Attempted to access dz(1,1000); index out of bounds because numel(dz)=1.

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Alex 2013-10-23

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回答： Walter Roberson 2013-10-23

Could you please check my codes? The error message "Attempted to access dz(1,1000); index out of bounds because numel(dz)=1." appears at the third last line.

k = 0.5;
t = 1;
n = 100;
M = 1000;
dt = t/n;
randn('state',101);
dz = randn()*sqrt(dt);
for m = 1:M
      for i = 0:n-1
            y((i+1),M) = exp(-k*(n*(dt))-i*(dt))*dz((i+1),M);
      end;
end;

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Answer 1

sixwwwwww 2013-10-23

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Dear Alex, here is your working code, I made some modifications(If i understood correctly what you want to do):

k = 0.5;
t = 1;
n = 100;
M = 1000;
dt = t/n;
dz = randn(1000)*sqrt(dt);
i = 0:n - 1;
for m = 1:M
    y(i+1, m(1:end)) = exp(-k*(n*(dt))-i*(dt))*dz(i+1,m(1:end));
end

In your code you initializing "dz" with 101x101 elemets however in your for loop you were trying to access element out of this range. I hope it is what you need. Good luck!

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Answer 2

Walter Roberson 2013-10-23

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The subexpression dz((i+1),M) on the first trip through the "for i" loop, is going to be trying to use dz((0+1),M) which is dz(1,1000) . But you initialized dz = randn()*sqrt(dt); which is only a single value not a vector of length 1000.

Perhaps you wanted

dz = randn(n, M)*sqrt(dt);

Caution: when you use a variable named "i", then when you refer to "i" in a formula, it becomes difficult to know whether you meant the variable from the for loop, or meant the built-in value for "i" as sqrt(-1). It is recommended that you do not use variables named "i" or "j".