I am new to stochastic analysis. I am working on a project that uses splitting methods to numerically solve SDEs. I need to solve a 2D system - the catch however is I need both components in the form:

$$ dv = (av+b)\;dt + cdW_t $$

Where $a$ and $b$ are function that do not depend on $v$ (they depend on other coordinates so in the splitting I treat them as constants). I know how to simulate paths from this SDE exactly.

I have an SDE with is not originally in this form so I transform it into this form using a coordinate change and Itô's lemma. However, when I perform this change, I get something like this:

$$ dv = (av+b)\;dt + c_1dW_1(t) + c_2dW_2(t) $$

because the coordinate change causes the brownian motion from the other component to appear in this equation. As a result, I need to combine these two brownian motion terms into a single one, so that my SDE is in the right form.

A formula for this appears in Rewriting sum of correlated Brownian Motions as a single brownian motion. It may be important to note that my Brownian motions are completely independent.

This formula says that:

$$ c dW(t) = c_1dW_1(t) + c_2dW_2(t) $$ where $$ c= \sqrt{c_1^2+2c_1c_2\rho+c_2^2} $$ where $\rho \; dt$ = $Cov(dW_1(t),dW_2(t))$.

This $\rho$ term is the problem for me. I need it in an explicit form for me to continue. What is explicit value of such a $\rho$. Shouldn't the covariance of two independent brownian motions be the same for all of them and therefore have some explicit value? Without an explicit value for row, my equation is not in the correct form and I cannot go forward.

I have an inkling that it might be 0 because my brownian motion terms are independent but I am not sure.

Also, is there a theorem or lemma that I can quote when applying this formula?


1 Answer 1


If the BMs $W_1$ and $W_2$ are independent then $\rho=0$, so you can replace your two terms with $\sqrt{c_1^2+c_2^2} dW$.

  • $\begingroup$ Also if you want some theorem to connect it to, it is simply because $W_{1}(t)$ and $W_{2}(t)$ are normally distributed with expected value $0$ and variance $t$. So the result follows from properties of the normal distribution. See for example en.wikipedia.org/wiki/… $\endgroup$ Jan 3, 2022 at 14:42
  • $\begingroup$ @user7924249 The fact that it's a Brownian motion is not entirely trivial, but the "if it's a BM it has to be this one" part is indeed easy to read off from properties of the normal distribution. $\endgroup$
    – Ian
    Jan 3, 2022 at 14:47

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