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If I have a random walk without drift the differences form a white noise process.

But what happens if I incorporate a drift $d$? Does this still hold true? I'm not sure because with the drift term d:

$$ y_{t} = d + y_{t-1} + \epsilon_{t} $$ $$ y_{t} - y_{t-1} = d + \epsilon_{t} $$

The right hand side of equation 2 can't be a white noise process (for instance its expected value would be $d \neq 0$.

Any help would be appreciated!

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    $\begingroup$ It is not since the white noise is of mean zero. $\endgroup$ – dfhgfh Jan 16 '13 at 11:36
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You are having a special case of a Ornstein-Uhlenbeck process; in your case the mean-reversion part is stable.

In general, an OU-model can be expressed as:

$dX(t) = \alpha[\theta - X(t)]dt + \sigma dB(t)$

  • $\theta$ being the process' optimum / "mean you are trying to revert back to"
  • $\sigma$ being the intensity of the process' random fluctuations
  • $\alpha$ being the strength of the selection towards the process' optimum $\theta$

As you see, the first term that you call "drift", is practically the "strength of mean reversion" towards $\theta$ and usually get larger the further away you are at time $t$ from $\theta$. In your example this term is fixed making $\theta$ and $\alpha$ are non-identifiable but still to call your increments white noise you need to subtract the "(deterministic) drift" $d$. So yeah, your intuition is right, $d$+ $\epsilon_t$ is not white noise, if $d=0$ isn't true. (In which case you have Wiener process)

I am getting to all the trouble to say this stuff cause the OU process in the only stationary Gauss-Markov stochastic process out there so if you are inclined you can really do fancy stuff stochastic-wise with it. :-D

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Not when $d\neq 0$; then $y_{t} - y_{t-1} -d = \epsilon_{t} $ is white noise (provided the errors $\epsilon$ are uncorrelated & their mean is zero).

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