Matematik simulation and monte carlo with applications in finance and mcmc phần 10 potx

h > pump:=procbeta0,x,t,alpha,delta,gam,replic,burn localbayeslam,bayeslambda, i,k,n,beta,a1,g1,j,jj,lam,lamval,lambda,lambdaval,s,lam5,tot; # beta0=initial beta value # x[i]=number of f

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Sample 500 variates starting with x0 ¼ 4 and a ¼ 0:5 Note the long burn-intime and poor exploration of the state space (slow mixing of states) due to thesmall step length.

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Sample 100 000 variates, starting with initial value of 0, to produce a

histogram showing their distribution, together with sample mean and

We are interested in determining survival probabilities for components

The age at failure of the ith component is Xi; i¼ 1; 2; These are i.i.d

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Weibull random variables with parameters and , where the joint priordistribution is

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negligible and that the mixing of states is quite reasonable Bayes estimates

are given for survival probabilities at ages 1000, 2000, and 3000 hours Of

course, if interval estimates are required it will be necessary to replicate these

runs It is instructive to see the effect of reducing the sample size (discard

some of the original 43 ages at failure) The posterior density should then

reflect the prior density more so, with the result that the acceptance

probability becomes higher and successive and values are less correlated

than previously This is good from the viewpoint of reducing the variance of

# n=number of data items in x;

# q=number of ages at which survivor probabilities are

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# Perform k iterations;

for i from 1 to k do;

r1:=evalf(rand()/10^12);r2:=evalf(rand()/10^12);r3:=evalf(rand()/10^12);r4:=evalf(rand()/10^12);

# Sample candidate point (ap,bp) and compute likelihood(L2)for (ap,bp);

# Decide whether to accept or reject candidate point;

if ln(r4)<L2-L1 then a1:=ap; b1:=bp; L1:=L2; end if;

# Enter survivor probs and alpha and beta values intoith row of C;

for j from 1 to q do;

24

35

y1:¼ 1000

y2:¼ 2000

y3:¼ 3000100020003000

24

35

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375

264

375

> for i from 1 to k do:

for j from 1 to q+2 do:

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> f2:=[seq(f[i,2],i=1 k)]:PLOT(CURVES(f2),TITLE("survivalprobability at 2000 hours against iteration

number:"),AXESSTYLE(NORMAL));

dat1:=[seq(op(2,f1[i]),i=1 k)]:Estimate_Prob_survive_1000_hours=describe[mean](dat1);

0.8 0.75

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beta values against iteration number 3200

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How do the Bayesian estimates compare with a classical analysis, usingmaximization of the likelihood function, subject to 1 < ? (The parameterspace is restricted to the space used for the prior This ensures that theasymptotic Bayes and likelihood estimates would be identical.) Below, theunrestricted likelihood function, L1, and its contour plot are computed Thishas a global maximum at ^¼ 0:97 Using NLPsolve from the optimizationpackage, the constrained maximum is at ^¼ 1; ^¼ 2201 This represents acomponent with a constant failure rate (exponential life) and an expectedtime to failure of 2201 hours.

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[ImportMPS, Interactive, LPSolve, LSSolve, Maximize,

Minimize, NLPSolve, QPSolve]

The data are from D.P Gaver and I.G O’Muircheartaigh (1987), Robust

empirical Bayes analyses of event rates, Technometrics 29, 1–15

The number of failures for pump i in time t½i ði ¼ 1; ; 10Þ; x½i, are assumed

to be i.i.d PoissonðitiÞ where iare i.i.d gamma½; and is a realization

from gamma [,gam] The hyperparameters ; , and gam have been

esti-mated as described in the text

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The procedure ‘pump’ returns, for each pump i, a list of simulated

i; ½i; i; ð þ xiÞ=ð þ tiÞ values The latter are individual Bayes estimates of i

"

The Gibbs sampling procedure ‘pump’ is as follows

h

> pump:=proc(beta0,x,t,alpha,delta,gam,replic,burn) localbayeslam,bayeslambda,

i,k,n,beta,a1,g1,j,jj,lam,lamval,lambda,lambdaval,s,lam5,tot;

# beta0=initial beta value

# x[i]=number of failures of pump in in time t[i],i=1 10

# alpha, delta, gam are hyperparameters

# replic= number of (equilibrium) observations used

# burn= number of iterations for burn in

# n=number of pumps=10

# lambdaval[k]= a list of lambda values for pump k

# lam[k]= a list of [iteration number, lambda value] forpump k

if j>=1 then lambda[k,j]:=[j,g1];lambdaval[k,j]:=g1;bayeslam[k,j]:=(alpha+x[k])/(beta+t[k]);end if

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0 :¼ 0:25replic := 2000burn := 200

Print the Bayes failure rates and plot their posterior and prior densities

Note that the prior becomes unbounded as the failure rate approaches zero,

so the failure rate axis does not extend to zero

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end do:

display(A);

‘Bayes estimate failure rate pump’, 1, ‘is’, 0.05809446755

.2 1

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Let n1;0; n0;1; n1;1 be the number of animals captured in the first only,

second only, and both episodes respectively Then the number of distinct

animals captured is nprime¼ n1;0þ n0;1þ n1;1and the total number of animals

captured in both episodes is ncap¼ n1;0þ n0;1þ 2n1;1 Let p¼ probability that

an animal has of being capured on an episode The prior distribution of p is

assumed to be U(0,1)

