Re: [Gretl-users] vectorized form of the following loop

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(a)googlemail.com&gt;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 > > _______________________________________________ > Gretl-users mailing list > Gretl-users(a)lists.wfu.edu > http://lists.wfu.edu/mailman/listinfo/gretl-users > _______________________________________________ Gretl-users mailing list Gretl-users(a)lists.wfu.edu http://lists.wfu.edu/mailman/listinfo/gretl-users