Problem 11.9

Nasser Abbasi

Conclusion

I ran the program 3 times, each time for different ranges of random variables. Each ranges matches the range specificed in problem 11.8 parts a,b and c.

So, the program asks the user for which part to run. If part a is selected, then the range used for the random numbers is 0..1, if part b is selected then the range is 0..infinity, and if part c is selected, then the range used is –infinity..infinity.

I use +- 100 to represent infinity.

So, this below shows the 3 runs. Each time it is run, it shows the 6 moments for N=10,100 and 10,000.

For comparison to part a, I see that the numerical approximation gets closer to analytical solution as N is increased. For N=10,000 for example, the first moment is 0.505851 compared to 0.5 analytical, and second moment is 0.333982 compared to 1/3 analytical.

For comparison to part b, the analytical solution gave LAMBDA for first moment and 2*LAMBDA^2 for second moment. If I assume LAMBDA=50, then at N=10,000 the program gave LAMBDA=50.585066, but for the second moment, I expected to get something close to 5,000 since this is what 2*50^2 is, but the program gave 3,339. Maybe the convergance is not fast in this case?

For comparison with part c, the analytical solution when mu=0 gave the first moment as 0 and the second moment as sigma^2. Running the program, I see for N=10,000 that the first moment is going to zero indeed.

» problem_11_9

solves problem 11.9 in book

Nasser Abbasi

Enter a or b or c parts from 11.8 to solve for numerically: 'a'

moment 1 for part a when N=10 =0.566857

moment 2 for part a when N=10 =0.527903

moment 3 for part a when N=10 =0.084884

moment 4 for part a when N=10 =0.197667

moment 5 for part a when N=10 =0.119564

moment 6 for part a when N=10 =0.108071

moment 1 for part a when N=100 =0.510070

moment 2 for part a when N=100 =0.313264

moment 3 for part a when N=100 =0.268097

moment 4 for part a when N=100 =0.201896

moment 5 for part a when N=100 =0.163267

moment 6 for part a when N=100 =0.141122

moment 1 for part a when N=10000 =0.505851

moment 2 for part a when N=10000 =0.333982

moment 3 for part a when N=10000 =0.251965

moment 4 for part a when N=10000 =0.203513

moment 5 for part a when N=10000 =0.169042

moment 6 for part a when N=10000 =0.138806

»

» problem_11_9

solves problem 11.9 in book

Nasser Abbasi

Enter a or b or c parts from 11.8 to solve for numerically: 'b'

moment 1 for part b when N=10 =56.685705

moment 2 for part b when N=10 =5279.026059

moment 3 for part b when N=10 =84883.977329

moment 4 for part b when N=10 =19766700.521324

moment 5 for part b when N=10 =1195638748.756388

moment 6 for part b when N=10 =108071460392.877720

moment 1 for part b when N=100 =51.006979

moment 2 for part b when N=100 =3132.637115

moment 3 for part b when N=100 =268096.906575

moment 4 for part b when N=100 =20189632.329101

moment 5 for part b when N=100 =1632665688.281873

moment 6 for part b when N=100 =141122467921.472350

moment 1 for part b when N=10000 =50.585066

moment 2 for part b when N=10000 =3339.821404

moment 3 for part b when N=10000 =251964.890258

moment 4 for part b when N=10000 =20351348.243054

moment 5 for part b when N=10000 =1690416560.356563

moment 6 for part b when N=10000 =138806093701.635190

»

» problem_11_9

solves problem 11.9 in book

Nasser Abbasi

Enter a or b or c parts from 11.8 to solve for numerically: 'c'

moment 1 for part c when N=10 =13.371410

moment 2 for part c when N=10 =3893.480889

moment 3 for part c when N=10 =-252285.690056

moment 4 for part c when N=10 =18688477.533827

moment 5 for part c when N=10 =-631285505.582327

moment 6 for part c when N=10 =44379320039.771805

moment 1 for part c when N=100 =2.013959

moment 2 for part c when N=100 =3310.739300

moment 3 for part c when N=100 =38405.540366

moment 4 for part c when N=100 =16473346.298195

moment 5 for part c when N=100 =-267896939.293912

moment 6 for part c when N=100 =150281297760.044340

moment 1 for part c when N=10000 =1.170132

moment 2 for part c when N=10000 =3347.817805

moment 3 for part c when N=10000 =736.245324

moment 4 for part c when N=10000 =20322193.440434

moment 5 for part c when N=10000 =40758028.654677

moment 6 for part c when N=10000 =140197535497.539060

»

function problem_11_9()

% solves problem 11.9 in book

% Nasser Abbasi

clear all; help problem_11_9;

% design:

% for each range from problem 11.8, we find

% a set of random numbers in the range.

% for part(a) in 11.8 the range is 0..1

% for part(b) in 11.8 range is 0..infinity

% for part(c) in 11.8 tange is -infinity..infinity

% Since I do not know how to find random number

% all the way to infinity, I picked +-1000 to

% represent infintify.

rand('state',0);

part = input('Enter a or b or c parts from 11.8 to solve for numerically: ');

switch part

case 'a'

upperRange = 1;

lowerRange = 0;

process(part,upperRange,lowerRange);

case 'b'

upperRange = 100;

lowerRange = 0;

process(part,upperRange,lowerRange);

case 'c'

upperRange = 100;

lowerRange = -100;

process(part,upperRange,lowerRange);

end

%%%%%%%%%%%%%%%%%%%%%%%

%

%

%%%%%%%%%%%%%%%%%%%%%%%

function process(part,upperRange,lowerRange)

numbers=[10 1e2 1e4];

for(k=1:length(numbers))

N=numbers(k);

for(i=1:6)

moment(i)=0;

for(j=1:N)

randomNumber = rand(1);

% map the random number to fall into the required actuall range.

%

% 1<--X------>0

%

% upper range<----X'-------------->lower range

%

newRange = upperRange-lowerRange;

randomNumber = randomNumber * newRange;

randomNumber = randomNumber + lowerRange;

moment(i) = moment(i)+ randomNumber^i;

end

moment(i) = moment(i)/N;

fprintf('moment %d for part %c when N=%d =%f\n',i,part,N,moment(i));

end

fprintf('\n');

end