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Let's say, that $B_t$, $t\geq0$ is standard Brownian motion (Wiener process). Let's define process $$X_t=B_t+B_{t^2}\text{, }t\geq0$$

I need to find its variance, covariance, find out if it's Gaussian and its stationarity.

I started from variance:

1) $EX_t=E(B_t+B_{t^2})=EB_t+EB_{t^2}=0$ ?

2) $cov(X_s,X_t)=E(X_sX_t)=E(X_s[X_t-X_s+X_s])=E(X_sX_t-X_{s^2}+X_{s^2})=\ldots$

any ideas how to show those things? How I can interpret sum of two Brownian motions?

Thanks for suggestions

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Hints:

  • You have correctly computed $E X_t$.

  • Recall that $\mathrm{Cov}(B_t,B_s) = t \wedge s$, and that $\mathrm{Cov}(\cdot,\cdot)$ is linear in both components.

  • Your answer for the variance and distributions will depend on whether $t<1$, $t=1$, or $t>1$.

  • Recall that Brownian motion has multivariate normal finite distributions. In particular $(B_t,B_{t^2})$ is bivariate normal (you compute the mean and covariance matrix (easy))

  • If $N$ is multivariate normal and $A$ is an affine transformation (like $(x,y) \to x + y$ cough cough)), then $AN$ is multivariate normal.

  • The sum of (jointly) stationary processes is stationary, can you show this?

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  • $\begingroup$ Got it! Thanks : ) $\endgroup$ – atomoutside Jun 13 '13 at 19:03
  • $\begingroup$ "The sum of (jointly) stationary processes is stationary" is true but slightly misleading since it does not apply to the present case. $\endgroup$ – Did Jun 14 '13 at 6:00

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