Khi λ được thực hiện, qt là một chức năng xác định với chu kỳ 2π và giai đoạn chuyển đổi ω λ nhưng qt là một chức năng ngẫu nhiên cũ trước. Chúng ta sẽ cần hai mối quan hệ lượng giác cơ bản. | . FILTERING 69 Now let A U 0 n 29. Imagine that we take a draw from this distribu- Figure n 2Phase shift. Solid cos i Dashed cos t n 2 . tion. Let the realization be A and form the time-series qt a cos w A . Once A is realized qt is a deterministic function with periodicity 2n and phase shift A but qt is a random function ex ante. We will need the following two basic trigonometric relations. Two useful trigonometric relations. Let b and c be constants and i be the imaginary number where if -1. Then cos b c cos b cos c sin b sin c eib cos b i sin b is known as de Moivre s theorem. You can rearrange it to get 7A eib e-ib . 7A eib e-ib cos b ------------- and sin b --------- -----. 2 2i 29You only need to worry about the interval 0 n because the cosine function is symmetric about zero-cos x cos x for 0 x n 70 CHAPTER 2. SOME USEFUL TIME-SERIES METHODS Now let b vt and c A and use to represent as qt a cos vt A cos vt a cos A sin vt a sin A . Next build the time-series qt q1t q2t from the two sub-series q1t and q2t where for j 1 2 qjt cos vjt aj cos Aj sin vjt aj sin Aj and V1 V2. The result is a periodic function which is displayed on the left side of Figure . Figure For 0 IV1 vN n qt pj i qjt where qjt cos vjt aj cos Aj sin vjt aj sin Aj . Left panel N 2. Right panel N 1000 The composite process with N 2 is clearly deterministic but if you build up the analogous series with N 100 of these components as shown in the right panel of Figure the series begins to look like a random process. It turns out that any stationary random process can be arbitrarily well approximated in this fashion letting N TO. . FILTERING 71 To summarize at this point for sufficiently large number N of these underlying periodic components we can represent a time-series qt as N qt cos ujt uj sin ujt vj j i where Uj aj cos Aj and Vj aj sin Aj E u2 u2 E ujUj 0 i j E v2 Ơ2 E vivj 0 i j. Now suppose that E uiVj 0 for all i j and let N
