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I need to express $E[\Delta x_{t}]$ and $\gamma_{\Delta X}(h)$ in terms of $E[x_{t}]$ and $\gamma_{X}(h)$. The first part of this is trivial and can easily be shown that $E[\Delta x_{t}] = \mu_{X} - \mu_{X}=0$. I am stuck on showing $\gamma_{\Delta X}(h)$ is only dependent on $h=s-t$.

Writing out the formula for autocovariance: $\gamma_{\Delta X}(h)=E[(\Delta x_{t} - \mu_{\Delta X})(\Delta x_{t+h} - \mu_{\Delta X})]$, this simplifies to $\gamma_{\Delta X}(h)=E[\Delta x_{t}\Delta x_{t+h}$] as $\mu_{\Delta X}=0$. It is here though that expanding and multiplying out does not obviously simplify into a function dependent on $\gamma_{X}(h)$.

Appreciate any help on how to approach this. Thanks.

Richard Hardy
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rst231
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  • What is your definition of "weakly stationary"? The last phrase in this post, "function dependent on," suggests this might be the crux of the matter, because that phrase doesn't characterize any kind of stationarity. BTW, because the first difference is a moving window operation, you can find a complete proof for strict stationarity at https://stats.stackexchange.com/questions/518527 -- but it applies almost without change to weak stationarity. – whuber Jun 28 '22 at 12:50
  • Weakly stationary would be $E[x_t]$ is not a function of time and $\gamma_{X}(h)=f(s-t)$. Since I assume $x_t$ is stationary, if I can express $\gamma_{\Delta X}(h)$ as a function of only $\gamma_{X}(h)$ then it must also satisfy this condition. – rst231 Jun 28 '22 at 15:19
  • Try a different strategy: directly using the definition, demonstrate $\gamma_{\Delta X}(s,t)$ depends only on $s-t.$ – whuber Jun 28 '22 at 15:25

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