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Let $x(t)$ be a Wiener process and let $\sigma_1^2 = var(x(1))$, $\sigma_0^2 = var(x(0))$. If $x(0) = 0$, then we know that $C(t,s) = \sigma_1^2 min(t,s)$ (please see Mean and covariance of Wiener process). I want to prove that if $x(0)$ is not zero and instead a r.v then $C(t,s) = \sigma_0^2 + min(t,s)(\sigma_1^2 - \sigma_0^2)$. Any ideas to start with?

eet
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1 Answers1

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Write $x(t)=B(t)+x(0)$ where $B$ is a Wiener process started at $0$ with Var$[B(1)]=\sigma_1^2-\sigma_0^2$ and $x(0)$ is independent of $B$. Now use bilinearity of covariance to obtain the claim.

Yuval Peres
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  • Thanks for the comment. I know that Wiener process has the property of independent and stationary increments. Hence, how do you tell B is independent of $x(0)$ alone? – eet Jan 08 '21 at 17:51
  • Note all we are using is independence of $B(1)=x(1)-x(0)$ from $x(0)$. This is part of the definition. – Yuval Peres Jan 08 '21 at 18:28
  • I do not get how to use bilinearity of the covariance. Could you write it more explicitly? – eet Jan 08 '21 at 18:47
  • Expanding Cov$(x(0)+B(s),x(0)+B(t))$ you get 4 terms, two of them are zero due to independence; The others give you what you expected. – Yuval Peres Jan 08 '21 at 23:17