2
$\begingroup$

To clarify, by a Simple Random Walk I mean

$$ Y_i = \begin{cases} -1 & prob = 1/2\\ 1 & prob = 1/2 \end{cases} $$ $$ X_t = \sum_{i=1}^t{Y_i} \quad \textrm{,}\,X_0 = 0 $$

and by Random Walk I mean

$$ X_t = X_{t-1} + \epsilon \quad ,\,X_0 = 0 \quad and \quad \epsilon \sim N(0, 1) $$

While I see how they differ, it is also my intuition that they follow similar patterns. Simple Random Walk is stationary, because in any interval what happens is irrelevant of the starting point and the distribution is the same as in any other interval. Also the values will not explode to $Y=X$ or $Y=-X$, because in the long run they will be contained in a cone shaped region (on the image below) with values $\sim N(0, \sqrt{t})$. This also means that the values will somewhat oscillate around $0$, which is quite stationary.

enter image description here

What is different about Random Walk, that it does not have the same properties and is not stationary?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ "Stationary" does not mean "oscillate about 0" (nor is the converse true), nor does it agree with your characterization of a "simple random walk" (usually known as a Binomial random walk), so the first thing to do is review the definition of stationarity. $\endgroup$
    – whuber ♦
    Commented Oct 8, 2021 at 13:26
  • 4
    $\begingroup$ Your notions of the meaning of stationary are incorrect. Your description of the non simple random walk is incorrect: you don't add the same $\varepsilon$ each time, you add a sequence of independent normal random variables. In other words, $X_t = X_{t-1}+\varepsilon_t$ where $\varepsilon_t$'s are iid $N(0,1)$. Note that even what you call the simple random walk can be expressed as $X_t = X_{t-1}+Y_t$. Neither of the random walks is stationary. $\endgroup$
    – Dilip Sarwate
    Commented Oct 8, 2021 at 13:32
  • $\begingroup$ @DilipSarwate I might have gone to far with that oscillation statement and I did specify the second random walk incorrectly. But both the statement about Simple Random Walk being stationary and the reasoning for it in any interval what happens is irrelevant of the starting point and the distribution is the same as in any other interval are not my ideas, but I have taken them from an MIT lecture. I have heard that there are a few definitions of stationarity, am I maybe confusing some of them, or is something else going on here? $\endgroup$
    – Mateusz
    Commented Oct 8, 2021 at 13:52
  • 5
    $\begingroup$ You may have misunderstood what the MIT lecture was saying, or the MIT lecture is dead wrong; it has been known to happen. Certainly your claimed definitions of stationarity will not be accepted on this forum. $\endgroup$
    – Dilip Sarwate
    Commented Oct 8, 2021 at 14:06

2 Answers 2

Reset to default
0
$\begingroup$

the video you link to is wrong unfortunately.

random walks have independent and stationary increments, but that doesnt mean that random walk itself is stationary

( as far as I understand the video, it mathematically defines stationarity of the increments but suggests that therefore random walk is stationary.

Both random walks you mentioned have stationary increments, but they are not themselves stationary.

Improve this answer
$\endgroup$
-1
$\begingroup$

Both of these models represent similar processes, they only differ in the distribution of the error (innovation, noise) term. If you were to plot the differences between consecutive observations in realizations of either process, you will observe independent residuals following the corresponding distribution.

The definition of stationarity you give here is not necessarily incorrect, as claimed in the comments, you most likely are (perhaps mistakingly) explaining strict stationarity. Strictly stationary processes are indeed defined as having the same joint probability distribution when shifted in time, given that the frame (interval) length is held constant. According to Brockwell and Davis' Introduction to Time Series Analysis and Forecasting (1996), a time series is said to be weakly stationary if i) its mean function is independent of time ii) its covariance function is independent of time and changes only with respect to the interval size.

Referring again to Brockwell & Davis (page 17 in my copy), it can be seen that the covariance of the series you gave above changes with respect to time, hence are non-stationary. Furthermore, both of these are unit root processes, as the roots of their corresponding characteristic equations are equal to 1.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ @Mateusz: stationarity needs more details because one can have weak stationarity or strong ( wide sense ) stationarity, so that's probably where the confusion is coming from. Also, when they talk about stationarity of the RW, they are referrring to the differenced $X_t$ process rather than the summed processes. So, stationarity is referrring to $X_t - X_{t-1}$ rather than $X_t$ $\endgroup$
    – mlofton
    Commented May 22, 2023 at 1:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.