How can I fix "Attempted to access Y(0); index must be a positive integer or logical." MATLAB for loop error?

Question

shackelferd 2015-9-7

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/240515-how-can-i-fix-attempted-to-access-y-0-index-must-be-a-positive-integer-or-logical-matlab-for-lo

回答： Walter Roberson 2015-9-7

I'm trying to use a for loop in MATLAB and I keep getting an error message. The instructions I'm following are to create a 10000X1 vector using a random number generator and label it "e", and then a second 10000X1 vector that only contains zeros. Then using a loop I have to simulate the process Y(t)=(1-a)*Y(t-1)*e(t) where a=0.1. The code I have been attempting to run is:

Y=zeros(10000,1);
e=randn(10000,1);
for t=1,2,...,10000
a=0.1;
Y(t)=(1-a)*Y(t-1)*e(t);
end

I keep getting an error message that says: "Attempted to access Y(0); index must be a positive integer or logical." Any help is appreciated

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2015-9-7

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/240515-how-can-i-fix-attempted-to-access-y-0-index-must-be-a-positive-integer-or-logical-matlab-for-lo#answer_191532

在 MATLAB Online 中打开

for t=1,2,...,10000

is not valid syntax. Please post your actual code.

Your formula has

Y(t)=(1-a)*Y(t-1)*e(t);

when t = 1, the Y(t-1) part is going to be trying to access Y(0). In MATLAB you cannot index an array at location 0.

I suspect the formula you need is

Y(t+1) = (1-a)*Y(t)*e(t);

with the "for" running from 1 to 9999. However, it is possible that instead you want the formula that you have now but with t starting from 2 in the loop.

The difference between the two formulas has to do with whether e(1) should ever be used and if so whether it should be used with Y(1) or with Y(2)