Molecular Note: Stationarity Verification Pipeline

Linked Atomic Concepts: Strict Stationarity, Weak Stationarity, Strict vs Weak Stationarity, Autocovariance Function (ACVF), Autocorrelation Function (ACF), IID Noise, White Noise (WN), IID vs White Noise

Scene-Setting

You’re given $X_t = 5 + 3(-1)^t Z_{t-1} Z_{t-2}$ where $\{Z_t\} \sim \text{IID}(0, \sigma^2)$. Is this stationary? If so, what are its ACVF and ACF? This is exactly the type of problem on MT2.

Concept Chain

Step 1: Compute $E(X_t)$ → check constant mean

If $E(X_t)$ contains $t$ → not stationary, stop. Otherwise proceed.

Step 2: Compute $\text{Var}(X_t)$ → check constant variance

If $\text{Var}(X_t)$ depends on $t$ → not stationary, stop. This is the step where $(-1)^t$ terms become dangerous — they can make variance time-dependent.

Step 3: Compute $\text{Cov}(X_t, X_{t+h})$ → check lag-only dependence

If the result depends on $t$ (not just $h$) → not stationary. If all three pass → weakly stationary.

Step 4: Extract ACVF and ACF

$\gamma(h) = \text{Cov}(X_t, X_{t+h})$ (the $t$-free expression from Step 3). $\rho(h) = \gamma(h)/\gamma(0)$.

Step 5: Classify the process

  • Is it white noise? (Check: $\gamma(h) = 0$ for $h \neq 0$?)
  • Is it IID? (Requires independence, not just uncorrelatedness — need distribution info)
  • Is it MA(q)? AR? Something else?

Worked Pipeline

Problem (MT2 Problem 4c): $X_t = 5 + 3(-1)^t Z_{t-1} Z_{t-2}$, $\{Z_t\} \sim \text{IID}(0, \sigma^2)$.

Step 1: $E(X_t) = 5 + 3(-1)^t E(Z_{t-1})E(Z_{t-2}) = 5 + 3(-1)^t \cdot 0 \cdot 0 = 5$. Constant. ✓

Step 2: $\text{Var}(X_t) = 9 \cdot \text{Var}((-1)^t Z_{t-1}Z_{t-2}) = 9 \cdot E(Z_{t-1}^2)E(Z_{t-2}^2) = 9\sigma^4$.

Note: $(-1)^{2t} = 1$, so the $(-1)^t$ factor squares away. Constant. ✓

Step 3: For $h \geq 1$: $\text{Cov}(X_t, X_{t+h}) = 9(-1)^t(-1)^{t+h} E(Z_{t-1}Z_{t-2}Z_{t+h-1}Z_{t+h-2})$.

Since $Z$’s are iid with zero mean, $E(Z_i Z_j Z_k Z_l) = 0$ unless indices pair up. For $h \geq 3$: no overlap → $\gamma(h) = 0$.

For $h = 1$: $E(Z_{t-1}Z_{t-2} \cdot Z_t Z_{t-1}) = E(Z_{t-1}^2)E(Z_{t-2})E(Z_t) = 0$ (unpaired $Z_t$ and $Z_{t-2}$).

For $h = 2$: $E(Z_{t-1}Z_{t-2} \cdot Z_{t+1}Z_t) = 0$ (no matching indices).

So $\gamma(h) = 0$ for all $h \neq 0$. Depends only on $h$. ✓

Result: Weakly stationary. ACVF = $\gamma(0) = 9\sigma^4$, $\gamma(h) = 0$ for $h \neq 0$. ACF = $\rho(0) = 1$, $\rho(h) = 0$ for $h \neq 0$. This is a white noise-like process (0-correlated) but it is not white noise in the standard sense (it has nonzero mean 5).

Exam Patterns

Pattern 1: “Check stationarity of $X_t = f(t, Z_t, Z_{t-1}, \ldots)$” (MT2 Problem 4)

  • What they’re testing: systematic 3-step verification, handling $(-1)^t$ or $t$-dependent coefficients
  • Key move: If there’s a $(-1)^t$ factor, track it carefully through $E(X_t)$ and $\text{Var}(X_t)$. It often squares away in variance but can survive in covariance.

Pattern 2: “Show $X_t + Y_t$ is stationary” (MT2 Problem 3)

  • Template: $X_t, Y_t$ uncorrelated stationary processes. Show sum is stationary, ACVF = sum of ACVFs.
  • Key move: $\text{Cov}(X_t + Y_t, X_{t+h} + Y_{t+h}) = \gamma_X(h) + \gamma_Y(h) + \underbrace{\text{Cov}(X_t, Y_{t+h})}_{=0} + \underbrace{\text{Cov}(Y_t, X_{t+h})}_{=0}$.

Pattern 3: “Show $X_t = a\cos(\theta t) + b\sin(\theta t)$ is stationary” (MT2 Problem 2)

  • Key move: $E(X_t) = 0$ (since $E(a) = E(b) = 0$). $\text{Cov}(X_t, X_{t+h})$: expand, use $E(a^2) = E(b^2) = 1$, $E(ab) = 0$, then apply angle addition to get $\cos(\theta h)$ — depends only on $h$.

Pattern 4: True/False stationarity statements (MT2 Problem 2 of Lec2)

  • “For any stationary process, $\text{Cov}(X_t, X_{t+h}) = 0$ for $h \neq 0$” → False (that’s white noise, not general stationary)
  • “White noise is stationary” → True (weakly)
  • “Stationary ⟹ white noise” → False
  • “IID with finite variance is special case of WN” → True
  • “Weakly stationary Gaussian ⟹ strictly stationary” → True

Edge Cases & Pitfalls

  • $(-1)^t$ coefficients: they square away in variance ($(-1)^{2t}=1$) but multiply in covariance ($(-1)^t \cdot (-1)^{t+h} = (-1)^{2t+h} = (-1)^h$) — this is a constant in $h$, so stationarity is preserved.
  • $t$-dependent coefficients that don’t cancel: $X_t = 1 + 3t + 2Z_tZ_{t-1}$ has mean $1 + 3t$ → not stationary immediately.
  • Don’t forget: mean being constant is necessary but not sufficient — must also check variance and covariance.