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The difference $\Delta X_t=X_t-X_{t-1}$ can remove a unit-root stochastic trend. A seasonal difference operator instead subtracts an observation a full seasonal period earlier. Neither operation automatically removes a changing variance.

Series 1 wanders over a changing level rather than fluctuating around a stable local mean. Its sample autocorrelation function is strongly positive and decreases very slowly. This is the usual diagnostic evidence for an ordinary unit root: an autoregressive polynomial containing $1-z$, with a zero at $z=1$, and a stationary model after first differencing. The plots support an integrated model, rather than specifying the number of its remaining stationary autoregressive or moving-average terms.

Series 2 has a pronounced oscillation with period about six observations. Its sample autocorrelation alternates between large positive and negative values with little damping: approximately positive at multiples of six and negative halfway between. Together with the changing amplitude, this suggests a conjugate pair of unit-circle zeros nearThe associated real autoregressive factor is $1-2\cos (\pi /3)z+z^2=1-z+z^2$. A targeted filter $1-B+B^2$ removes this pair; the broader seasonal difference operator $1-B^6$ also contains it but introduces additional differencing factors. This is the oscillatory unit-root diagnosis from an undamped sample autocorrelation.

Thus Series 1 suggests a zero at 1; Series 2 suggests a conjugate pair on the unit circle at a seasonal frequency. These are model diagnoses, not deductions of exact roots from a finite sample. A stationary model very close to a unit root can look similar, and an undamped periodic covariance can also arise from a stationary random sinusoid. The figure does not identify exact orders or prove nonstationarity by itself.

Put $c_t=\cos (2\pi t/S)$ and assume ${\sigma }^2>0$. ThenThus $Y$ generally has a periodic mean, and $Z$ generally has a periodic variance. The edge cases allowed by $S\in {\mathbb{N}}^{\ast }$ matter: $Y$ is stationary when $S=1$, because the mean is then constant, and $Z$ is stationary for $S=1$ or $S=2$. In the latter case multiplying iid centered Gaussian white noise by $(-1{)}^t$ leaves its iid distribution unchanged. For every $S\ge 3$, $c_0^2=1$ but $c_1^2<1$, so $Z$ is nonstationary. For $S>1$, $Y$ is nonstationary because $c_t$ is not constant.

Use the seasonal difference operator ${\Delta }_S=1-B^S$. Since $c_t=c_{t-S}$,This is a stationary moving-average model of iid Gaussian white noise. Its covariance is $2{\sigma }^2$ at lag $0$, $-{\sigma }^2$ at lags $\pm S$, and zero otherwise.

For the variance-modulated process the same operation givesIts variance is $2{\sigma }^2{c}_t^2$ and its covariance at lag $S$ is $-{\sigma }^2{c}_t^2$. Therefore it remains nonstationary for $S\ge 3$, and the operation doubles the marginal variance at nonzero seasons. Seasonal differencing does not remove periodic variance: it removes a periodic mean, but the periodic variance generally remains. Practical alternatives are a periodic model or seasonal variance standardization; at seasons with $c_t=0$ the observations are deterministically zero, so division by $c_t$ is not possible there. For the already stationary special cases $S=1,2$, filtering preserves stationarity.

A periodically correlated process with period $S$ has its mean and two-time covariance unchanged when both times are shifted by $S$. For $Z$, the mean is zero andHence $Z$ is a periodically correlated process. Its covariance period can be smaller than $S$: when $S$ is even, $c_{t+S/2}^2=c_t^2$, so $S/2$ is already a covariance period.

Seasonality is systematic repetition at a calendar period. It can appear in a deterministic mean, in a periodic variance, or in dependence across seasons. A seasonal difference operator removes a fixed periodic mean but seasonal differencing does not remove periodic variance. A periodic-looking path alone does not prove a nonstationary process.