Sample Size Requierement for Monte Carlo - …salserver.org.aalto.fi/vanhat_sivut/Opinnot/Mat-2.4108/pdf-files/...Sample Size Requierement for Monte Carlo - simulations ... culating

HELSINKI UNIVERSITY OF TECHNOLOGY

Department of Engineering Physics and Mathematics

Systems Analysis Laboratory

Mat-2.4108 Independent Research Projects in Applied Mathematics

Sample Size Requierement for Monte Carlo - simulationsusing Latin Hypercube Sampling

Anna Matala

60968U

20.5.2008

1

Contents

1 Introduction 3

2 Monte Carlo Integration 3

3 Latin Hypercube Sampling (LHS) 5

3.1 Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3.2 Variance analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3.3 Estimate of variance . . . . . . . . . . . . . . . . . . . . . . . . . 8

4 Sample size 8

5 Examples 12

5.1 Example of sampling with lognormal distribution . . . . . . . . . 12

5.2 Example of multinormal distribution . . . . . . . . . . . . . . . . 15

5.3 Example of binary variables . . . . . . . . . . . . . . . . . . . . . 19

5.4 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

6 Conclusions 23

2

1 Introduction

The goal of probabilistic fire risk analysis is to discover the probabilities of the

harmful consequences of fire. The variating initial and boundary conditions,

that lead to growth of fire, are taken into account statistically. The conse-

quences are represented as a limit state function. The probability is calculated

by integrating the limit state function over multidimensional joint distribution.

In practice, the integration is made by using Monte Carlo method, because the

physical models of fire are non-linear and often really complex. In Probabilis-

tic Fire Simulator (PFS, developed by VTT) [1] the sampling can be done by

Simple Random sampling (also known as Monte Carlo sampling) or Latin Hy-

percube sampling.

The purpose of this work is to study the accuracy of Latin Hypercube sampling

and to find a simple manner to evaluate the sample size. Estimates and ex-

periences are searched from literature. Finally, the results are tested in simple

simulations.

2 Monte Carlo Integration

Random sampling has been used for solving numerical problems as early as 1777

when Comte de Buffon made his experiment by dropping needle to the board

with parallel lines with same distance [2]. If the length of the needle is L and

the distance between lines d (d>L), the probability that the needle intersects a

line is

p =2Lπd. (1)

Later Laplace realized that this method could be used for estimating the value

of π.

Modern Monte Carlo (SRS) method was born in 1940s when Stanislaw Ulam,

John von Neumann and others started to use random numbers to examine

physics from the stochastic perspective. One of the most effective use of Monte

Carlo method is to evaluate definite integrals which analytically would be too

difficult to find [3]. If the problem is

3

y =∫ b

a

f(x)dx, (2)

the solution can found by ’hit-or-miss’ (acceptance-rejection) method: A box is

drawn around the function from a to b and from 0 to y0 (y0>f(x)). After this,

N random numbers are taken from a uniform distribution and plotted (as in

Figure 1). A number of points that fall below f(x) are then counted and marked

with N0. Now the estimate of integral can be calculated using equation

yest =N0

N(y0(b− a)). (3)

The estimate is getting more and more precise when N is increased and will

converge to the correct answer when N →∞. In this example, x had only one

dimension for simplicity, but the method can be successfully applied to multi-

dimensional problems as well.

Figure 1: ’Hit or miss’ method in integration of curve f(x).

4

3 Latin Hypercube Sampling (LHS)

3.1 Method

The Latin Hypercube Sampling [4]-[6] is a type of stratified Monte Carlo sam-

pling. All the areas of the sample space are represented by input values. To

generate a sample size N from K variables x=[x1, x2, ..., xK ]T with probability

density function f(x), the procedure is as follows. The range of each variable is

portioned into N non overlapping intervals of equal probability 1/N (Figure 2).

From each interval one value is selected randomly according to the probability

density of the interval. The N values of x1 are paired in a random manner with

values of x2 (Figure 3), these pairs are then paired similarly with values of x3

and so on, until N samples of n variables are formed. So created cube has NK

cells which cover the sample space. Denote the input vector X=[x1,...,xN ].

Figure 2: Density function of variable xi divided to N=5 non overlapping inter-

vals.

5

Figure 3: Pairing x1 with x2 when N=5.

