loop starting from a specific time

I first had to calculate two time paths y_permanent and y_temporary, the codes are the following ones:

y_temporary=zeros(101,1);
psi=0.0001*randn(101,1);
y_temporary(1)=100;

for t=1:29
   y_temporary(t+1)=y_temporary(t)*(a+0.0001*psi(t,1));

end

for t=30:32
   y_temporary(t+1)=y_temporary(t)*(a-0.2);

end

for t=33:100
 y_temporary(t+1)=y_temporary(t)*(a+0.0001*psi(t,1));

end

The path for the permanent change was:

y_permanent=zeros(100,1);
y_permanent(1)=100;

for t=1:100;
    y_permanent(t+1)=y_permanent(t)*(1.01+0.0001*psi(t,1)); %alpha is now 1.01 
end;

Now, my problem is, that I have to figure out a loop wich starts at time t=33 and stops, when the temporary shocks value is twice as high as the value of the permanent shock, counting the number of periods it needs to do so.

I tried several for loops, while loops and combination of a while and if loop but with no success.

If someone could give me a hint how to easily create such a loop, I would be so glad! I'm desperately looking for an answer.

Best Rico

Answer 1

[Geoff Hayes]

Rico - if a period is just one iteration of your for loop (so since your arrays have 101 elements, then there are 101 periods), then you could try the following

 numPeriods  = 0;
 criteriaMet = false;
 for k=33:length(y_temporary)

     if y_temporary(k)>=2*y_permanent(k)
         % the temporary shock value is at least twice that of the
         % permanent shock, so we are done
         criteriaMet = true;
         break;
     end

     numPeriods = numPeriods + 1;
 end

Note how we use break to exit the for loop once the criteria has been met. We use criteriaMet to determine if we ever did find a period where the temporary shock was at least twice that for the permanent.

[Rico]

Dear Geoff

Thank you very much for your professionel response.

I first had to extend the periods and then I used your code. In the end, it worked perfectly, so thank you very much!!!

Have a nice evening

Best Rico

[Geoff Hayes]

Glad it worked out, Rico!