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Consider the time series ->

$X(t) = 2 + 3t + Z(t) $

where Z(t) are gaussian white noises from $\mathcal{N}(0,1)$.

  1. is $X(t)$ stationary - why or why not?
  2. is $Y(t) = X(t) - X(t-1)$ stationary, why or why not?
  3. let $V(t)= \frac{1}{2q+1}\sum_{j=-q}^q X(t-j)$.
    What is the mean and auto-covariance function of $V(t)$.

My approach is that: I know a stationary process is on in which the statistical properties of a given series is constant, such as constant mean, auto co variance etc. I know that the expected or mean of the noise component is zero. How do i compute the expectation of X(t). How do i show the statastical properties are constant or not constant?

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If this is homework, please add the homework tag. – Dilip Sarwate Feb 6 '12 at 15:45
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your previous questions show that you now the way to ask. would you fix the formulation of the current question? – Ilya Feb 6 '12 at 15:46
For (1), ask yourself the following questions: First, what is the expectation of X(t) for a given t? Second, what does "stationary" mean? – Daniel McLaury Feb 6 '12 at 16:00
@MikeWierzbicki I wonder if the editing and LaTEXing has changed the meaning of part 3 of the question. Perhaps the sum was supposed to be $$V(t) = \frac{1}{2q+1}\sum_{j=-q}^q X(t-j)$$ to give the running average of $2q+1$ values. The way it is written now $\sum X(t)-j$ makes $V(t)$ equal to $X(t)$, right? – Dilip Sarwate Feb 6 '12 at 16:29
The part 3 of the question has been edited, it actually is X (t-j) – Probabilityman Feb 6 '12 at 17:24
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1 Answer

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The series is not stationary because there is a linear trend which means that the process average is changing with time.

Y(t) is just Z(t)-Z(t-1) . So the linear trend is remove. It now has mean 0 because Z has mean 0. Z is stationary because it is Gaussian white noise.

Expressed in terms of past Ys Y(t)= Y(t-1)-Y(t-2)+Y(t-3)-...+Z(t). It has a constant mean but to be statioanry it must have a constant variance too. Does it?

V(t) = ∑$_j$ Z(t-j)/(2q-1) +(2q+1)(2+3t)/(2q-1) - ∑$_j$ j/(2q-1) But the sum of j from -q to q is 0 so we get

V(t) = ∑$_j$ Z(t-j)/(2q-1) + (2q+1)(2+3t)/2q-1). From this you should be able to compute the mean function and autocovariance function+

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