3.2 Variance analysis

Let h(x) be the transformation function of input x [4],[7]. The estimate of

expectation value E(h(x)) is

h = N−1N∑j=1

h(xj). (4)

The variance of the estimate depends on the sample method used. For Simple

Random sampling the variance is

var(h) = N−1var(h(x)) (5)

and for the Latin Hypercube sampling

var(h) = N−1var(h(x)) +N − 1N

cov(h(x1), h(x2)). (6)

Thus the LHS sampling decreases the variance if and only if cov(h(x1),h(x2))<0.

In their article McKay et al. [4] proved that the covariance is negative whenever

h(X) is monotonic in each of its components. However, in many problems the

monotonicity does not hold. Stein [7] showed 1987 that when N → ∞, the

covariance term is asymptotically non positive. Define function

gk(xk) =∫h(x)

K∏i=1,i6=k

dFi(xi) (7)

6

and {XjN}(where j=1,...,N and N=1,2,...). Stein’s Theorem 1 shows that if

E(h2)<∞ and N →∞,

cov(h(X)1N , h(X)2N ) =K

NE(h)2 −N−1

K∑k=1

∫gk(x)2dFk(x) + o(N−1). (8)

By Jensens’s inequality,

∫gk(x)2dFk(x) ≥ (

∫gk(x)dFk(x))2 = (Eh)2, (9)

so the highest-order term in expansion of cov(h(x1),h(x2)) is non positive. That

means, that

limN→∞Ncov(h(x1), h(x2)) ≤ 0. (10)

This can be written easier if we define

ha(x) =K∑k=1

gk(xk)− (K − 1)E(h) (11)

and

r(x) = h(x)− ha(x). (12)

The function ha(x) is the best additive fit to h(x) which means

∫r2(x)dF (x) ≤

∫(h(x)−

K∑k=1

hk(xk))2dF (x). (13)

Now an approximation of variance can be written as N →∞,

var(N−1N∑j=1

h(XjN )) = N−1

∫r(x)2dF (x) + o(N−1). (14)

This is relevant whenever N is much larger than K, so the sample size is greater

than the number of variables.

7

3.3 Estimate of variance

A simple way to get an estimate to the variance of average is called replicated

Latin hypercube sampling [7]. The idea is to produce several independent LHS

samples and then compare the variances between samples. If the number of

samples is α and size of them M, N=α·M. If

hi = M−1M∑j=1

h(Xij), (15)

then an unbiased estimator for the variance is

var(α−1α∑i=1

hi) =1

α(α− 1)[α∑i=1

h2

i − (∑αi=1 hiα

)2]. (16)

This is exactly same as the sample variance of mean for normal sampling. Now

if α is too small, the estimate of variance is not precise. On the other hand,

increasing α while N is fixed increases this estimator of variance. However, as

long as ratio MK is large, the increase in variance is small.

4 Sample size

In literature almost no recommendations or estimates for the sample size of LHS

sampling were proposed. However, it is possible the study the examples and ap-

ply them to fire-Monte Carlo simulation. In fire simulations, the most limiting

factor is time. Often the samples are simulated by Fire Dynamics Simulator

(FDS) and that makes the Monte Carlo simulation quite slow. As an example

of order of magnitude, already 1000 samples can require computation time of

several months. Typically number of variables is less than 10. All the following

examples are about the LHS sampling unless else is mentioned.

Stein [7] gives one example where ratio MK =10. In his example N=1000, M=200,

K=20 and α=5. For fire simulation, this could mean 5 independent samples of

size 100 makes total sample size of 500. In his other example there are 6 vari-

ables and the estimator of variance is calculated by 100 independent replications

of samples of size 100. Now MK would be nearly 17. This is unnecessary high

value and clearly impossible for fire simulation. However, according to the first

8

example it is possible to get reliable results by using values α ≥ 5 and MK > 10.

Jianfang Jia et al. [8] give an example of NF-κB signaling pathway model. They

did global sensitivity analysis using 64 independent variables and 1000 samples,

so the ratio of samples and variables is 15.6. Applied to 10 variables, this ex-

ample would suggest sample size 156.

As mentioned before, the latin hypercube sampling converges at least as fast

as simple random sampling if N is big enough. Swidzinski et al. [9] go further

stating that LHS could be nearly five times more effective in yield estimation

than traditional sampling methods. They justify it by their study at the Texas

A&M. In the tests, they generated 10 iterations of samples of size 200 using

both LHS and simple random sampling. After that, sample size of 1000 was

generated 10 times using simple random sampling. The results showed that

variation in the yield estimate during the 10 iterations were similar when LHS

sample size was 200 and simple random sample was 1000. The theory of simple

random sampling is better known, so this gives suggestive size of LHS samples.

