vectorizing for-loops for autoregressive or other recursive processes

Ginters Bušs ginters.buss at gmail.com
Thu Jul 21 12:23:48 CEST 2011


On Thu, Jul 21, 2011 at 1:18 PM, Ginters Bušs <ginters.buss at gmail.com>wrote:

> Dear all,
>
> I am inspired by the recent Allan's comment that I can vectorize
> for-loops.
>
> Say, I have a for-loop for AR(2) process
> y(t+1)=f1*y(t)+f2*y(t-1)+epsilon(t+1) with a degenerate disturbance term:
>
> f1=1.2; f2=-0.5;
> n=100000;
> c=1:n;
> c(2)=f1*c(1)+c(2);
> for i=3:n;
>     c(i)=c(i)+f1*c(i-1)+f2*c(i-2);
> end
>
> Allan proposed trying
>
> f1=1.2; f2=-0.5;
> n=100000;
> b=1:n;
> b(2)=f1*b(1)+b(2);
> b(3:n) = b(3:n)+f1*b(2:n-1)+f2*b(1:n-2);
>
>
> but the result is different.  Then, I thought I can try to use ode; say,
> for AR(1) process y(t+1) = a*y(t)+ epsilon(t+1) it would look like
>
> deff("yp=a_function(k,y)","yp=a*y+sigma*u(k)")
> y0=0;
> a=.9;
> n=100000;
> u=1:n;
> y=ode("discrete",y0,1,1:n,a_function);
>
> but it turns out that for-loop is a bit faster than this ode code.
>
> Another idea would be using Wold representation of (only stationary) AR
> process by rewriting the above AR(2) process as
>
> y(t+1) = ((1-f1*L - f2*L^2)^(-1))*epsilon(t+1)
>
> where (1-f1*L - f2*L^2) is a lag polynomial, L defined as Ly(t)=y(t-1)
>
> and trying to get the series representation of  (1-f1*L - f2*L^2)^(-1) but
> I'm stuck here, and this would work only for a stationary process.
>
> Do you have any ideas/experience with rewriting for-loops for AR or other
> recursive processes more efficiently than the above first code?
>
>
>
>
>
>
>
> Just clarifying that I'm using a degenerate disturbance term here for easy
comparison; in real application, it would be some stochastic process.
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