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));endfor t=30:32 y_temporary(t+1)=y_temporary(t)*(a-0.2);endfor 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
Best Answer