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$\newcommand{\E}{\mathrm{\E}}$ $\newcommand{\Cov}{\mathrm{Cov}}$ $\newcommand\abs[1]{\left|#1\right|}$ Consider a Wiener process $W=(W_t,t\in\mathbb{R}_+)$. I want to calculate $\Cov(W_t,W_s),s\leq t$. The text I am reading says the following:

(...) and $W$ is a Wiener process with the $W_t$ jointly Gaussian, $W_0=0$, conditional means $\mathrm{E}(W_t\mid W_s)=W_s$ for $s\leq t$, which implies $\Cov(W_s,W_t)=\abs{t-s}$.

However, I managed to obtain the following: $$ \begin{align} \Cov(W_s,W_t)&=\Cov(W_s-W_0,W_t-W_s+W_s-W_0)\\ &=\Cov(W_s-W_0,W_s-W_0)+\Cov(W_s-W_0,W_t-W_s)\\ &=s+0 \end{align} $$

Am I doing something wrong? What is it? How can I get the quoted result?

Thanks for helping! :D

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  • $\begingroup$ Which text is this? Are you sure $W_t$ in this snippet refers to the Wiener process? The quoted covariance is not correct (consider what happens when $t=s$). $\endgroup$ Commented Aug 19, 2016 at 6:56
  • $\begingroup$ Hey @JuhoKokkala I updated the question with the remaining part of the text (it comes from a lecture pdf). I think it is clear now that it refers to the Wiener process. What do you think? $\endgroup$ Commented Aug 19, 2016 at 15:44

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There is an error in the quoted text. This becomes obvious when one considers $t=s$: \begin{equation} \mathrm{Var}(W_t) = \mathrm{Cov}(W_t,W_t) "=" |t-t| = 0, \end{equation} but the Wiener process does not have zero variance. Any other process cannot have that covariance function, either, since if the variances of $W_t$ and $W_s$ are zero, then their covariance must be zero, too.

The calculation in the question is correct if $s\leq t$ (as was assumed).

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I'm trying to discern how the book made this error. In general $\Cov(W_s,W_t)=\min(s,t)$ for arbitrary $s,t\geq 0$, and if you recall the nifty formula:

$$\min(s,t)=\frac{s+t-|s-t|}{2},$$

then

$$\Cov(W_s,W_t)=\frac{s+t-|s-t|}{2}.$$

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  • $\begingroup$ Thanks for pointing that out. The original pdf seems to be missing a big chunk of the equation in this case. $\endgroup$ Commented Aug 19, 2016 at 18:56

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