Find the brownian motion associated to a linear combination of dependant brownian motions

I have $$N$$ correlated standard one-dimensional Brownian motions $$W_1,\ldots,W_N$$ with correlation matrix $$\rho$$ and I consider the process $$Z_t \equiv \sum_{i=1}^N \mu_i (t) W_t$$ where the $$\mu_i$$ are deterministic functions that are at least piecewise linear. How could I find a function $$\mu$$ such the process $$Y_t$$ defined by $$\mu(t) Y_t = Z_t$$ would be a standard Brownian motion ?

• Should the first occurrence of $\mu_i$ be $\lambda_i$ ?. I am confused as to the role of $\lambda_i$, it is described but never used again. – Alex C Apr 15 '19 at 2:38
• Do you know Levy's characterization? – Gordon Apr 15 '19 at 15:02
• I corrected OP's typo – Olorin Apr 15 '19 at 15:09
• @ujsgeyrr1f0d0d0r0h1h0j0j_juj Thank you – 11house Apr 15 '19 at 16:06
• @Gordon A looked about it but I don't see the connection with my problem – 11house Apr 15 '19 at 16:07

Let's calculate the quadratic covariation : $$d \langle Z,Z \rangle_t = \left(\sum_{i=1}^N \mu_i(t)^2 + 2 \sum_{1\leq i < j \leq N} \mu_i (t) \mu_j (t) \rho_{i,j}\right) dt$$ where $$\rho_{i,j}$$ is the instantaneous correlation between $$W_{i}$$ and $$W_{j}$$. So if we define $$\alpha (t) \equiv \sum_{i=1}^N \mu_i(t)^2 + 2 \sum_{1\leq i < j \leq N} \mu_i (t) \mu_j (t) \rho_{i,j}$$ and $$W_t \equiv \frac{1}{\sqrt{\alpha(t)}} Z_t$$ we see that $$d \langle W,W \rangle_t = dt.$$ As $$W$$ in $$0$$ is almost surely equal to zero as the $$W_i$$'s are, the only last hyposthesis to check to be able to apply Lévy's characterization (of continuous Brownian motion) theorem is that $$W$$ is a continuous martingale. It is obviously a martingale, and its continuity depends on the $$\mu_i$$'s which are piecewise linear but not necessarily continuous.
• The martingality of $W_t$ can be a problem, as the weights must be in a particular form for this to hold. – Gordon Apr 17 '19 at 11:43
In general, you are not able to find such $$\mu(t)$$ such that $$Y=\{Y_t, t \ge 0\}$$, defined by $$\mu(t) Y_t = Z_t$$, is a martingale, unless all $$\mu_i(t)$$ are scalar multiples of the same positive function.
In fact, note that \begin{align*} Y_t &=\frac{1}{\mu(t)}Z_t\\ &\equiv \sum_{i=1}^N \hat{\mu}_i(t) W_i(t) \end{align*} For $$0\le s \le t$$, \begin{align*} E\left(Y_t \,|\,\mathcal{F}_s \right) &=E\left(\sum_{i=1}^N \hat{\mu}_i(t) W_i(t) \,|\,\mathcal{F}_s \right)\\ &=E\left(\sum_{i=1}^N \hat{\mu}_i(t) \left(W_i(t)-W_i(s)\right) + \sum_{i=1}^N \hat{\mu}_i(t) W_i(s) \,|\,\mathcal{F}_s \right)\\ &=\sum_{i=1}^N \hat{\mu}_i(t) W_i(s). \end{align*} Then, for $$Y$$ to be a martingale, $$\hat{\mu}_i(t)$$, for $$i=1, \ldots, N$$, are constants. In other words, $$\mu_i(t) = \alpha_i\, \mu(t)$$, where, $$\alpha_i$$, for $$i=1, \ldots, N$$, are constants, and $$\mu(t)$$ is a positive function.