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Good morning, I have a huge problem that I can't solve, I hope someone can help me. I have the following process: $X=(X_t)$ with $t ∈N$ that is a time-discrete stochastic process defined as $X_t=0.5X_{t-1}+e_t$ with $e_t$ i.i. distributed following a normal with parameters N(0,B) each t=1,2,... with B>0. I have to verify:

  • 1) if $E(X_t)$ cannot be computed each $t>=0$; I have computed $E(X_t)$ and I find that is equal to $0.5E(X_{t-1})$, but I don't know if this shows that is possibile to compute $E(X_t)$.
  • 2) X is not weakly stationary: I have to prove that $E(X_t)$ is costant, but for me is not, and that $cov(s,s+h)=cov(t,t+h)$ but I have no idea how to compute it (this point is very important, I hope you will give me some advice in computing cov)
  • 3)$Cov(X_t,X_{t-1})$ is different from B, same problem of point 2, I don't know how to compute cov
  • 4) X is strictly stationary: I have to prove that X's at time $t_i$ are distributed as X's at time $t_i+h$ Thank to everyone, your help will be very fruitful.
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    Hi and welcome to the site. It would be advisable to learn do some typesetting and layout as if the question is easy to read it increases the chance to get positive responses. You seem to already be familiar with Latex / mathjax. That is a very good start. – 2017-02-04
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    Hi, yes I know Latex, but I have to learn how to order properly the lines! – 2017-02-04
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    What is the definition of $X_0$? – 2017-02-04
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    These are all the disposable informations, I don't know the definition of $X_0$. – 2017-02-04

1 Answers 1

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Answers could be found in books on time-series analysis. The equation $X_t=0.5X_{t-1}+e_t$ is an auto-regressive model of order 1. The notation used is AR(1).

In general, if you do not specify $X_0$, then it is not possible to compute the required quantities. Rephrasing the question: Does there exist a stationary process that satisfies the recursive equation? If so, compute its variance and auto-correlation.

Define $Y_t:=\sum_{i=0}^\infty 0.5^je_{t-j}$ where $e_j\sim N(0,\sigma_{e}^2)$ iid. Then $Y$ satisfies the recursive equation $Y_t=0.5Y_{t-1}+e_t$. The process $Y$ is strictly stationary as well. Note that $E[Y_t]=0$. For the variance you have $var(Y_t)=\sum_{j=0}^\infty 0.5^{2j}\sigma_{e}^2 = \frac43\sigma_{e}^2$. $$cov(Y_t,Y_{t-1})=cov(0.5Y_{t-1}+e_t,Y_{t-1})=0.5 cov(Y_{t-1},Y_{t-1})+cov(e_t,Y_{t-1})=0.5 \frac43\sigma_{e}^2+0$$

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    Hi Galan, I understand your point. Could you explain me why $E(Y_t)=0$? If this is true I have also proved that the process is weakly stationary since its expectation is equal to zero (costant) and $cov(s,s+h)=cov(t,t+h)=0$, am I right? The main problem for me is to demonstrate if it is also strictly stationary, I haven't find any examples and I don't know how to procede. Thank you very much. – 2017-02-07
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    $Y_t$ was defined as $Y_t=\sum_{i=0}^{\infty}0.5^j\epsilon_{t-j}$. Since each $\epsilon$ are mean zero, $Y_t$ would also be mean zero. It is true that $cov(s, s+h)=cov(t,t+h)$, but why do you say they are zero? I have shown in the answer that if $h=1$ then, covariance is $\frac23\sigma_{\epsilon}^2$. Strictly stationary means that statistics are invariant with respect to shift in time. No matter at what time you consider $Y$, it is a infinite sum of iid random variables. So statistics remain invariant. – 2017-02-07
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    Ok, now it's clear. I have one last problem, in my problem I haven't any information on the definition of $X_t$, I have to consider, as you have done, as an infinite sum and, if the answer is yes, why? Unfortunately I am a beginner in stochastic processes and even the simpler thing results difficult to me. – 2017-02-07
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    If you had specified $X_0$, then you could have computed mean, variance, covariance etc. You have not specified anything about $X_0$. This problem comes from time series analysis. This infinite sum is a way of assuring stationarity. – 2017-02-07
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    As a side note: If for example, you had $X_t = \phi X_{t-1}+\epsilon_t$, with $\phi \geq 1$ this procedure would not have worked. Look for causality in AR models. – 2017-02-07
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    So, in my point 1, I have that expectation is equal to $0.5X_{ t-1}$ I must conclude that I can't compute expectation? – 2017-02-07
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    If you do not specify $X_0$, then you can not compute. However, if you consider the stationary version defined in terms of infinite sum of mean zero variables, then you obviously have zero for $E[X_t]$. – 2017-02-08