Exam SRM — Time Series Practice Flashcards

**Answer: (C).** Weak stationarity requires a constant mean, a constant variance, and an autocovariance that depends only on the lag $k$, not on $t$. White noise satisfies all three: $E[\epsilon_t]=0$, $\text{Var}(\epsilon_t)=\sigma^2$, and $\text{Cov}(\epsilon_t,\epsilon_s)=0$ for $t\neq s$. The others fail: a random walk has $\text{Var}(Y_t)=t\sigma^2$ (grows with $t$); a deterministic trend has $E[Y_t]=3+0.5t$ (mean changes); a random walk with drift fails on both mean and variance; a seasonal series has a mean that changes with the season. Choosing (B) reflects the common error of thinking a fixed, predictable trend is "stable" enough to be stationary — but a changing mean alone breaks stationarity.

**Answer: (C).** White noise requires zero autocorrelation: $\text{Cov}(\epsilon_t,\epsilon_s)=0$ for all $t\neq s$. A non-zero lag-1 autocovariance of $0.4\sigma^2$ directly violates this — the series carries serial dependence, so it is not white noise. (A) and (B) are properties white noise should *have*, so satisfying them is consistent with white noise, not evidence against it. (D) normality is not required (white noise need not be Gaussian). (E) $\rho_5=0$ is exactly what white noise predicts, so it is consistent, not contradictory. The trap is picking a property that white noise is supposed to satisfy; only (C) is an actual violation.

**Answer: (A).** Deviations $d_t=Y_t-\bar Y$: $-1,\,1,\,0,\,2,\,-2$. Numerator $\sum_{t=2}^{5} d_t d_{t-1}=(1)(-1)+(0)(1)+(2)(0)+(-2)(2)=-1+0+0-4=-5$. Denominator $\sum_{t=1}^{5} d_t^2=1+1+0+4+4=10$. $\hat\rho_1=\dfrac{-5}{10}=-0.50$. The negative value reflects the up-down zig-zag of the series. Forgetting the sign on the last cross-product $(-2)(2)$ — counting it as $+4$ instead of $-4$ — gives numerator $+3$ and the wrong $+0.30$ distractor (D).

**Answer: (B).** The three classic remedies target different defects: differencing removes a stochastic *trend* (changing mean from a unit root); detrending removes a deterministic *trend*; seasonal differencing removes a repeating *seasonal* pattern. None of those fix a non-constant *variance*. A variance that grows with the level is stabilized by a **variance-stabilizing transform** — a log or power (Box-Cox) transform — applied before any differencing. Choosing (A) or (C) confuses a variance problem with a mean/trend problem; those leave the changing spread untouched.

**Answer: (D).** The fitted trend line is $\hat Y_t=\hat\beta_0+\hat\beta_1 t=12.0+2.5t$. At $t=11$: $\hat Y_{11}=12.0+2.5(11)=12.0+27.5=39.5$. Reporting only the slope term $2.5(11)=27.5$ and forgetting the intercept gives distractor (A). Plugging in the number of periods *ahead of the last observation* ($11-8=3$) instead of the absolute time index, $12.0+2.5(3)=19.5$, is another common slip; the correct input is the absolute $t=11$.

**Answer: (A).** In a multiplicative model the seasonal forecast multiplies the trend by the seasonal index: $\hat Y=\text{trend}\times S_q=500\times 0.80=400$. Dividing instead of multiplying, $\dfrac{500}{0.80}=625$, gives distractor (E) — that operation is *deseasonalizing* an actual, not forecasting. Treating the index as additive, $500-80=420$ (reading $0.80$ as $-80$), gives (B). With an index below $1$, Q1 is a below-average quarter, so the seasonal forecast should sit below the trend of $500$.

**Answer: (B).** Additive seasonal indices must sum to $0$. The raw deviations sum to $15-6-13+8=4$, so each index is reduced by the mean $\dfrac{4}{4}=1$. Adjusted Q1 index $=15-1=14.0$. Check: $14-7-14+7=0$. Forgetting to normalize and reporting the raw $15$ gives distractor (D); adding instead of subtracting the correction, $15+1=16$, gives (E). Only the normalized $14.0$ keeps the indices summing to zero.

**Answer: (B).** To deseasonalize in a multiplicative model, divide the actual by its seasonal index: $\dfrac{312}{1.20}=260$. This strips the seasonal effect, leaving a trend-comparable figure. Multiplying instead, $312\times1.20=374.4$, gives distractor (E) — that is the wrong direction. The trend value $240$ (A-adjacent) and the raw actual $312$ (D) are not the deseasonalized figure. Note $260$ exceeds the trend $240$, indicating this month ran above its own trend even after removing seasonality.