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$aX_{t/9}-bX_{t/4}$, knowing that $X_t$ is a Wiener Process.

This was already twice in the exams I failed, very likely it will be in todays exam in 2 hours. Can somebody help me with this one?

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Cov(Z(s),Z(t))=c.min(s,t) yields ab=0. And a=0 yields a solution while b=0 yields the other. –  Did May 22 '12 at 13:39

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Let $Z_t = a X_{t/9} - b X_{t/4}$. Note that $Z_t$ is a Gaussian process, being a sum of two Gaussian processes. It is therefore determined by its mean function and its covariance function. In order for $Z_t$ to be a Wiener process, we should establish $\mathbb{E}(Z_t) = 0$ and $\mathbb{Cov}(Z_s, Z_t) = c^2 \min(s, t)$, as pointed out by @Didier.

The mean function is easy $\mathbb{E}(Z_t) = a \mathbb{E}(X_{t/9}) - b \mathbb{E}(X_{t/4)} = a \cdot 0 - b \cdot 0 = 0$. Now turn to compute the covariance function: $$ \begin{eqnarray} \mathbb{Cov}(Z_s, Z_t) &=& a^2 \mathbb{Cov}(X_{s/9}, X_{t/9}) + b^2 \mathbb{Cov}(X_{s/4}, X_{t/4}) \\ &&- a b \mathbb{Cov} \left( X_{s/9}, X_{t/4} \right) - a b \mathbb{Cov} \left( X_{s/4}, X_{t/9} \right)\\ &=& \frac{a^2}{9} \min(s,t) + \frac{b^2}{4} \min(s,t) - a b \left( \min\left( \frac{s}{9}, \frac{t}{4}\right) + \min\left( \frac{t}{9}, \frac{s}{4}\right) \right) \end{eqnarray} $$ Since $\min(s/9,t/4) \not= \frac{1}{36} \min(s,t)$ for all $s$ and $t$, $a b$ must vanish.

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