With what a and b this IS a Wiener Process $aX_{t/9}-bX_{t/4}$, 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?
 A: 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.
