Exam Pattern: Difference Operators and Seasonality

Problems involving the backshift and difference operators applied to processes with trend and/or seasonal components.

Problem Type 1: Smallest $d$ to Eliminate Seasonality

Setup: $\{X_t\}$ has seasonal components with periods $d_1, \dots, d_k$. Find the smallest $d$ such that $\nabla_d X_t$ has no seasonal components.

Answer: $d = \text{lcm}(d_1, \dots, d_k)$.

Why: $\nabla_d$ eliminates a seasonal component of period $p$ iff $p \mid d$ (since $s_t = s_{t+p}$ implies $\nabla_d s_t = s_t - s_{t-d} = 0$ iff $d$ is a multiple of $p$). The LCM is the smallest $d$ divisible by all periods.

Example: Periods 12 and 28. $\text{lcm}(12, 28) = 84$.

Problem Type 2: $\nabla_{d_1 d_2}$ vs $\nabla_{d_1 + d_2}$

Given seasonal periods $d_1, d_2$ with $d_1 \neq d_2$:

(i) $\nabla_{d_1 d_2} X_t$ is a constant?

  • TRUE. Since $d_1 \mid d_1 d_2$ and $d_2 \mid d_1 d_2$, both seasonal components are eliminated. If there’s no trend, the result is the random component applied with the operator.
  • If $X_t = s_t^{(d_1)} + s_t^{(d_2)}$ (pure seasonal, no trend), then $\nabla_{d_1 d_2} X_t = 0$ (constant).

(ii) $\nabla_{d_1 + d_2} X_t$ is a constant?

  • FALSE in general. $d_1 \nmid (d_1 + d_2)$ and $d_2 \nmid (d_1 + d_2)$ in general.
  • Counterexample: $d_1 = 2, d_2 = 3$. Then $d_1 + d_2 = 5$. Neither 2 nor 3 divides 5, so neither seasonal component is eliminated.

Problem Type 3: Design a Causal Filter

Setup: $\{X_t\}$ has seasonal periods $d_1, d_2$ and polynomial trend of degree $k$. Design a causal filter with few nonzero coefficients that eliminates both seasonality and trend.

Method: Compose operators:

  1. $\nabla^{k+1}$ eliminates degree-$k$ polynomial trend (need $k+1$ applications of $(1-B)$)
  2. $\nabla_{d_1}$ and $\nabla_{d_2}$ eliminate seasonality

Combined: $\nabla^{k+1} \nabla_{d_1} \nabla_{d_2}$ applied to $X_t$.

Express in $B$-operator form: $(1-B)^{k+1}(1-B^{d_1})(1-B^{d_2})$

Example: Periods 5 and 7, quadratic trend ($k=2$).

  • Filter: $(1-B)^3 (1-B^5)(1-B^7)$
  • This is causal (all powers of $B$ are non-negative)

Key Formulas

OperatorEliminates
$\nabla = (1-B)$Constant trend (degree 0 becomes 0)
$\nabla^p = (1-B)^p$Polynomial trend of degree $< p$
$\nabla_d = (1-B^d)$Seasonal component of period $d$
$\nabla^2 = 1 - 2B + B^2$Check: $\nabla^2 X_t = X_t - 2X_{t-1} + X_{t-2}$