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Category: Science and Technology
Date Submitted: 11/08/2012 03:45 AM
c/d Financial models
clearvars; n=10^5; z=randn(1,n); varz=1;
mu_z=0; %i indicates which function you use (g_i)
for i=1:6
for k=1:n
a(k)=1/n*exp(z(k)-1/2)*g(i,z(k));
b(k)=1/n*g(i,(z(k)+1));
end
A(i)=sum(a); %sample mean
B(i)=sum(b); %sample mean
%--stda(i)=1/sqrt(n)*sqrt(sum((n*a-A(i)).^2));
%--stdb(i)=1/sqrt(n)*sqrt(sum((n*b-B(i)).^2));
Vargzb(i)=varz*(dg(i,mu_z))^2;%delta method
Vargza(i)=varz*(exp(mu_z-1/2)*dg(i,mu_z)+exp(mu_z-1/2)*g(i,mu_z))^2; %product rule
end
A
B
stda=(1/sqrt(n))*sqrt(Vargza); %stda alt
stdb=(1/sqrt(n))*sqrt(Vargzb); %stdb alt
CIAl=[A-1.96*stda]
CIAr=[A+1.96*stda]
CIBl=[B-1.96*stdb]
CIBr=[B+1.96*stdb]
overlap=[(CIAr>CIBl)]
diffa=mean(CIAr-CIAl)
diffb=mean(CIBr-CIBl)
%--stda
%--stdb
%USED: 1) Central Limit Theorem
%2)Hints: to compute the approximate confidence interval, recall that the
%standard deviation of the average of n independent identically distributed
%variables is given by n^ (-1/2) times the standard deviation of a single
%variable.
%3)The [single variable] -> What is the variance of (exp(z(k)-1/2)*g(z(k))
%given that the variance of z(k)=1. ?
%Answer: http://www.math.umt.edu/patterson/549/Delta.pdf
%DELTAMETHOD If Y=f(X), var(Y) is approx. given by Var(Y)=[f'(mu_x)]^2*Var(X)
OUTPUT:
A = 1.0078 2.0095 4.4859 0.6065 0.3417 0.1826 (A,B are the numbers
B = 1.0027 2.0109 4.5002 0.6068 0.3396 0.1858 to be calculated for g1,g2..,g6)
CIAl = 1.0040 2.0095 4.4783 0.6065 0.3379 0.1792 (lower endpnt CI95% for A)
CIAr = 1.0115 2.0095 4.4934 0.6065 0.3455 0.1860 (upper endpntCI95%for A)
CIBl = 0.9965 2.0109 4.4940 0.6006 0.3396 0.1858
CIBr = 1.0089 2.0109 4.5064 0.6130 0.3396 0.1858
overlp = 1 0 0 1 1 1 (Do intervals overlap)
diffa = 0.0062 diffb = 0.0062 (CI length)