TS Module 7 Stationary mixed processes
(The attached PDF file has better formatting.)
Time Series Practice Problems Stationarity
We model stationary time series by the techniques in the textbook. Most real world time series are not stationary, so take differences or logarithms to make them stationary.
Your student project make regress one time series on another or de-trend a time series to make it stationary. For example, a time series of average claim severity is not stationary.
Average claim severity in real dollars (detrended) may be stationary.
The residuals of a regression of average claim severity on inflation may be stationary.
*Question 1.2: Stationary
China’s economy is expanding rapidly. Inflation is 5.3% per annum and the number of cars is increasing 4.5% per annum. Chinese car makers are designing safer autos, and accident frequency is declining 20% per annum. Claim severity as a percentage of the auto’s value is not changing.
Which of the following time series is most likely to be stationary?
A. Annual auto insurance claim frequency
B. Annual auto insurance claim severity in deflated values.
C. Number of cars insured each year
D. Average car value each year
E. Total loss costs for all vehicles each year
Answer 1.2: B
A stationary process has a constant mean; it does not increase or decrease. In this scenario,
The number of cars insured increases 4.5% per annum.
The average car value and the claim severity increase 5.3% a year with inflation. In real (deflated) values, average claim severity may not be changing.
Claim frequency decreases 10% a year with car safety.
Total loss costs change by 1.045 × 1.053 / 1.200 – 1 = -8.30%
*Question 1.3: Autocorrelations
The absolute value of ñ2 is always at least as great as the absolute value of ñ–3 for all but which of the following time series?
A. MA(1)
B. MA(2)
C. AR(1)
D. ARMA(1,1)
E. AR(3)
Answer 1.3: E
For MA(1) and MA(2), ñ3 = 0.
For AR(1), AR(3), and ARMA(1,1), the value of ñk declines exponentially from the latest period with its own ARMA parameter. If the decline starts with Period 2, then ñ3 < ñ2. For a stationary process ñ3 = ñ–3.
*Question 1.4: Stationarity
Let be the first differences of the time series .
If = 1.1 yt-1 + åt,
which of the following time series is stationary?
Answer 1.4: E
Choice A: This is an increasing autoregressive process with ö > 1, which is not stationary.
Choice B: ln(yt) = ln(1.1) + ln(yt-1); this is a random walk with a drift of ln(1.1), which is not stationary.
Choice C: Δ(yt) = 1.1 yt-1 – yt-1 = 1.1 yt-1 – 1.1 yt-2 = 1.1 × Δyt-1. This time series has the same pattern as the initial time series, though the values are smaller.
Choice D: Since the time series in Choice C has the same pattern as the initial time series, the time series in Choice D has the same pattern as the time series in Choice B.
Choice E: Choice B is a random walk, so Choice E is a white noise process, which is stationary.
*Question 1.5: Stationary Process
Which of the following is stationary?
A. An autoregressive process with ö1 = ö2 = ½.
B. An autoregressive process with ö1 = ö2 = –1.
C. An autoregressive process with ö1 = ½, ö2 = , ö3 = ¼, ö4 = 1/5, ö5 = 1/6.
D. A moving average process with è1 = ½, è2 = , è3 = ¼, è4 = 1/5, è5 = 1/6.
E. A moving average process of infinite order with èj = (–1)j × ½: è1 = –½, è2 = ½, è3 = –1/2, è4 = ½, è5 = –½, …
Answer 1.5: D
Jacob: Why is Statement A not stationary? Isn’t this a simple average of the past two figures? Shouldn’t this converge to a value between the most recent two figures?
Rachel: Consider the time series Yt = ä + ½ Yt-1 + ½ Yt-2 + å
If ä > 0, the forecast for period t as t is + .
If ä < 0, the forecast for period t as t is – .
This process is like a random walk with a drift.
Jacob: What if ä = 0, so the process has no drift?
Rachel: We examine the variance of the forecasts. We consider first an autoregressive process with φ1 = 1 and φ2 = 0. Suppose the variance of the error term is ó2.
The variance of the one period ahead forecast is ó2.
The variance of the two periods ahead forecast is ó2 + ö12 × the variance of the one period ahead forecast = 2 × ó2.
The variance of the three periods ahead forecast is ó2 + ö12 × the variance of the two periods ahead forecast = 3 × ó2.
The variance of the T periods ahead forecast is ó2 + φ12 × the variance of the T-1 periods ahead forecast = T × ó2. As T ➝ ∞, the variance ➝ ∞.
Jacob: What about the scenario where φ1 = φ2 = ½?
Rachel: The process has more pieces, but the reasoning is the same.
Another way to grasp the intuition is to repeat the reasoning above for φ1 = 0 and φ2 = 1.
The only change is one additional lag, so the variance of the T periods ahead forecast is ó2 + φ12 × the variance of the T-2 periods ahead forecast.
If T is 1 or 2, the variance is ó2.
If T is 3 or 4, the variance is 2ó2.
In general, the variance is ó2 × ceiling(½T).
The scenario where φ1 = φ2 = ½ is the average of the scenarios where
the scenario where ö1 = 1 and ö2 = 0
the scenario where ö1 = 0 and ö2 = 1