Progressively measurable process.

measure-theorystochastic-calculusstochastic-differential-equations

Let $b: [0,T] \times \Omega \times \mathbb{R} \rightarrow \mathbb{R}$ with the properties:

  1. For all $ x \in \mathbb{R}$ the process $(t,\omega) \mapsto b(t,\omega, x)$ is progressively measurable.

  2. There exits $ C > 0 $ such that for all $\omega, t, x_1, x_2 $ we have
    $$ | b(t,\omega, x_1) – b(t,\omega, x_2) | \leq C | x_1 – x_2 |. $$

Let $X = (X_t)_{t \in [0,T]}$ be progressively measurable. I want to show that the process
$$ (t, \omega) \mapsto b(t,\omega, X_t(\omega) ) $$
is progressively measurable. This should follow from 1. and 2. but I was not able to prove it. I hope someone can help me out.

Edit: I actually need a more general result. Let $\mathcal{P}_2(\mathbb{R})$ be the space of measures on $\mathbb{R}$ with finite second moment. Equip this space with the Wasserstein metric $W^2.$ Then, $\mathcal{P}_2(\mathbb{R})$ is a Polish space.

Let $b: [0,T] \times \Omega \times \mathbb{R} \rightarrow \mathbb{R}$ with the properties:

  1. For all $ x \in \mathbb{R}, \mu \in \mathcal{P}_2(\mathbb{R})$ the process $(t,\omega) \mapsto b(t,\omega, x, \mu)$ is progressively measurable.

  2. There exits $ C > 0 $ such that for all $\omega, t, x_1, x_2 , \mu_1, \mu_2$ we have
    $$ | b(t,\omega, x_1, \mu_1) – b(t,\omega, x_2, \mu_2) | \leq C \big( | x_1 – x_2 | + W^2(\mu_1, \mu_2) \big). $$

Let $X = (X_t)_{t \in [0,T]}$ be progressively measurable. Let $P(X_t)$ be the push forward measure of $X_t$. I want to show that the process
$$ (t, \omega) \mapsto b(t,\omega, X_t(\omega), P(X_t) ) $$
is progressively measurable.

I tried to imitate the proof of the easier case. However, the problem seems to be that I cannot split $\mathcal{P}_2(\mathbb{R})$ into disjoint sets of "small radius". Do you have an idea how to deal with this?

Best Answer

Consider $b_n(t,\omega,x):=b(t,\omega,k/n)$ for $x\in[k/n,(k+1)/n)$ and integer $k$, where $n=1,2,\ldots$. The corresponding process $$ X^{(n)}_t(\omega):=b_n(t,\omega,X_t(\omega))=\sum_k b(t,\omega,k/n)1_{[k/n,(k+1)/n)}(X_t(\omega)) $$ is a countable sum of progressive process, and so is progressive. And because of the uniform Lipschitz condition you have imposed, you have the uniform convergence: $$ \lim_n\sup_{0\le t\le T}|X^{(n)}_t(\omega)-X_t(\omega)|=0. $$ It follows that $X$ is progressive as well.

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