Eduard Hofer [10] studied computationally intensive models, where sample sizes

were 100 as maximum and the samples were generated by Simple Random sam-

pling. As the distributions of parameters are usually not known, the fractiles of

5% and 95% can not be immidiately obtained, and hence Monte Carlo simula-

tions are used. The resulting distributions are often long-tailed so the fractile

estimates are highly dependent of the sample size N. Therefore, the possible

impact of the sampling error needs to be quantified. That can be done by cal-

culating statistical tolerance limits (u(%), v(%)) where u is the percentage of

the combined influence of the quantified uncertainties and v is the confidence

level of it. In other words, with confidence level v(%) we can say that u(%) of

the samples are inside the tolerance limits. The tolerance interval is chosen to

be from the smallest to the largest observation in the sample.

Hofer gives an example: Two sided statistical tolerance limits (95%, 95%) for

Simple Random sample require only 93 model runs. This is a result of non-

parametric statistics and depends only on the percentages u and v, not on the

number of uncertainties taken into account. Besides, it does not assume any

particular type of underlying distribution. This result is general and really

promissing from the point of view of fire simulation, especially because LHS

9

Table 1: Sample sizes for two-sided nonparametric tolerance limits. 1-α is the

confidence coefficient and q is the fraction of the population that lies between

the smallest and the largest observation of the sample.q 0.5 0.7 0.750 0.8 0.85 0.9 0.95 0.975 0.98 0.99

1-α

0.5 3 6 7 9 11 17 34 67 84 168

0.7 5 8 10 12 16 24 49 97 122 244

0.75 5 9 10 13 18 27 53 107 134 269

0.8 5 9 11 14 19 29 59 119 149 299

0.85 6 10 13 16 22 33 67 134 168 337

0.9 7 12 15 18 25 38 77 155 194 388

0.95 8 14 18 22 30 46 93 188 236 473

0.975 9 17 20 26 35 54 110 221 277 555

0.98 9 17 21 27 37 56 115 231 290 581

0.99 11 20 24 31 42 64 130 263 330 662

0.995 12 22 27 34 47 72 146 294 369 740

0.999 14 27 33 42 58 89 181 366 458 920

sampling can be expected to show less variability than simple random sampling.

The solution for this example can be found from Table 1 originally extracted

from Conover’s book ([11]: Appendix A6). The table shows the sample sizes

required according some confidence coefficients and population fractions. If the

Table 1 does not offer direct answer, it is possible to make an approximation

N ≈ 14χ(1−α)

1 + q

1− q+

12

(r +m− 1), (17)

where r is the number of observation of lower bound of tolerance limit and

(N + 1 −m) the number of observation of upper bound. The sample must be

assorted to ascending order. r=m=1 is commonly used, so the lower bound is the

smallest observation and the upper bound the largest. χ1−α is the 1−α quantile

of a chi-square random variable with 2(r+m) degrees of freedom. Revising the

example before causes the same result:

N ≈ 14· 9.488 · 1 + 0.95

1− 0.95+

12

(1 + 1− 1) = 93.008.

However, the statistical tolerance limits cannot be computed from LHS sample.

The reason is that the sample estimates of the 95% fractile can be significantly

10

below the true 95% fractile for long-tailed distributions. Without statistical

tolerance limits it is not possible to know how likely this shortcoming is and

therefore the analysis has to be made using simple random sampling. Of course,

after finding sample size N, it can be considered as upper limit of real LHS sam-

ple size needed.

11

5 Examples

5.1 Example of sampling with lognormal distribution

The following example is about samples of lognormal distribution. Lognormal

distribution is chosen because it is important in fire simulations.

Lognormal variables can be written y=ex where x is normally distributed with

mean α and variance β. Density function of lognormal distribution [12] is

f(y) =1

βy√

2πe− (ln(y)−α)2

2β2 (18)

with mean µ = eα+β2/2 and variance σ2 = e2α+β2(eβ

2 − 1). The mean and

variance of ln(y) (α, β2 respectively) can be calculated backwards using following

equations:

α = ln(µ)− 12ln(

σ2

µ2+ 1) (19)

and

β2 = ln(σ2

µ2+ 1). (20)

Monte Carlo simulations were made with sample sizes N=50, 100, 500, 1000,

5000 and 10000, using both Simple Random sampling (SRS) and Latin Hyper-

cube sampling (LHS). The idea was to compare mean and variance of samples.

