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 ?
$\begingroup$
$\endgroup$
5
-
2$\begingroup$ 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. $\endgroup$– Alex CApr 15, 2019 at 2:38
-
$\begingroup$ Do you know Levy's characterization? $\endgroup$– GordonApr 15, 2019 at 15:02
-
$\begingroup$ I corrected OP's typo $\endgroup$– OlorinApr 15, 2019 at 15:09
-
$\begingroup$ @ujsgeyrr1f0d0d0r0h1h0j0j_juj Thank you $\endgroup$– 11houseApr 15, 2019 at 16:06
-
$\begingroup$ @Gordon A looked about it but I don't see the connection with my problem $\endgroup$– 11houseApr 15, 2019 at 16:07
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
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.
-
$\begingroup$ The martingality of $W_t$ can be a problem, as the weights must be in a particular form for this to hold. $\endgroup$– GordonApr 17, 2019 at 11:43
$\begingroup$
$\endgroup$
1
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.
-
$\begingroup$ Thanks @gk94. This is a similar question to the one that you have pointed out the mistake. The answer below also has the same issue. Indeed, it is easy to make such mistakes. $\endgroup$– GordonSep 7, 2019 at 0:06