Let notcap¼ N-nprime This is the number of animals in the population

that were not captured in either of the episodes We will use Gibbs sampling to

estimate the posterior distributions of notcap and p and the Bayes estimate of

# burn=number of iterations for burn in

# iter=number of (equilibrium) iterations used

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l1:=[seq(op(2,pa[i]),i=1 500)]:# a list of p-valuesl2:=[seq(op(2,notcapa[i]),i=1 500)]:# a list of notcapvalues

l3:=[seq((1-l1[i])^2,i=1 500)]:# a list of (1-p)^2values

histogram(l1,title="distribution of sampled pvalues",labels=["p","density"]);

Mean=evalf(describe[mean](l1));

StdDev=evalf(describe[standarddeviation](l1));

histogram(l2,title="distribution of samplednotcapvalues",labels=["notcap","density"]);

Mean=evalf(describe[mean](l2));

StdDev=evalf(describe[standarddeviation](l2));

Bayes_estimate_of_number_not_captured=250*evalf(describe[mean](l3));

p-value against iteration number 0.4

0.36 0.32 0.28 p

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distribution of sampled p values

density

1612840

p

Mean= 0.3277568452StdDev= 0.02987484843

distribution of sampled notcapvalues 0.03

0.025 0.02 0.015 0.01 0.005 0

notcapdensity

Mean= 114.5420000StdDev= 14.69966789Bayes_estimate_of_number_not_captured= 113.2008414

Here, slice sampling is explored as applied to the generation of variates from a

truncated gamma distribution with density proportional to x1 exon

sup-port½a; 1Þ where a > 0

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> restart:with(stats):with(stats[statplots]):

h

> truncatedgamma:=proc(alpha,a,x0,iter,burn)localx,r1,r2,r3,u1,u2,u3,lower,upper,pa,i,ii,tot;

Mean:=evalf(describe[mean](l1));StdDev:=evalf(describe[standarddeviation[1]](l1));

burn :¼ 100iter :¼ 500

:¼ 3

a :¼ 9x0 :¼ 10

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truncated gamma variate against iteration number 14

distribution of sampled x values 0.8

Mean :¼ 10:16067140StdDev :¼ 1:147220533

In order to obtain an estimated standard error for the mean of the truncated

gamma, the sample standard deviation is calculated of 100 independent

replications (different seeds and starting states), each comprising 200

obser-vations after a burn-in of 100 obserobser-vations In the code below the seed is

changed for each new replication It is not sufficient merely to use the last

random number from one replication as the seed for the next replication

This would induce positive correlation between the two sample means

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for j from 1 to replic do:

seed:=seed+113751:

randomize(seed):

x0:=a-ln(evalf(rand()/10^12))/1.04; #an initial state is

a random observation from an approximation to thetruncated gamma estimated from the histogram abovepa:=truncatedgamma(alpha,a,x0,iter,burn):

time1:=time()-t1;

t1 :¼ 3:266replic :¼ 100seed :¼ 6318540iter :¼ 200

:¼ 3

a :¼ 9burn:=100Mean :¼ 10:205022061Std error1 :¼ 0:0178901125time1 :¼ 7:077

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:¼ 3

a :¼ 9iter :¼ 100Mean :¼ 10:27322534Std error2 :¼ 0:1155931003time2 :¼ 5:750

100 independent truncated gamma variates

Iteration number

variate

13 12 11 10 9

For comparison of mixing 100 successive truncated variates from Gibbs

sampling are displayed below

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Mean:=evalf(describe[mean](l1)):StdDev:=evalf(describe[standarddeviation[1]](l1)):

burn :¼ 100iter :¼ 100

:¼ 3

a :¼ 9x0 :¼ 10

truncated gamma variate against iteration number 12.5

12 11.5 11 10.5 10 9.5 9

Iteration number variate

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:¼ 0:54

:¼ 1:11gam :¼ 2:20

0 :¼ 0:25replic :¼ 2000burn :¼ 200

xa[6]:="lambda[6]":xa[7]:="lambda[7]":xa[8]:="lambda[8]":xa[9]:="lambda[9]":xa[10]:="lambda[10]":

A:=array(1 5,1 2):

for i from 1 to 5 do:

A[i,1]:=display({

statplots[histogram](v2[2*i-1]),plot(u/k,lambda=ms[2*i-1]*0.9 m[2*i-1]*1.1,labels=[xa[2*i-1],"prior and posteriordensities"],labeldirections=[HORIZONTAL,VERTICAL])},tickmarks=[4,2]):

A[i,2]:=display({

statplots[histogram](v2[2*i]),plot(u/k,lambda=ms[2*i]*0.9 m[2*i]*1.1,labels=[xa[2*i],"prior and posterior

densities"],labeldirections=[HORIZONTAL,VERTICAL])},tickmarks=[4,2]):

end do:

display(A);

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.1e2

0.

.3 2 1

.2e2

0.

.6 4 2

.1e2

0.

.2 1

prior and posterior densities

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