PHYSICS 115/242 Homework 5, Solutions X = and the mean and variance of X are N times the mean and variance of 12/N y, so

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1 PHYSICS 5/242 Homework 5, Solutions. Central limit theorem. (a) Let y i = x i /2. The distribution of y i is equal to for /2 y /2 and zero otherwise. Hence We consider µ y y i = /2 /2 σ 2 y y 2 i y i 2 = X = ydy = 0 /2 /2y 2 dy = 3 2 ( 2 2 N N y i, i= ) 3 = 2. and the mean and variance of X are N times the mean and variance of 2/N y, so µ X X = N 2 N µ y = 0, σ 2 X X 2 X 2 = N 2 N σ2 y =. (b) SOURCE CODE double sum, sum2, sum3, sum4, sum5, sum6, x, xadd; int i, j, m, seed; printf ("sample size =?, seed =? "); scanf("%d %d", &m, &seed); srand(seed); sum = 0; sum2 = 0; sum3 = 0; sum4 = 0; sum5 = 0; sum6 = 0; for (i = 0; i < m; i++)

2 x = 0; for (j = 0; j < 2; j++) x = x + rand() / (RAND_MAX +.0); x = x - 6; xadd = x; sum += xadd; xadd *= x; sum2 += xadd; xadd *= x; sum3 += xadd; xadd *= x; sum4 += xadd; xadd *= x; sum5 += xadd; xadd *= x; sum6 += xadd; sum = sum /m; sum2 = sum2 /m; sum3 = sum3 /m; sum4 = sum4 /m; sum5 = sum5 /m; sum6 = sum6 /m; printf (" Took N = 2\n "); printf ("sample size = %8d \n", m) ; printf (" k= %8.4f\n k=2 %8.4f\n k=3 %8.4f\n k=4 %8.4f\n k=5 %8.4f\n k=6 %8 sum, sum2, sum3, sum4, sum5, sum6); OUTPUT sample size =?, seed =? Took N = 2 sample size = 0000 k= k= k= k= k= k= Note that this is basically correct except that the 6th moment, and to a lesser extent the 4th moment, are a bit small. This is because I only took N=2 numbers in the sum, which does not give quite enough weight in the tail to be a true Gaussian, and the higher moments need the tail to be correct. Taking N = 00, still with a sample size of 0000, I got the following results: 2

3 Took N = 00 sample size = 0000 k= k= k= k= k= k= which is better. (c) 242 students We can understand better the results in the previous part by explicitly calculating the moments on the distribution for finite N. It is convenient to define y i = 2(x i 2 ) so y i = 0, yi 2 =. For the 4-th moment we have X 4 = N (y 2 +y 2 + +y N ) 4 = ( N y4 +3 ) y 2 2, N where we used that the only non-zero terms are those where all four indices are equal and those where two of the indices are equal and the other two are both equal to a different value. There are N terms where all indices are equal, and 3N(N ) terms where there are two equal pairs (this comes from their being N(N )/2 ways of choosing two distinct indices and there are 4 C 2 = 4!/(2!) 2 = 6 arrangements for these indices). Now y 2 =, and y 4 = 9 5 σ4. Hence we have X 4 = 9 5 ( N +3 ) N = 3 6 5N. For N = 2 this gives 2.9, in better agreement with the numerics. For the sixth moment, similar combinatorical arguments give X 6 = N (y 3 +y 2 + +y N ) 6 = N 2 y6 + 5 ( ) ( y 4 y 2 +5 )( 2 ) y 2 3 N N N N = ( 27 5 N N 5 ) N 2 5 N + 30 N 2 = 5 8 N N, 2 where we used that y 6 = 27/7. For N = 2 this gives , rather than 5, which agrees much better with the numerics. 2. Lorentzian Distribution 3

4 (a) To generate random numbers with the distribution P(x) = π, ( < x < ), +x2 as discussed in class we generate a uniform random number r in the integral from 0 to and compute x as a function of r, i.e. x(r), where the inverse function is given by x r(x) = P(x )dx = [ ( tan x π )], π 2 so x = tan [ π ( r 2)]. Note that, in the algorithm, we have to make sure that the values r = 0 and are excluded since the tan will blow up for these values. (b) In the numerics I took N = SOURCE CODE int N = 0000, i, nless; double t, x; srand(time(null)); printf (" N = %d\n ", N); nless = 0; for (i = 0; i < N; i++) while() t = rand() / (RAND_MAX +.0); if (t!= 0 && t!= ) break; // exclude t exactly 0 or x = tan(m_pi * (t - 0.5)); if (fabs(x) < ) nless += ; printf (" Fraction with x < = %0.4f\n", (float) nless / (float) N); 4

