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In the following post it is trying to explain why the lag-1 autocorrelation is negative after first differencing a stationary time series.

Why does differencing time-series introduce negative autocorrelation

It argues that $\psi_j$ are decreasing and less than one. However, if we consider ARMA(1,1) with $\theta = 0.5$ and $\phi = 0.9$, above claim violates and lag-1 auto correlation is not negative. Because $\psi_j = 1.4(0.9)^{j-1}$ for $j \geq 1$.

Does anyone know whether the explanation given in this post holds for specific cases?


Edited:

If $\{X_t\}$ is a mean zero stationary time series with autocovariance function $\gamma(h)$. Let $Y_t = X_t - X_{t-1}$.

\begin{align*} Cov(Y_t, Y_{t-1}) & = Cov(X_t - X_{t-1}, X_{t-1} - X_{t-2})\\ & = \gamma(1) - \gamma(2) - \gamma(0) + \gamma(1)\\ & = -\gamma(0)[1 - 2\rho(1) + \rho(2)]. \end{align*}

Variance of $Y_t$: \begin{align*} Var(Y_t) & = 2[\gamma(0) - \gamma(1)]\\ &= 2\gamma(0)[1 - \rho(1)] \end{align*}

where $\rho(h) = \frac{\gamma(h)}{\gamma(0)}$.

Then autocorrelation function of $Y_t$ is \begin{align*} Corr(Y_t, Y_{t-1}) &= \frac{Cov(Y_t, Y_{t-1})}{\sqrt{Var(Y_t)Var(Y_{t-1})}}\\ & = -\frac{1 - 2\rho(1) + \rho(2)}{2(1 - \rho(1))}. \end{align*}

\begin{align*} -\rho(1) & < \rho(2) < \rho(1)\\ 1 - \rho(1) & < 1 + \rho(2) < 1 + \rho(1)\\ 1 - 3\rho(1) & < 1 - 2\rho(1) + \rho(2) < 1 - \rho(1)\\ \frac{1 - 3\rho(1)}{2(1 - \rho(1))} & < \frac{1 - 2\rho(1) + \rho(2)}{2(1 - \rho(1))} < \frac{1}{2}\\ -\frac{1}{2} & < -\frac{1 - 2\rho(1) + \rho(2)}{2(1 - \rho(1))} < -\frac{1 - 3\rho(1)}{2(1 - \rho(1))}\\ -\frac{1}{2} & < Corr(Y_t, Y_{t-1}) < -\frac{1 - 3\rho(1)}{2(1 - \rho(1))} \end{align*}

So, I think it is not always the case that the lag-1 autocorrelation of the differenced series is negative. But, if it is negative it can't exceed -0.5, which sometimes used to identify over-differenced series.

Any feedback about this procedure?

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