distributed

Importance Sampling The error estimate in Monte Carlo is proportional to the variance of the integrand: ( - 2 ) 1/2 n 1/2 How can we reduce the variance? b Introduce a function p(x) such that  p(x)dx = 1 a b And rewrite F =  [f(x)/p(x)] p(x) dx a

Importance Sampling Evaluate the integral by sampling according to p(x): F n = (1/n)  f(x i )/p(x i ) Choose p(x) to minimize variance of f(x)/p(x) i.e try to make f(x)/p(x) slowly varying since a constant has zero variance

1 eg. F =  exp(-x 2 ) dx = Sample [0,1] uniformly n F n  n Choose p(x)= A exp(-x) and sample [0,1] again n F n  n The variance of the integrand is reduced by about a factor of 4 which means that fewer samplings are needed to obtain the same accuracy.

c Program Importance Sampling n=10000 h=1. a=0. b=1. sum=0. psum=0. sum2=0. psum2=0. m=2 do 4 i=1,n ww=r250(idum) x=a+b*ww y=-log(1.-ww+ww*exp(-1.)) g=exp(-y*y) p=(1.-ww+ww*exp(-1.))/(1.-exp(-1.)) f=exp(-x*x) sum=sum+f psum=psum+g/p sum2=sum2+f*f psum2=psum2+(g*g)/(p*p) sig2=sum2/i -(sum/i)*(sum/i) sig=sqrt(sig2) psig2=psum2/i -(psum/i)*(psum/i) psig=sqrt(psig2) if((i-(i/10**m)*10**m).eq.0) then write(6,10) 1.*i,sum/i,sig,psum/i,psig m=m+1 else continue end if 10 format(1x,f10.0,3x,f10.5,3x,f10.5,3x,f10.5,3x,f10.5) 4 continue stop end

The above ideas can be used to simulate many different types of physical problems Random walks and polymers Percolation Fractal growth Complexity and neural networks Phase Transitions and Critical Phenomena

Monte Carlo Methods Random numbers generated by the computer are used to simulate naturally random processes many previously intractable thermodynamic and quantum mechanics problems have been solved using Monte Carlo techniques how do we know is the random numbers are really random?

Random Sequences A sequence of numbers r 1,r 2,… is random if there are no correlations among the numbers in the sequence however most random number generators yield a sequence in which each number is used to find the succeeding one according to a well defined algorithm the most widely used random number generator is based on the linear congruential method

where a,c and m are integers the notation y= z mod m means that m is subtracted from z until 0  y <m the process is characterized by the multiplier a, the increment c and the modulus m since m is largest integer generated by this method, the maximum possible period is m Given a seed x 0, each number in the sequence is determined by the one-dimensional map

Example a=3 c=4 m=32 and x 0 =1 produces x1=(3 x 1 + 4) mod32 = 7 x2=(3 x 7 + 4) mod32 = 25 x3=(3 x 25 +4) mod32= 79mod32=15 and so on …. 1,7,25,15,17,23,9,31,1,7,25,…. period is 8! Rather than the maximum of 32

Random Sequences If we choose a, c and m carefully then all numbers in the range from 0 to m-1 will appear in the sequence to have the numbers in the range 0  r <1, the generator returns x m /m which is always < 1 there is no necessary and sufficient test for the randomness of a finite sequence of numbers we need to consider various tests an obvious requirement for a random number generator is that its period be much greater than the number of random numbers needed in a specific problem

Sequences A way of visualizing the period is to consider a random walker and plot the displacement as a function of the number of steps N when the period of the random number generator is reached the plot will begin to repeat itself consider a=899, c=0, m=32768 with x 0 =12 Sequence

Correlations We can check for correlations by plotting x i+k as a function of x i if there are any obvious patterns in the plot then there are correlations correlations

Uniformity test

Correlations One way to reduce sequential correlation and to lengthen the period is to mix or shuffle two different random number generators statistical tests should be performed on random number generators for serious calculations

Exploration It has been claimed that the logistic map in the chaotic region is a good random number generator test this for yourself x i+1 = 4x i (1 - x i ) will make the sequence uniform