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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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    your previous questions show that you now the way to ask. would you fix the formulation of the current question?2012-02-06
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    For (1), ask yourself the following questions: First, what is the expectation of X(t) for a given t? Second, what does "stationary" mean?2012-02-06
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    @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?2012-02-06
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    The part 3 of the question has been edited, it actually is X (t-j)2012-02-06
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    @DilipSarwate Ah sorry, yes I think you're correct. It was previously written as (Xt - j) and I didn't even think about whether it should be X(t)-j or X(t-j). The latter makes much more sense.2012-02-06
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    @Probabilityman Oops, sorry about that.2012-02-06
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    Hi i am sorry i was away for a while. i know that a stationary process is one in which the statistical properties such as mean, co variance etc. will remain constant over time. I know that the expectation of zt(noise) is zero, i am not sure how i go about computing the expectation of the above series. Could someone help me proceed?2012-02-06
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    Hi could someone help me proceed with my question as i am stuck.2012-02-07
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    @Probabilityman: Instead of posting a new copy of your question, you should *edit the current one instead*. The "edit" button is on the lower left of the post. This "bumps" the question to the front page again, so you don't have to worry about whether people will see it. I have deleted your duplicate question and incorporated the change you made into this, the original version of the question.'2012-02-07
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    @ZevChonoles thank you for that. I am new to this forum so i am not aware of the rules here.2012-02-07
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    Sooo... In a nutshell, you know that $X(t)=2+3t+Z(t)$ and $E(Z(t))=0$ and you do not know how to deduce the value of $E(X(t))$. What course is this question, homework for?2012-02-07
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    I would like to answer your questions but would need your support. My first question: Do you mean *stationary* or also *quasi-stationary*?2013-07-05

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