Proton: Stationarity Judgment Drill
Goal: Given a process definition, rapidly determine stationarity and (if stationary) compute ACVF/ACF. Simulates MT2-style problems.
Problem 1
$\{Z_t\} \sim \text{IID}(0, \sigma^2)$. Let $X_t = Z_t^2$.
Is $\{X_t\}$ weakly stationary?
Answer 1
Check 1 — $E(X_t) = E(Z_t^2) = \sigma^2$. Constant. ✓
Check 2 — $\text{Var}(X_t) = E(Z_t^4) - \sigma^4$. This depends on $E(Z_t^4)$ (the kurtosis), but since $Z_t$ are identically distributed, $E(Z_t^4)$ is the same for all $t$. Constant. ✓
Check 3 — $\text{Cov}(X_t, X_{t+h}) = E(Z_t^2 Z_{t+h}^2) - \sigma^4$. For $h \neq 0$: $Z_t$ and $Z_{t+h}$ are independent → $E(Z_t^2 Z_{t+h}^2) = E(Z_t^2)E(Z_{t+h}^2) = \sigma^4$ → $\gamma(h) = 0$.
Result: Weakly stationary. In fact it’s an uncorrelated (white noise-like) process with mean $\sigma^2$.
Note: this is not a standard white noise process because $E(X_t) = \sigma^2 \neq 0$.
Problem 2
$X_t = t \cdot Z_t$ where $\{Z_t\} \sim \text{WN}(0, \sigma^2)$.
Answer 2
Check 1 — $E(X_t) = t \cdot E(Z_t) = 0$. Constant. ✓
Check 2 — $\text{Var}(X_t) = t^2 \sigma^2$. Depends on $t$. ✗
Not stationary. Stop here. No need to check covariance.
Problem 3
$X_t = (-1)^t Z_t$ where $\{Z_t\} \sim \text{IID}(0, \sigma^2)$.
Answer 3
Check 1 — $E(X_t) = (-1)^t \cdot 0 = 0$. ✓
Check 2 — $\text{Var}(X_t) = (-1)^{2t}\sigma^2 = \sigma^2$. ✓ (the $(-1)^t$ squares away)
Check 3 — For $h \neq 0$: $\text{Cov}(X_t, X_{t+h}) = (-1)^t(-1)^{t+h}E(Z_t Z_{t+h})$. Since $Z_t \perp Z_{t+h}$: $= (-1)^{2t+h} \cdot 0 = 0$. Depends only on $h$ (trivially zero). ✓
Weakly stationary. In fact this is WN$(0, \sigma^2)$ — it has the same ACVF as white noise.
Moreover, since the $Z_t$ are iid, the $X_t = (-1)^t Z_t$ are also independent (each is a deterministic function of a different independent $Z_t$), so this is actually IID noise (with alternating sign, but identically distributed because $(-1)^t Z_t$ and $Z_t$ have the same symmetric distribution if $Z_t$ is symmetric; if not symmetric, the marginal distributions alternate and it’s not identically distributed → not strictly stationary).
Subtle trap: If $Z_t$ has a non-symmetric distribution, then $X_1 = -Z_1$ and $X_2 = Z_2$ have different distributions. In that case $\{X_t\}$ is weakly stationary but not strictly stationary. This illustrates the Strict vs Weak gap.
Problem 4
$X_t = 0.5X_{t-1} + Z_t + 2$, $\{Z_t\} \sim \text{WN}(0, 1)$.
Answer 4
Rewrite as $X_t - \mu = 0.5(X_{t-1} - \mu) + Z_t$ where $\mu$ satisfies $\mu = 0.5\mu + 2$ → $\mu = 4$.
Let $Y_t = X_t - 4$. Then $Y_t = 0.5Y_{t-1} + Z_t$ — this is AR(1) with $|\phi| = 0.5 < 1$.
$Y_t$ is causal and stationary. Therefore $X_t = Y_t + 4$ is stationary with mean 4.
$\gamma_X(h) = \gamma_Y(h) = 0.5^h \cdot \frac{1}{1-0.25} = 0.5^h \cdot \frac{4}{3}$.
Key move: Don’t be confused by the constant. Absorb it into the mean.
Problem 5 (MT2 Problem 1 variant)
$X_t = 3 + X_{t-1} + Z_t$, $X_0 = 0$, $\{Z_t\} \sim \text{IID}(0, 4)$.
(a) Mean and ACVF? Stationary? (b) Is $\nabla X_t$ stationary?
Answer 5
(a) $X_t = 3t + \sum_{j=1}^t Z_j$.
$E(X_t) = 3t$ — depends on $t$. Not stationary.
$\text{Var}(X_t) = 4t$ — also depends on $t$.
$\gamma_X(t, t+h) = \text{Cov}(X_t, X_{t+h}) = \text{Cov}(\sum_{j=1}^t Z_j,\;\sum_{j=1}^{t+h}Z_j) = \sum_{j=1}^t \text{Var}(Z_j) = 4t$ — depends on $t$, not just $h$.
(b) $\nabla X_t = X_t - X_{t-1} = 3 + Z_t$.
$E(\nabla X_t) = 3$ (constant). $\text{Var}(\nabla X_t) = 4$ (constant). $\text{Cov}(\nabla X_t, \nabla X_{t+h}) = \text{Cov}(Z_t, Z_{t+h}) = 0$ for $h \neq 0$.
Stationary (in fact, it’s a constant plus iid noise).