One of the properties of the Wiener process is that its increment is stationary. Which means for time $t$ and $s$ with $0\leq s\leq t$

$$ W_t - W_s \sim W_{t-s} \sim N(0,t-s) $$

This basically says that the difference/Wiener increment is distributed normally with mean $0$ and variance $t-s$. Suppose now I have another sequence of stochastic variables $Y = \{Y_1,Y_2,...,Y_n\}$, what are the rules/conditions that I need to impose, so that $Y$ is also normally distributed and has stationary increments? Namely, $Y$ also satisfies

$$ Y_{t}-Y_{s}\sim Y_{t-s}\sim N(0,t-s) $$

? Naively, I thought that one could simply define such a sequence of stochastic variables $Y$ which maintains such a property. Am I wrong or is there a way to flesh this out rigorously?


1 Answer 1


The sufficient condition is the next one.

Consider $\xi_i \sim N(0, \sigma_1^2)$, $1 \le i \le n-1$, $\eta \sim N(0, \sigma_2^2)$. Suppose that $(\xi_1, \ldots, \xi_{n-1}, \eta)$ has normal distribution with independent components. Hence $(Y_1, \ldots, Y_n) = (\eta, \eta+\xi_1, \eta+\xi_1+\xi_2, \ldots, \eta+\xi_1 + \ldots + \xi_{n-1})$ has normal distribution and stationary increments.

  • $\begingroup$ Thank you for your response. Can you elaborate further on the last step? In particular, how did you show that $Y$ is a sequence of $\eta + \xi$? Also, what should the new distribution of $Y$? Should it be $Y \sim N(0,\eta^{2})$? How can I show this rigorously? $\endgroup$
    – kowalski
    Commented Oct 31, 2022 at 16:11
  • $\begingroup$ @kowalski I suppose that $$Y = (Y_1, \ldots, Y_n) = (\eta, \eta+\xi_1, \eta+\xi_1+\xi_2, \ldots, \eta+\xi_1 + \ldots + \xi_{n-1}).$$ I speak about sufficient condition, so I just put by definition that $Y$ has this form. The distribution of $Y$ is normal distribution with zero mean and some covariance matrix. But $Y$ can't be $N(0, \eta^2)$ at least because the covariance matrix can't be random. The covariance matrix may be found easily. $\endgroup$ Commented Nov 1, 2022 at 11:08

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