The lognormal distribution used had mean 27.4 and variance 16, which result

(equations (19) and (20)) α=3.3000 and β2=0.0211. The sample means and

variances are calculated using equations [13]

x =1N

N∑i=1

xi (21)

and

s2 =1

N − 1

N∑i=1

(xi − x)2. (22)

This is valid for both SRS and LHS sampling [4].

12

Figure 4: Sample means as function of sample size N.

In Figures 4 and 5 are shown sample means and variances with SRS and LHS

sampling. In simple case of only one variable, the LHS sampling is better with

every N. The differences between the correct and estimated mean and variance

values are listed in table 2.

Table 2: Difference of mean and variance.N ∆µ (SRS) ∆σ2 (SRS) ∆µ (LHS) ∆σ2 (LHS)

50 0.169231071 3.549920858 0.039925728 0.068954216

100 0.333796062 2.874371726 0.007375298 0.155918985

500 0.235102161 0.102067908 0.000254559 0.032888354

1000 0.071753115 0.424214694 0.000419834 0.003795222

5000 0.012967127 0.153119925 0.000180192 0.004282639

10000 0.023052468 0.054337911 0.0000689647 0.00197263

13

Figure 5: Sample variances as function of sample size N.

14

5.2 Example of multinormal distribution

Consider next a situation where variable X depends on five independent, normal

distributed variables (with parameters µi, σi). The x is defined as

x =5∑i=1

xi, (23)

where xi ∼N(µi,σi). The parameters used in this example are listed in table 3

and the simulations were made with N=50, 100, 500, 1000 and 5000. Sum of nor-

mal distributed variables is also normal distrbuted with parameters µ =∑µi

and σ =√∑

σ2i . In this example, the mean µ is 16.5 and standard deviation

σ=5.7879 (variance 33.4998). In Figures 6 and 7 are shown the sample mean

and variance with different sample sizes. It seems that LHC is closer to real

parameter values than SRS with each N.

Table 3: Parameters of variables.i µi σi

1 1 0.5

2 2.5 2

3 -4 5

4 10 2

5 7 0.5

Study now the distributions of samples using the Pearson’s chi-square test. That

compares sample distribution F(x) to some known distribution G(x). The null

hypothesis is H0: F(x)=G(x). To calculate the p-value, the sample has to be

divided in m categories. The χ2 is

χ2 =m∑i=1

(Oi − Ei)2

Ei, (24)

where Oi is frequency in category i and Ei=NP, the expected frequency of cate-

gory i (calculated from known distribution). The degree of freedom is df=m-l -1,

where l is the number of parameters in known distribution. The prerequisites

are N>50, Ei >2 for all i and only 20 % of categories have Ei <5. If p-value is

less than chosen risk level, H0 will be refuted. The risk level is normally 5 %.

15

Figure 6: Mean of sample as function of sample size N.

Null hypothesis is now

H0: F(x)=N(16.5,5.7879).

Divide distribution into 8 classes that are listed in table 4. The probabilities

come from the standard normal distribution table, where z = x−µσ . It can be

seen that now both prerequisites are filled, because when N≥ 50, sample size is

expected to be >5 in every cathegory.

The χ2-values with different sample sizes are presented in table 5. Now df =

8− 2− 1 = 5. In χ2-table the maximum value to be accepted with risk level 5

% is 11.07. In this example, all the χ2-values are in the acceptance region, so

all the samples normally distributed with parameters µ = 16.5 and σ = 5.7879.

In this case, with five independent variables, sample size 50 was enough in both

SRS and LHS sampling.

16

Figure 7: Variance of sample as function of sample size N.

Cathegory x P

I [-∞,9.09) 0.1003

II [9.09,12) 0.1174

III [12,14) 0.1159

IV [14,16) 0.1305

V [16,18) 0.1385

VI [18,20) 0.1231

VII [20,23) 0.1429

VIII [23, ∞) 0.1314

Table 4: Categories of normal distribution.

17

Table 5: χ2 values.N χ2(SRS) χ2(LHS)

50 10.37789289 9.148392622

100 10.21332038 2.156814332

500 3.388478928 4.64044467

1000 6.533513491 5.451614406

5000 3.706511017 6.611654417

10000 2.131056413 0.857279318

18

5.3 Example of binary variables

Consider a nuclear power plant where in case of fire the failure time t1 of some

important component is lognormally distributed with parameters µ1 and σ1.

