for loop - Error Attempted to access y(30); index out of bounds because numel(y)=2.

Dear all

I have to compute a time path over t=1:100 periods. I have to assume a temporary shock in t=30:32 so I tried this and used different for loops:

y(1)=100; %(starting value of y) a=1.02;

    for t=1:29;
        y(t+1)=y(t)*(a+0.0001*randn(1,1));
        for t=30:32         (here there is the temporary shock)
            y(t+1)=y(t)*(a-0.2);
        end;
        for t=33:100
            y(t+1)=y(t)*(a+0.0001*randn(1,1));
        end;
    end;

Unfortunately, it gives me the error: Attempted to access y(30); index out of bounds because numel(y)=2.

I really tried different things but couldn't get the loop fixed.

I already thank you for answering my question.

Answer 1

[Julia]

Hi,

the best is to define y before you use it.

y=zeros(100,1)

I think your loops are nested in the wrong way. You set t = 1, calculate y(2), enter the next loop and calculate y(31), y(32) and y(33) and then enter the last loop to calculate the entries up to y(100). You do all of this for t = 2 to t = 29. I suggest you execute one loop after the other, this should save a lot of time.

[Rico]

Wow, thank you Julia for your immediate response! It worked perfectly when I used seperate loops, thank you!

Answer 2

[Torsten]

y=zeros(101);

b=randn(101,1);

y(1)=100;

for t=1:29

   y(t+1)=y(t)*(a+0.0001*b(t,1));

end

for t=30:32 %(here there is the temporary shock)

   y(t+1)=y(t)*(a-0.2);

end

for t=33:100

   y(t+1)=y(t)*(a+0.0001*b(t,1));

end

Best wishes

Torsten.

[Rico]

Torsten, thank you very much for your quick reply! I tried yours and it worked perfectly as well.

Have a nice day

best Rico