Laplace approximation,binomial distribution

hosein farahani

2015 1 28

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8 次查看（30 天）

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    % code
clear
clc
close all
n = input ('Please Enter Number of Exams : ');
p = input ('Please Enter Probability : ');
eta=n*p;
sigma=sqrt(n*p*(1-p));
for x=1:n*100;
a=rand(1,n);
b=0;
for i=1:n
    if(a(i)>=1-p)
        b=b+1;
    end
end
c(x)=b;
d(x)=binornd(n , p);
e(x)=(1./(sqrt(2*pi*x*p*(1-p)))).* exp(-(((x+1)*p-(x*p)).^2)./(2*x*p*(1-p)));
e(x)=sigma*e(x)+eta;
end
histfit(c);
axis([0 n 0 inf])
M = mean(c);
varc=var(c);
disp(['var is : ', num2str(varc)]);
disp(['Mean is : ', num2str(M)]);
figure
histfit(d);
axis([0 n 0 inf])
M2 = mean(d);
varc2=var(d);
disp(['var is : ', num2str(varc2)]);
disp(['Mean is : ', num2str(M2)]);
figure
histfit(e,100);
axis([0 n 0 inf])
M3 = mean(e);
varc3=var(e);
disp(['var is : ', num2str(M3)]);
disp(['Mean is : ', num2str(M3)]);

Hi to All. i using this code for generating random numbers with binomial distribution and plotting it. the first way and the binornd is true but i very confused with third histfit. the e(x) is formula of Laplace approximation in this link : De Moivre Laplace theorem

output is :