Itô process and covariance of two Brownian motion

Question

I'm a novice in studying the stochastic different equation, and didn't know whether I have describe the question correctly.

Here is the question:

Suppose $$\begin{array}{rcl} \frac{du_t}{u_t}&=&a_udt+b_udz_t^u\\ \frac{dv_t}{v_t}&=&a_vdt+b_vdz_t^v\\ \end{array}$$ where $a_u,a_v,b_u,b_v$ are constants and $\operatorname{Cov}(dz_t^u,dz_t^v)=\rho dt$. Let $x_t=u_tv_t$ and $y_t=u_t/v_t$. Describe $dx_t/x_t$ and $dy_t/x_t$.

My thoughts are shown as below: $$\begin{array}{rcl} dx_t&=&v_t\cdot du_t+u_t\cdot dv_t+du_t\cdot dv_t\\ \Rightarrow \frac{dx_t}{x_t}&=&\frac{du_t}{u_t}+\frac{dv_t}{v_t}+\frac{du_t\cdot dv_t}{u_t\cdot v_t}&\\ &=&(a_udt+b_udz_t^u)+(a_vdt+b_vdz_t^v)+(a_udt+b_udz_t^u)\cdot(a_vdt+b_vdz_t^v)&\\ &=&(a_u+a_v)dt+b_udz_t^u+b_vdz_t^v+b_udz_t^u\cdot b_vdz_t^v&\\ \end{array}$$

Here comes the problems:

1. I don't know how to simplify the expression of $dx_t/x_t$, and

2. What is the meaning of the condition $\operatorname{Cov}(dz_t^u,dz_t^v)=\rho dt$ and how to make use of it.

Can someone kindly give me some instructions about the question?

Answer 1

See Ito's Lemma, specifically its multivariate form. This presentation from MIT actually directly addresses your above problem (and goes a bit further). See slides 7 - 9.

Answer 2

1. Why should there be a simpler expression for $dx_t/x_t$? Your calculations are correct and as long as we do not have any further information on the coefficients, we cannot expect to simplify this further.
2. The condition $\text{cov}(dz_t^u,dz_t^v) = \varrho \, dt$ is supposed to provide an information about the correlation of the Brownian motions $(z_t^u)_{t \geq 0}$ and $(z_t^v)_{t \geq 0}$. Honestly, I can stand this notation; it doesn't make sense at all to write it this way. So let me explain what the meaning of this is: For two Brownian motions which are not independent, we would like to describe the dependence of the two processes - this is done by introducing the so-called correlation coefficient $\varrho$. If $\varrho=0$, then the Brownian motions are independent, otherwise they are not. Rigorously, $\varrho$ is chosen such that $$M_t := z_t^u \cdot z_t^v - \varrho t$$ is a martingale. In particular, it holds that $$\text{cov}(z_t^u,z_t^v) = \mathbb{E}(z_t^u \cdot z_t^v) = \varrho t.$$ For Itô's lemma this means that whenever we "multiply" the stochastic differentials $dz_t^u$ and $dz_t^v$, then it equals $\varrho \, dt$, i.e. $$dz_t^u \, dz_t^v = \varrho \, dt.$$ Note that it is not obvious that this equality holds, it has to be proven.