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Consider the following time-series process

$$y_{t+1}=y_{t}-x_t+\epsilon_{t+1},$$

where $\epsilon_{t+1}\sim N(0,\sigma^2)$ (and i.i.d. over time) and $x_t$ is a stationary process. I am interested in forward solutions of $y_t$. Specifically, we can iterate this forward to obtain

$$y_t=\mathbb{E}_t\sum_{j=0}^{\infty}x_{t+s}+\lim_{s\rightarrow\infty}\mathbb{E}_ty_{t+s},$$

assuming $\mathbb{E}_ty_{t+s}$ is well-defined. How is the term $\lim_{s\rightarrow\infty}\mathbb{E}_ty_{t+s}$ typically evaluated? Also any references to related material are welcome.

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    $\begingroup$ This is a random walk with a deterministic drift, a very well known process. Lookup in interweb. You can re-write it as $\Delta y_t=-x_{t-1}+\varepsilon_t$. $\endgroup$
    – Aksakal
    Oct 3, 2017 at 13:05
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    $\begingroup$ @Aksakal, if $x_t$ is a stationary process, then generally the drift is stochastic, not deterministic, which might make it harder to find. $\endgroup$ Oct 3, 2017 at 13:21
  • $\begingroup$ @RichardHardy, he didn't say what is the process for $x_t$. also OP doesn't seem to be interested in estimation $\endgroup$
    – Aksakal
    Oct 3, 2017 at 13:34
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    $\begingroup$ @Aksakal, check out the second paragraph, there it says $x_t$ is a stationary process. I did not mean estimation, I meant finding the related literature (poor formulation on my side). $\endgroup$ Oct 3, 2017 at 14:54
  • $\begingroup$ @RichardHardy, right, but what is that process? If it's AR, then the whole thing can look like Kalman filter. $\endgroup$
    – Aksakal
    Oct 3, 2017 at 15:54

1 Answer 1

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As pointed out in the comments (hat tip to Aksakal and Richard Hardy), if you condition on the sequence $\mathbf{x} \equiv \{ x_t | t \in \mathbb{Z} \}$ then the sequence of interest is a random walk with with this conditioning sequence giving drift terms. You can write your model in the form of an adjusted random walk as:

$$\Delta y_t = -x_{t-1} + \epsilon_t \quad \quad \quad \quad \quad \epsilon_t \sim \text{IID N}(0, \sigma^2).$$

Now, if $\mathbf{x}$ is even weakly stationary (you stipulate in your question that it is stationary) then you must have a common mean $\mathbb{E}(x_t) = \mu$ so you get $\mathbb{E}(\Delta y_t) = -\mu$, which then gives $\mathbb{E}(y_{s+t}) = \mathbb{E}(y_s) - t \mu$. The limiting term of interest to you is zero if $\mu=0$ and it diverges if $\mu \neq 0$.


Typically, we would not try to write a random walk in the form you are using, since the limit you are trying to use as a reference point usually diverges. Instead we would write the value of the time-series relative to a reference point at an arbitrary "starting time". Using the starting point at time $s$ we get the equation:

$$y_{s+t} = y_s + \sum_{i=1}^t \Delta y_{s+i} = y_s - \sum_{i=0}^{t-1} x_{s+i} + \sum_{i=1}^t \epsilon_{s+i},$$

which gives the conditional distribution:

$$y_{s+t} | y_s, \mathbf{x} \sim \text{N} \bigg( y_s - \sum_{i=0}^{t-1} x_{s+i}, \ t \sigma^2 \bigg).$$

If you are willing to stipulate a joint distribution for the values in the sequence $\mathbf{x}$ you can then use the law of total probability to find the conditional distribution of $y_{s+t} | y_s$.

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