I must be having some memory problems!

My example is wrong if you test it you will see that the results differ. I ment to warn you of that but wrote exactly the opposite (missed the "not" word): (but results are NOT the expectED ones)

Please double check the script.

Hélio


On Sun, Mar 30, 2014 at 6:14 PM, Artur T. <artur.tarassow@googlemail.com> wrote:
Thank you both Jack and Hélio for your suggestions! Both example work well.
Jack's nice version is about 10% less time consuming than my original example (at least in my procedure here).

Best,
Artur


2014-03-29 16:08 GMT+01:00 Hélio Guilherme <helioxentric@gmail.com>:

This was my attempt (but results are the expect ones):

set seed 732237


scalar T = 100
scalar b1 = -0.2
scalar b2 = 0.5
matrix u = mnormal(T,1)
matrix dx = mnormal(T,1)
matrix dy = zeros(T,1)
matrix y = zeros(T,1)

loop i=2..T -q
    dy[i] = b1*y[(i-1)] + b2*dx[i] + u[i]
  
    y[i] = y[i-1] + dy[i]
endloop

#Vectorized
set seed 732237
matrix uu = mnormal(T,1)
matrix dxx = mnormal(T,1)
matrix w = zeros(T,1) # same as initial y

result = ( uu = u ) ? "Equal" : "Different"
printf "uu and u matrix are %s\n", result

w[2:T,1] = w[1:(T-1),1] +  b1*w[1:(T-1),1] + b2*dxx[2:T,1] + uu[2:T,1]

result = ( w = y ) ? "Equal" : "Different"

print result

matrix z = w ~ y

print z

---
I now that Jack already gave a better answer.

Hélio


On Sat, Mar 29, 2014 at 12:40 PM, Artur T. <artur.tarassow@googlemail.com> wrote:
Hello,

I was thinking how to implement the following code snippet into a vectorized form, but I couldn't figure out how to do this effectively. The issue I face is the lagged value of y which changes over the loop. Does anybody have an alternative way to run this thing?

<hansl>
scalar T = 100
scalar b1 = -0.2
scalar b2 = 0.5
matrix u = mnormal(T,1)
matrix dx = mnormal(T,1)
matrix dy = zeros(T,1)
matrix y = zeros(T,1)

loop i=2..T -q
    dy[i] = b1*y[(i-1)] + b2*dx[i] + u[i]
   
    y[i] = y[i-1] + dy[i]
endloop
<\hansl>

Thanks in advance.
Artur

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