# Burn-in period for random walk

We are trying to make simulation experiment involving a common stochastic trend, that is described by the random walk (or $I(1)$ process) $Y_t = Y_{t-1} + \varepsilon_t$, where innovations $\varepsilon_t$ ~ $N(0,1)$. However when could we be sure that the past innovations are more or less reasonably included into the stochastic trend? Is there any good proxy for random-walk's burn-in period?

I've looked for R's burn-in default suggestion for $AR(p)$ part: `ceiling(6/log(minroots))`, meaning that in case of unit root we get infinity here and roughly tried to take close to unity root 1.0001 (actually it is the same like taking 60000 at once). So any reasonable suggestions or rule of thumbs you do use in practice?

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Could you add a bit more context please? e.g. what R function and package are you using? – onestop Jan 4 '11 at 19:50
A link explaining why random-walk needs a burn-in also would help. – mpiktas Jan 4 '11 at 20:12
Well we will use diffinv() directly, but the burn-in default could be found in arima.sim() function. – Dmitrij Celov Jan 4 '11 at 20:39

## 1 Answer

Burn-in doesn’t make sense here. The random walk you describe does not have a stationary distribution.

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Well, I do know that the process is not stationary. Burn-in period's purpose is to be more or less confident that the starting history values are already present in the stochastic trend, in the sense that we may assume that we have started from the long long past. It is not clear why I have to go from zero if I don't want to. And else it would be interesting to hear some suggestions about the same questions when the process is random walk with drift component. – Dmitrij Celov Jan 5 '11 at 7:12

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If I understand correctly, you want to simulate a stochastic trend. Why not initialize $Y_0$ to a random value, say a sample from $N(0,\sigma^2)$ with $\sigma^2$ large? – vqv Jan 5 '11 at 18:22
One more comment. The problem with burn-in is that, in addition to being non-stationary, the process is not ergodic. So what you are asking for is not really possible. You will always be able to infer (roughly) what the initial value of the process is. – vqv Jan 5 '11 at 18:39

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Suppose $Y_t = \alpha Y_{t-1} + \epsilon_t$ where $\epsilon_t$ are i.i.d. $N(0,1)$ and $|\alpha| \leq 1$. The conditional distribution of $Y_t$ given $Y_0$ is $N(\alpha^t Y_0, \sigma_{\alpha,t}^2)$, where $\sigma_{\alpha,t}^2 \leq t$ depends only on $\alpha$ and $t$. If $\alpha = 1$, note that $Y_t$ given $Y_0$ remains centered at $Y_0$ even as $t \to \infty$. In that sense, you can always infer $Y_0$. On the other hand, if $|\alpha|< 1$ then the conditional mean of $Y_t$ given $Y_0$ tends to 0 as $t\to\infty$ so that asymptotically $Y_t$ is conditionally independent of $Y_0$. – vqv Jan 6 '11 at 15:42

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If $|\alpha| = 1$, then $\sigma_{\alpha,t}^2 = t$. If $|\alpha| < 1$, then $\sigma_{\alpha,t}^2 = (1-\alpha^t) / (1-\alpha)$. So although you can infer $Y_0$ from $Y_t$ when $\alpha = 1$ the standard error will be of the order $\sqrt{t}$. I think to better answer your question you need to be less specific and explain what you will use $Y_t$ for. – vqv Jan 6 '11 at 15:49
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