## Solutions to Exam 1 Matching Problems for STAT 626

### H. Joseph Newton Professor of Statistics and Dean of Science Texas A&M University

Time series data sets come in a variety of types in terms of their frequency content:
• Low frequency content only.
• High frequency content only.
• Both high and low frequency content.
• Cycles with possibly some low or high (or both) frequency content too.
• None of the above, that is, white noise or close to white noise.

These are easy to see in the log spectra and harder to see in the data sets themselves, although I try to include only data sets where the frequency content type can be seen.

As a general strategy for doing the matching problems, I suggest looking at the data sets for the most obvious non-cyclic cases and eventually finishing by looking at all the cyclic cases and ordering them according to the number of cycles and then ordering the log spectra in terms of the frequency of the peak, remembering that the highest number of cycles corresponds to the highest frequency.

#### The 1996 Exam 1

1. Series A has both high and low so it must be 7.
2. F and H are both low but F has less high so it must be 4 and so H is 10. is 4 which makes F 10.
3. B is white looking so it's 6.
4. E and G are high so they must be 3 and 5 but G looks higher so it is 3 and E then is 5.
5. The cyclic series from longest to shortest cycles are C, J, D, I and the spectra from lowest to highest frequency are 1, 8, 2, 9. I do these by counting cycles and looking at the frequency axis on the spectra to read off the peak frequency.

Thus we have A-7, B-6, C-1, D-2, E-5, F-4, G-3, H-10, I-9, J-8.

#### The 1997 Exam 1

1. A and C are both high and low so they must be 1 and 7 but 7 has stronger low and weaker high so it must be A which means C is 1.
2. J is the only high so it must be 10.
3. B is the only white looking series so it must be 4.
4. G is the only low so it is 9.
5. The cyclic ones from fewest to most cycles are I-3, F-14, E-19, H-22, and D-26, so they must be 3 (.04), 2 (.15), 5 (.20), 6 (.23), and 8 (.27).

Thus we have A-7, B-4, C-1, D-8, E-5, F-2, G-9, H-6, I-3, J-10.

#### The 1998 Exam 1

1. A is the only high and low so it is 4.
2. B and I are low so they are 1 and 6, but B has more high so B is 6 and I is 1.
3. E is the only white looking series so it must be 3.
4. D is the only high so it is 8.
5. The cyclic series from fewest to most cycles are H, J, C, F, and G, while the spectra from lowest frequency to highest are 5, 7, 2, 9, 10.

Thus we have A-4, B-6, C-2, D-8, E-3, F-9, G-10, H-5, I-1,J-7.

#### The 2000 Exam 1

1. D and G are both low but D is a stronger low so D and G are 9 and 1.
2. A is the only high so it is 7.
3. F is both low and high so it is 2.
4. H is the only non-cyclic series so it must be 5.
5. The cyclic series from fewest to most cycles are J, I, E, B, C ( approximately 12, 16, 18, 20, 26 cycles), so they are 3, 4, 8, 6, and 10, respectively.

Thus we have A-7, B-6, C-10, D-9, E-8, F-2, G-1, H-5, I-4,J-3.