draw histogramm of X(i)
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ad lyn
el 25 de Ag. de 2021
Comentada: the cyclist
el 25 de Ag. de 2021
v=9.91256303526217e-3;
x(1)=3.442619855899;
x(128)=0;
for i=2:128
a(i)=exp(-0.5*x(i-1)^2)+(v/x(i-1));
x(i)=sqrt(-2*log(a(i)))
y(i)=nrmlpdf(x(i))
for i=2:128
zigr(i)=(x(i)/x(i-1))
end
r=x(1);
for i= i:1:128
if i==0
u0 = 2*rand()-1;
if(abs(u0)<zigr(i))
X(i)=u0*x(i);
if(i==0)
X(i) = tail( r);
z(i)=u*x(i);
f0=exp(-0.5*(x(i)^2)-(x(i)^2));
f1=exp(-0.5*(x(i+1)^2)-(x(i+1)^2));
if(f1+rand()*(f0-f1)<1.0)
X(i)=z(i);
end
end
end
end
end
end
hist(X)
2 comentarios
the cyclist
el 25 de Ag. de 2021
I don't think we can run your code, because nrmlpdf is not a standard MATLAB function.
Also, you did not actually ask a question. What do you need? Be specific.
the cyclist
el 25 de Ag. de 2021
I've edited your code here (mostly lining up the code blocks), and changed your nrmlpdf function into an inline function, for convenience.
nrmlpdf = @(x) exp(-x.^2/2);
v=9.91256303526217e-3;
x(1)=3.442619855899;
x(128)=0;
for i=2:128
a(i)=exp(-0.5*x(i-1)^2)+(v/x(i-1));
x(i)=sqrt(-2*log(a(i)));
y(i)=nrmlpdf(x(i));
for i=2:128
zigr(i)=(x(i)/x(i-1));
end
r=x(1);
for i= i:1:128
if i==0
u0 = 2*rand()-1;
if(abs(u0)<zigr(i))
X(i)=u0*x(i);
if(i==0)
X(i) = tail( r);
z(i)=u*x(i);
f0=exp(-0.5*(x(i)^2)-(x(i)^2));
f1=exp(-0.5*(x(i+1)^2)-(x(i+1)^2));
if(f1+rand()*(f0-f1)<1.0)
X(i)=z(i);
end
end
end
end
end
end
hist(X)
You are getting your error because your code never reaches the line where X is defined. The heart of the problem seems to be where you have a second for loop over the same variable:
for i= i:1:128
This is a very confusing line of code, and certainly doesn't do what you intend. I'm guessing you probably want a second looping variable j, that loops from the current value of i to 128.
But I don't understand the indexing of that section of code, and am not sure which index values should be i, and which should be j.
Furthermore, you have the statement
if i==0
and MATLAB is never going to enter that section, because neither looping variable is ever equal to zero.
Respuesta aceptada
Más respuestas (1)
Steven Lord
el 25 de Ag. de 2021
And I second the cyclist's question for more information about the difficulty you're experiencing when you try to create the histogram.
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