Time Series: Question on Stationarity

  • #1
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Hi,

I have a question on stationarity in time series.

I basically understand the concept, I think. However, I don't understand why the lag should affect the joint distribution.

For example, the joint distribution of <Yt, Yt+a> should be the same as the joint distribution of <Yp, Yp+a>. If now the a were to increase, the joint distributions should still be the same. Is that correct?

If Yp+a is changed to some other value, say Yp+b, then surely the joint distribution would still be the same in a stationary time series. Does anyone know if this is correct?
 
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  • #2
I'm not sure precisely what you're asking, but the joint distribution of Y_t and Y_{t+1} might be different from the joint distribution of Y_t and Y_{t+2}. Suppose you have a random walk, where at each step you can go up with probability 1/2, and down with probability 1/2. Then the joint event Y_t = 0, Y_{t+1} = 0 has zero probability, since you never stay in the same place on two consecutive steps. But the event Y_t = 0, Y_{t+2} = 0 has probability 1/2 * P(Y_t=0).

Actually, reading around it seems that you usually talk about stationarity in terms of the _difference_ Y_{t+a} - Y_t, not their joint distribution. In that case as well, Y_{t+1}-Y_t might be distributed differently from Y_{t+2}-Y_t.
 
  • #3
roadworx said:
Hi,

I have a question on stationarity in time series.

I basically understand the concept, I think. However, I don't understand why the lag should affect the joint distribution.

For example, the joint distribution of <Yt, Yt+a> should be the same as the joint distribution of <Yp, Yp+a>. If now the a were to increase, the joint distributions should still be the same. Is that correct?

If Yp+a is changed to some other value, say Yp+b, then surely the joint distribution would still be the same in a stationary time series. Does anyone know if this is correct?
In simplest terms a stationary process has a joint distribution which depends on the difference of the times, but not on the times themselves.
 

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