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I'm aware of the simple 'proof' that shows random walks with a normal error term are non-stationary in original form and stationary in first-difference form but what happens if the errors have a different distribution? I'm looking into modelling these errors under different distributions, but am concerned that if I am using the principles of first-differencing to model, this is actually pointless under different distributions.

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    $\begingroup$ Stationarity is a "non-parametric" concept, it does not depend on any specific distribution. Random walks are non-stationary whatever distribution their building block white noise (what you call the "error") follows - and their first difference is stationary likewise. $\endgroup$ – Alecos Papadopoulos Mar 6 '14 at 23:47
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    $\begingroup$ Central limit theorem is a powerful thing. Random walks with disturbance terms that have finite variance will in the limit look a lot like normal random walks. If the disturbance terms do not have finite variance, then things get trickier. $\endgroup$ – DavidShor Jan 5 '15 at 22:04
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Normality is used in random walk processes for a couple of reasons.

The main reason is the divisibility of the time step. As you know the sum of normal variables, is a normal variable. So, if you have a process $$y_{t+\Delta t}=y_t+e_t$$, where $e_t\sim\mathcal{N}(0,\sigma^2)$, then $$y_{t+x\Delta t}=y_t+\xi_t$$, where $\xi_t\sim\mathcal{N}(0,x\sigma^2)$. Here, $$E[y_t]=y_0$$ and $$Var[y_t]=x\sigma^2$$

What is nice about this property is that now you have a good theoretical argument to expect the normally distributed errors. Consider this, if you are looking for monthly revenues of a store, one could argue that it's comprised of daily revenues. Then applying CLT to them you get normally distributed monthly revenues. So, in any situation where your time interval is divisible, and the shocks are independent, you'd argue that the errors to be normal: $$\sigma^2_{mon}=30\sigma^2_{day}$$

Also, think about this. If your process has non-normal errors, then while time passes, the sum of non-normal errors will become normal due to CLT. So, you'll end up with different distributions of errors for different time intervals. Let's say your monthly errors are Gamma distributed, but your annual revenues will be more or less normal. This is inconvenient, and strange. I'll ask where did the Gamma come from? Why didn't daily error converge to normal monthly error? etc.

There's a class of distributions which are stable, i.e. like Normal distribution they add up to themselves. You can build the processes with them which are divisible.

Also, look at Levy processes. Poisson and Normal are Levy processes. The combination called jump-diffusion is used widely inn finance. These are applied to modeling of discontinuous processes.

I wrote this answer to draw your attention to the rationale for using Normal distribution, so that you consider this when thinking about using other distributions. Having said this all, for differencing to work normality of errors in your unit-root process is not required. It'll work for other reasonable distributions too. However, non-normal random walk processes bring up other issues, as I indicated above. So, it's better to first consider Wiener or Levy processes for random walks, then only venture outside if these are not appropriate.

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For a random walk $Y_t=Y_{t-1}+\epsilon_t$ with iid (but not necessarily normal) $\epsilon_t\sim(0,\sigma^2)$, the first difference is

$$\Delta Y_t=\epsilon_t$$

whose mean is

$$E(\Delta Y_t)=E(\epsilon_t)=0.$$

The variance is

$$Var(\Delta Y_t)=Var(\epsilon_t)=\sigma^2$$

and all autocovariances are, using iid-ness, 0. Hence, we see that the first differenced process is stationary without appealing to normality anywhere. As @DavidShor points out, assuming $\sigma^2<\infty$ is important (although not for using a CLT here!).

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  • $\begingroup$ You may want to add a line to show that $Y_i$ is non-stationary, and show the evolution of $Y_i$ given the starting point $Y_0$. $\endgroup$ – Aksakal Feb 23 '15 at 16:46
  • $\begingroup$ Nonstationarity of $Y_t$ is nicely explained in your own post below, @Aksakal :-) : we see that $Var(Y_t)$ depends on $t$ ($x$ in the notation in that answer) and hence is not constant. Hence, $Y_t$ is not stationary. $\endgroup$ – Christoph Hanck Feb 24 '15 at 4:57

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