5 OUTPUT N = 0000 Fraction with x < = The exact answer is π +x 2 dx = π [ tan x ] = π ( π 4 + π = 4) 2. The numerical result agrees to that within about %. (This error is reasonable since it is of order / N.) (c) The variance is infinite since x 2 P(x) const. for x ±. Furthermore, the mean is also not well defined. To see this note that to define integrals to infinity we have to ensure that the integral tends to a well defined value as the limit tends to infinity. Here, both the upper and lower limits are infinite, and we have to take do a limiting procedure for both of them, i.e. x = lim L M π L M x dx = lim +x2 L M [ ln(+x 2 ) ] L 2π = M lim L M π ln(l/m), so the answer depends on how the limits L,M are taken. For the central limit theorem to apply, the distribution of the individual random variables must have a well-defined mean and variance. Hence it is does not apply to the Lorentzian distribution. 3. Monte Carlo Integral with Error Bars. (a) I estimated the integral from an average over N values of x. To get the error bar on this estimate I computed the standard deviation of the N function values and divided by N as discussed in class. SOURCE CODE double f(double x) return log(x); double a =, b = 2, x, fun, ans_av, ans_err, exact; 5

6 int N = 0000, i; srand(time(null)); for (i = 0; i < N; i++) // Sum over the N x-values x = a + (b - a) * rand() / (RAND_MAX +.0); // get x-value fun = f(x); ans_av += fun; ans_err += + fun*fun; ans_av = ans_av / N; // Normalize ans_err = ans_err / N; ans_err = sqrt(fabs(ans_err - ans_av*ans_av)/(n - )); exact = * log(2.0); // error printf (" %d points \n", N); printf (" ans = %0.5f +/- %0.5f\n", ans_av, ans_err); printf (" exact = %0.5f \n", exact); OUTPUT 0000 points ans = / exact = (b) The exact result is given by I = 2 ln(x)dx = [xlnx x] 2 = 2ln2 2 (ln ) = 2ln2 = The above answer agrees within the exact value to within about one standard deviation, as expected. 4. Multi-dimensional Monte Carlo Integral (a) Clearly 0 x2 i dx = /3 and, for i j, x 0 0 ix j dx i dx j = /4. Hence, expanding out the square of the integrand in the question we have, for N variables, I = N 3 +N(N ) 4. Setting N = 0 gives I = =

7 (b) SOURCE CODE double a = 0, b =, x, term, term2, ans_av, ans_err, exact; int N = 0000, i, j; srand(time(null)); for (i = 0; i < N; i++) // Sum over the N x-values term = 0; for (j = 0; j < 0; j++) x = a + (b - a) * rand() / (RAND_MAX +.0); // get x-value term += x; term2 = term * term; ans_av += term2; ans_err += term2 * term2; ans_av = ans_av / N; // Normalize ans_err = ans_err / N; ans_err = sqrt(fabs(ans_err - ans_av*ans_av)/(n - )); exact = 55.0/6.0; // error printf (" %d points \n", N); printf (" ans = %0.5f +/- %0.5f\n", ans_av, ans_err); printf (" exact = %0.5f \n", exact); OUTPUT ~/courses/5/homework debussy.ucsc.edu => a.out 0000 points ans = / exact = Here N is the number of points in the Monte Carlo average. The Monte Carlo answer 7

8 agrees with the exact result to within the error computed from the Monte Carlo calcuation. 5. Random Walk SOURCE CODE int tmax, t, TSTORE=000, N, irun; double x, xav[tstore], xav2[tstore]; float r; printf (" tmax =? No. of runs =? " ); scanf("%d %d", &tmax, &N); if (tmax > TSTORE) exit; srand(time(null)); for (t = 0; t < tmax; t++) xav[t] = 0; xav2[t] = 0; for (irun =0; irun < N; irun++) x = 0; for (t = 0; t < tmax; t++) r = rand() / (RAND_MAX +.0); if (r < 0.5) x = x + ; else x = x - ; xav[t] += x; xav2[t] += x*x; for (t = 9; t < tmax; t+=0) xav[t] /= N; xav2[t] /= N; 8

9 printf (" %5i %0.4f %0.4f \n", t+, xav[t], xav2[t]); OUTPUT These results are obtained by averaging over 0000 walks each of length 200 t <x(t)> <x^2(t)> i.e. the mean is close to zero and the mean square is close to t as expected. 9