The failure can be prevented by the fire service if they arrive before the failure

happens. The arriving time of the fire service (t2) is lognormally distributed

with parameters µ2 and σ2. Now the interesting factor would be to know what

is the probability that the failure can be prevented, or that t2 < t1. This is

actually the probability p(A) of Bernoulli distribution, where X=1 if A happens

and X=0 if not. In lognormal distribution negative values are not allowed, so

the probability of P(t2 <t1) has to be calculated from joint distribution. The

joint lognormal distribution is

f(t1, t2) = f(t1)f(t2) =1

β1t1√

2πe− (ln(t1)−α1)2

2β21 · 1

β2t2√

2πe− (ln(t2)−α2)2

2β22 , (25)

where α and β are calculated from equations (19) and (20). Now we are inter-

ested in the probability of t2 < t1, and that can be found by integrating

F (t2 < t1) =∫ ∞

0

∫ t1

0

f(t1)f(t2)dt2dt1. (26)

This can be integrated numerically e.g. by matlab using function called ’trapz’.

In this example µ1=12, σ1=3, µ2=13 and σ2=5, so α1=2.455, β2=0.246, α2=2.496

and β2=0.371. The distributions are plotted in Figure 8. Expected probability

is calculated by integrating the joint distribution, and has value of 0.463.

Confidence intervals for Bernoulli distribution are calculated as

p± zα/2

√p(1− p)

n, (27)

where 1-α is the confidence level (often 95 %), zα/2 ∼N(0,1), p is the estimated

probability and n is the sample size. This is valid only for simple random sam-

pling. In m independent repetitions, the real value of p should be inside the

confidence intervals m·p times in an average.

Repeat the simulations 14 times (averages as a function of sample size N are

shown in Figure 9), and calculate the confidence intervals as in equation (27).

19

Figure 8: Lognormal distributions f1(α1,β1) and f2(α1,β2).

The idea is now to compare the confidence intervals of Monte Carlo samples

to similar calculated intervals of Latin Hypercube samples. These intervals are

not valid confidence intervals of LHS sample, but comparing these intervals can

give more information. Choose 95 % confidence level, so Z0.025 = 1.96. The

confidence intervals were calculated to each simulation, and then counted the

times the real value of p was included in the confidence interval in 14 repetitions.

The results are listed in table 6. It would be interesting to test null-hypothesis

H0: P(’The real value of p is inside the 95 % confidence interval in LHS sam-

pling’)=0.95, but that is not possible for two reasons. First, to make that test,

m should be big. The rule is that

mp ≥10

m(1-p)≥10.

The estimates are near 0.95 so the minimum m should be 200 which is too much

considering that with PFS (Probabilistic Fire Simulator) the maximum variable

number is 15 and for each p two variables are needed. This would lead to nearly

30 simulations for each N. However, with smaller number of repetitions it is

possible to get an idea of the result of the test. Another problem is, that the

test variable is

20

N n (SRS) n (LHS)

25 13 14

50 12 14

100 12 14

500 14 14

1000 12 14

5000 13 14

10000 13 14

Table 6: n is a number of times that p is inside the confidence intervals in 14

repetitions.

z =p− p√

p(1− p)/m, (28)

which goes to infinity if p=1. In this case the null-hypothesis would be always

rejected.

After 14 repetitions the estimate of probability is p=1 with each N in LHS

sampling. The test variable is ∞ and the null-hypothesis rejected. For LHS

sampling, the 95 % confidence intervals are significantly smaller than for SRS

sampling, so with every N the LHS sampling gives at least as good results as

the SRS.

21

Figure 9: Averages of 14 independent repetitions.

22

5.4 Discussion

To determine the sample size is not a simple task. Many guides have focused

on the Simple Random sampling, but only one common rule was found from

the literature for Latin Hypercube samples: It gives at least as good results as

Simple Random sampling when the sample size is big enough. The sample size

required may depend on the distribution, number of variables or something else.

The only way to see if the sample has converged, is to make more simulations

and see if the result stays similar. Even this does not work for latin hypercube

sampling, where the sample size should be known beforehand.

In all three examples the latin hypercube samples were closer to the real value

for each N. For five independent and normal distributed variables sample size

50 was enough to say that the sample distribution did not differ from the orig-

inal distribution significantly. In the last example it was seen that the samples

generated with LHS had smaller confidence intervals, so the results of LHS sam-

pling are more accurate.

6 Conclusions

The target of this work was to study the accuracy of Latin Hypercube sampling

and find some advices how to choose the sample size N. In literature, very few

directions were given, and in the tests it became clear that general rules are

difficult to find. However, both the literature and examples refer to the fact

that Latin Hypercube samples converge with smaller sample size than simple

random samples. The sample sizes needed in simple random sampling can be

used as upper limits for the Latin Hypercube sample size.

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