# linear combination of two Wiener processes

I have a question concerning the linear combination of two Wiener processes (please see http://en.wikipedia.org/wiki/Wiener_process for a definition). Let $W$ and $\tilde{W}$ be two Wiener processes and $a,b \in \mathbb{R}$.

Here is my question: Which conditions must $W, \tilde{W}, a$ and $b$ satisfy, that $Z:= aW + b\tilde{W}$ is a Wiener process?

My first guess would be: The processes and scalars must satisfy the following conditions:

(a) The increments $\{ W_{t_i} - W_{s_i} , \tilde{W}_{t_i} - \tilde{W}_{s_i} ~|~ 1\leq i \leq N \}$ of the processes must be independent for every $N \in \mathbb{N}$ and every choice of $t_i > s_i \geq t_{i-1} > s_{i-1} \geq 0$.

(b) $a^2 + b^2 = 1$

In this case we have that $Z_0 = 0$, the paths of $Z$ are a.s. continuous and the increments of $Z$ satisfy the independence condition. The increments $Z_t - Z_s$ are normally distributed with expectation zero and variance $\sigma^2 = s-t$. This follows from:

$Z_t - Z_s = a(W_t - W_s ) + b( \tilde{W}_t - \tilde{W}_s) \sim \mathcal{N}(0, (a^2+b^2)(t-s))\sim \mathcal{N}(0, t-s)$

Thus we have checked that $Z$ is a Wiener process. Unfortunately the conditions above seem to be very restircting. Can one do better? With best regards, Mat.

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You have found necessary conditions for $Z$ to be a Wiener process, thus no one can do better. Necessity means that if such conditions are not verified, then $Z$ is not a Wiener process. So, you have to have $(a,b)\in \Bbb S^1$ - here I assume that you meant $W,\tilde W$ to be independent.
You are only left to check that this condition is also sufficient for $Z = aW+b\tilde W$ to be a Wiener process. This is easy - you just need to check that $Z$ satisfies 3 properties (which you almost already did looking for the necessary conditions).