TS Module 14 Model diagnostics
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Time series practice problems model diagnostics
*Question 14.1: Selecting the model
The sample autocorrelations for lags of 1, 2, 3, 4, and 5 from a time series of 900 observations are
25%, 24%, 24.5%, 23.5%, 22.5%
Which of the following is the most likely model (of the following five) for the time series?
A. non-stationary
B. white noise
C. stationary moving average of order 1
D. stationary autoregressive of order 1
E. stationary ARMA(1,1)
Answer 14.1: A
Statement B: If this were a white noise process, the sample autocorrelations would be close to zero, with a standard deviation of 1/30 = 3.3%. The observed autocorrelations are too high.
Statement C: A stationary moving average process of order 1 has white noise autocorrelations for lags 2 and higher. They would be close to zero with a standard deviation of 3.3%.
Statement D: A stationary autoregressive process of order 1 has geometrically declining autocorrelations for lags 2 and higher; there is no decline in the observed autocorrelations.
Statement E: An ARMA(1,1) model is similar to an AR(1) model for the autocorrelations.
Statement A: For a non-stationary process, the autocorrelations may stay high.
*Question 14.2: Diagnostics
An actuary examines a the stock price of ABC Corporation for the 252 trading days in 20X7. How might the actuary test the hypothesis that the stock prices form a process yt = 1.01 × yt-1 × ó, where ó is a lognormally distributed random variable?
A. Assume the time series is a white noise process and test if the residuals are normally distributed.
B. Assume the logarithm of the time series is a white noise process and test if the residuals are normally distributed.
C. Assume the first differences of the logarithm of the time series is a white noise process and test if the residuals are normally distributed.
D. Assume the second differences of the logarithm of the time series is a white noise process and test if the residuals are normally distributed.
E. Diagnostic tests are used if the error term is normally distributed; thay are not used if the error term in the original time series is lognormally distributed.
Answer 14.2: C
The first differences of the logarithms of the original time series would be a white noise process with a normally distributed error term.
*Question 14.3: Autocorrelation Significance
We are testing whether the observed autocorrelations of a time series are significant at the 5% level. We have 1,067 observations, and the critical t value at a 5% level of significance is 1.96. The null hypothesis is that the autocorrelations are zero.
A sample autocorrelation is significant at the 5% level of significance if its absolute value is larger than which of the following?
A. 0.011
B. 0.030
C. 0.022
D. 0.060
E. 0.090
Answer 14.3: D
If the expected autocorrelations are zero, the observed values (sample autocorrelations) are randomly distributed about zero. Their standard deviation is 1/√T = 1/1,067½ = 0.031.
1.96 standard deviations is 1.96 × 0.031 = 0.060
Jacob: If an observed sample autocorrelation is greater (in absolute value) than 0.060, do we assume that the true autocorrelation is non-zero?
Rachel: If we examine 100 sample autocorrelations, we expect 5 of them to be greater (in absolute value) than 0.060. If 5 or fewer are greater than 0.060, we assume they are all zero, and these sample autocorrelations stem from fluctuations. Even if 7 or 8 of them are greater than 0.060, we might attribute this to chance. But if the first 3 autocorrelations are greater than 0.060 (especially if they are much greater than 0.060) and none of the subsequent sample autocorrelations are greater than 0.060, we might assume a MA(3) process.
*Question 14.4: Bartlett’s test
A time series of 144 observations, t=1, 2, …, 144, has ∑ t =0 and =
324 for k = 0
108 for k = 1
96 for k = 2
84 for k = 3
72 for k = 4
60 for k = 5
48 for k = 6
36 for k = 7
24 for k = 8
12 for k = 9
We are testing the null hypotheses ρk = 0 against the alternative hypotheses ρk ≠ 0 using a separate test for each value of k. What is the lowest value of k for which we would not reject the null hypothesis at a 5% level of significance, for which the critical z-value is about 2?
A. 2
B. 4
C. 6
D. 8
E. 9
Answer 14.4: C
We find the lowest value of k for which the z-value is less than 2.
z = √T, where = .
(W / 324) × √144 = 2 ➾ W = 324 × 2 / 12 = 54.000
For k = 6, W = 48 < 54