Free SOA Exam SRM (Statistics for Risk Modeling) Time Series Models Practice Questions

First differencing computes $\Delta Y_t = Y_t - Y_{t-1}$, the change between consecutive observations. This is commonly used to remove trends and achieve stationarity.

(D) is incorrect because taking ratios is a different transformation (related to returns in finance, or log differencing).
(C) is incorrect because the cumulative sum is the inverse of differencing, not differencing itself.
(E) is incorrect because this is a smoothing operation (moving average), not differencing.
(A) is incorrect because squaring values is a power transformation, unrelated to differencing.

The process $Y_t = 5 + 0.7Y_{t-1} + \epsilon_t$ is an AR(1) process with $\phi_1 = 0.7$. Since $|\phi_1| = 0.7 < 1$, this process is stationary. Its mean is $\mu = 5/(1 - 0.7) = 16.67$ and its variance is $\sigma^2/(1 - 0.7^2)$, both constant over time.

(B) is incorrect because the term $0.5t$ is a deterministic linear trend, causing the mean to increase with time.
(C) is incorrect because a random walk is non-stationary; its variance $t\sigma^2$ grows without bound.
(A) is incorrect because $\text{Var}(Y_t) = t^2 \sigma^2$ increases with time.
(E) is incorrect because the quadratic trend $10 + 2t + 0.3t^2$ causes the mean to change over time.

Step 1 — Unconditional variance:
$$\gamma(0) = \frac{\sigma^2}{1 - \phi^2} = \frac{10}{1 - 0.81} = \frac{10}{0.19} = 52.6316$$

Step 2 — $\phi^5 = 0.9^5 = 0.59049$

Step 3 — Autocovariance at lag 5:
$$\gamma(5) = \phi^5 \gamma(0) = 0.59049 \times 52.6316 = 31.08$$

(A) $52.63$ is $\gamma(0)$ itself — the lag-0 autocovariance, not lag 5.
(B) $47.37$ computes $\gamma(1) = \phi \gamma(0) = 0.9(52.63) = 47.37$, which is the lag-1 autocovariance.
(D) $27.10$ uses $\phi^5 = 0.5(0.9)^4 = 0.5(0.6561)$, incorrectly halving the first power.
(E) $5.90$ computes $\sigma^2 \phi^5 = 10(0.59) = 5.90$ without dividing by $(1-\phi^2)$ first — confusing $\gamma(5)$ with $\sigma^2 \phi^5$.