I'm working on basic code that generates data from AR and VAR processes; the code generates enough observations to dampen the effect of the initial conditions. For example, if I want to generate 30 observations, the code generates 300 observations and only returns the final 30.

In an answer to my question about this code , a user recommended that instead of generating $N$ observations and only returning the last $T$, I pass a parameter that specifies

the desired decay of the initial condition.

I'm not sure how to interpret this. Take a simple univariate AR(1) process as an example:

\begin{align} y_t = \rho y_{t-1} + u_t \end{align} where $\lvert \rho \rvert < 1$, $u \sim N(0, \sigma_u^2)$, and $y_0 = 0$. What does it mean to talk about the desired decay of this initial condition? I couldn't find any references that discussed this. For example, is this referring to calculating some number of observations after which the initial condition has decayed by, say, 90%? What exactly does that mean?

The actual processes I'm working with are slightly more involved, e.g.

\begin{align} \begin{bmatrix} y_{1,t} \\ y_{2,t} \end{bmatrix} &= \begin{bmatrix} 0.02 \\ 0.03 \end{bmatrix} + \begin{bmatrix} 0.5 & 0.1 \\ 0.4 & 0.5 \end{bmatrix} \begin{bmatrix} y_{1,t-1} \\ y_{2,t-1} \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0.25 & 0 \end{bmatrix} \begin{bmatrix} y_{1,t-2} \\ y_{2,t-2} \end{bmatrix} + \begin{bmatrix} u_{1,t} \\ u_{2,t} \end{bmatrix} \end{align}

but I'm just trying to get a general idea for now.

  • $\begingroup$ I'm starting to wonder if there are any references about this concept, if it's even a valid concept. After weeks of searching and no responses either on here or on Code Review, I'm increasingly skeptical. $\endgroup$ – Michael A Sep 10 '15 at 21:16
  • 1
    $\begingroup$ I think the user is just commenting on the choice of $N=10T$ in your code. There, "10" is a magic constant. I think he's saying you should instead calculate $N$ as a function of the desired "decay" of the initial condition. For example, in your AR(1) process, $y_t = \rho^{t-1}y_0 + f(\mathbf{u}_{1,\ldots,t})$. Waiting until $t=1+\log(0.01)/\log(\rho)$ would give you $y_t=0.01y_0+f(\mathbf{u}_{1,\ldots,t})$, washing out the effect of $y_0$ to one hundreth of its value, for example. $\endgroup$ – Creosote Sep 13 '15 at 14:06
  • $\begingroup$ For more complicated processes, rather than solve analytically, you might want to run a second process with no noise term in the recurrence relation. Keep stepping both processes until the variables in the no-noise process appear to have converged to steady-state values. $\endgroup$ – Creosote Sep 13 '15 at 14:18

I just remembered I'm supposed to give answers as an Answer, not a Comment, so I'll use this opportunity to offer some code in R that simulates your example process with a dynamically-determined burn-in period. Red points are burn-in; blue are steady-state.

In case you can't run R, here's an example of the output. a graph

And if you can run R, here's the code.

# parameters of the process
# rho is 0.99; each u_t is N(0,0.02**2)
rho = 0.99
u = function() rnorm(1,mean=0,sd=0.02)

# Y is the actual process, Z the no-noise cousin
# start each process at 1,1

# say a sequence is "converged" if the last three
# terms are within a factor of 1+-1e-6 of each other
converged = function(Z) {
  n = length(Z)
  if (n<3) F else { L=log(Z[(n-2):n]); mu=mean(L); max(abs(L-mu))<1e-6; }

# step until converged
while (!converged(Z1) | !converged(Z2)) {
  t = t + 1
  Y1[t] = 0.02 + 0.5*Y1[t-1] + 0.1*Y2[t-1] + u()
  Z1[t] = 0.02 + 0.5*Z1[t-1] + 0.1*Z2[t-1] # no noise
  Y2[t] = 0.03 + 0.4*Y1[t-1] + 0.5*Y2[t-1] + 0.25*Y1[t-2] + u()
  Z2[t] = 0.03 + 0.4*Z1[t-1] + 0.5*Z2[t-1] + 0.25*Z1[t-2] # no noise

# step the same amount again for "steady state" data
t_burn = t
while (t <= 2*t_burn) {
  t = t + 1
  Y1[t] = 0.02 + 0.5*Y1[t-1] + 0.1*Y2[t-1] + u()
  Y2[t] = 0.03 + 0.4*Y1[t-1] + 0.5*Y2[t-1] + 0.25*Y1[t-2] + u()

# plot the results; red=burn-in period, blue="steady state" data
add_points = function(Y) {
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  • $\begingroup$ This is quite helpful; thank you. Your explanation of the decay makes perfect sense, and I'll look over the code over the next couple of days. $\endgroup$ – Michael A Sep 14 '15 at 13:18

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