Sum of Ito processes

Question

A stochastic processes $X_t$ is called an Ito process if $X_t$ is of the form $$X_t = X_0 + \int_0^tu_sdB_s + \int_0^t v_sds,$$ where $B_s$ is a Brownian motion, and $u_s, v_s$ are square integrable and adapted to the filtration generated by $B_s$.

Now, suppose I have 2 Ito processes driven by two independent Brownian motions, say $$X_t = X_0 + \int_0^tu_sdB_s + \int_0^t v_sds,$$ $$\tilde{X}_t = \tilde{X}_0 + \int_0^t\tilde{u}_sd\tilde{B}_s + \int_0^t \tilde{v}_sds,$$ where $B_s$ and $\tilde{B}_s$ are two independent Brownian motions.

We may regard $(X_t,\tilde{X}_t)$ as a two dimensional Ito processes, and by Ito's lemma $X_t + \tilde{X_t}$ is also an Ito process (one dimensional).

My question is, how can we write $X_t + \tilde{X}_t$ as an Ito process? Specifically, I want to produce the following form $$X_t + \tilde{X}_t = X_t + \tilde{X}_0 + \int_0^t\bar{u}_sd\bar{B}_s + \int_0^t \bar{v}_sds.$$ So that, $\bar{B}_s$ may depend on $B_s$ and $\tilde{B}_s$. It is quite clear that $\bar{v}_t = v_t + \tilde{v}_t$. My question really concern $\bar{B}_t$ and $\bar{v}_t$.

Answer 1

I am sorry I don't have the answer but I just got interested in this question too. I think "m7e" is not responding something satisfactory and I came across equations where the author seems to sum Itô processes with different driving Brownian motions. Given his calculus, it looks like the following holds:

if $$dX_t = a_tdW_t^{(a)} + b_tdW_t^{(b)} $$ then by definition $X_t-X_0=\int_0^ta_sdW_s^{a} + \int_0^tb_sdW_s^{b} $, and the author writes $$dX_t = a_tdW_t^{(a)} + b_tdW_t^{(b)} = \sqrt{a_t^2+b_t^2} dW_t $$ where $W_t$ is brand new Wiener process. This means that: $$\int_0^ta_sdW_s^{a} + \int_0^tb_sdW_s^{b} = \int _0^t \sqrt{a_s^2+b_s^2}dW_s $$ and $W_t$ is only one-dimensional. I'll come back to you if I find the right theorem for this!

Answer 2

Quick add-up to build upon my previous answer, I think we can go a little further: $dX_t = a_tdW_t^{(a)} + b_tdW_t^{(b)} = \sqrt{a_t^2+b_t^2} dW_t \Leftrightarrow dW_t = \frac{a_t}{\sqrt{a_t^2+b_t^2}}dW_t^{(a)} + \frac{b_t}{\sqrt{a_t^2+b_t^2}}dW_t^{(b)}$ by the rules of Ito calculus. This mean that the brownian motion we are looking for is simply: $W_t = \int _0^t \frac{a_s}{\sqrt{a_s^2+b_s^2}}dW_s^{(a)} + \int _0^t \frac{b_s}{\sqrt{a_s^2+b_s^2}}dW_s^{(b)}$.

So proving that this is a brownian motion is the answer to the question. It is easy to prove that its expectation is $0$ and variance is $t$ if these integrals are independant random variables, which is a good start. I'm digging in the direction of Lévy's characterization theorem to really prove it is a brownian motion

Answer 3

Building off Clement's solution, we can close up the proof by using a multidimensional version of Ito's isometry formula -- for instance, the one linked here (Lemma D.1 of this, but unfortunately stated there without proof).

Using Clement's notation and assuming that $a_t$ and $b_t$ are both adapted to the joint sigma algebra generated by $\{W_t^{(a)},W_t^{(b)}\}$, we can view Clement's integral as an integral of exactly the right sort:

$$\int\frac{a_s}{\sqrt{a_s^2+b_s^2}}\;dW^{(a)}_s + \int\frac{b_s}{\sqrt{a_s^2+b_s^2}}\;dW^{(b)}_s = \int \mathbf{M}_s\cdot d\mathbf{W}_s,$$

where $\mathbf{M}_s$ is the vector $\frac{1}{\sqrt{a_s^2+b_s^2}}[a_s, b_s]$ and $\mathbf{W}$ is the 2-D Brownian cut out by $W^{(a)}$ and $W^{(b)}$. The formula given by the link above would then provide

$$\mathbb{E}\bigg\|\int_{t}^{t'} \mathbf{M}_s\cdot d\mathbf{W}_s\bigg\|^2=\mathbb{E}\int_{t}^{t'} \bigg\|\mathbf{M}_s\bigg\|^2_\text{frobenius}ds= t'-t.$$

Luckily, the martingale representation theorem saves the day here, as the only continuous local martingale satisfying the above is a Brownian motion.

Answer 4

Usually one allows the driving Brownian motion to be multidimensional, and, correspondingly, $u$ to be vector-valued. With this convention you simply set $\bar B=(B,\tilde B)$ and $\bar u=(u,\tilde u)$.