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

Time series modeling on SOA Exam SRM covers ARIMA processes, stationarity testing, autocorrelation analysis, forecasting methods, and model identification and selection criteria (AIC, BIC).

121 Questions
72 Easy
30 Medium
19 Hard
2026 Syllabus

Sample Questions

Question 1 Easy
Differencing a time series means computing:
Solution
C is correct.

First differencing computes ΔYt=Yt−Yt−1\Delta Y_t = Y_t - Y_{t-1}, the change between consecutive observations. This is commonly used to remove trends and achieve stationarity.
Question 2 Medium
Which of the following time series is stationary?
Solution
D is correct.

The process Yt=5+0.7Yt−1+ϵtY_t = 5 + 0.7Y_{t-1} + \epsilon_t is an AR(1) process with ϕ1=0.7\phi_1 = 0.7. Since ∣ϕ1∣=0.7<1|\phi_1| = 0.7 < 1, this process is stationary. Its mean is μ=5/(1−0.7)=16.67\mu = 5/(1 - 0.7) = 16.67 and its variance is σ2/(1−0.72)\sigma^2/(1 - 0.7^2), both constant over time.
Question 3 Hard
An AR(1) process has ϕ=0.9\phi = 0.9 and σϵ2=10\sigma^2_{\epsilon} = 10. Compute the unconditional variance γ(0)\gamma(0) and the autocovariance at lag 5, γ(5)=ϕ5γ(0)\gamma(5) = \phi^5 \gamma(0). What is γ(5)\gamma(5)?
Solution
C is correct.

Step 1 — Unconditional variance:
γ(0)=σ21−ϕ2=101−0.81=100.19=52.6316\gamma(0) = \frac{\sigma^2}{1 - \phi^2} = \frac{10}{1 - 0.81} = \frac{10}{0.19} = 52.6316

Step 2 — ϕ5=0.95=0.59049\phi^5 = 0.9^5 = 0.59049

Step 3 — Autocovariance at lag 5:
γ(5)=ϕ5γ(0)=0.59049×52.6316=31.08\gamma(5) = \phi^5 \gamma(0) = 0.59049 \times 52.6316 = 31.08
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