How could one think of taking the Ito differntial of an expectation or measure theortic integral?
In particular, I know how an Ito process $D_t$ evolves ($dD_t = \mu dt + \sigma dW_t$) and that it equals an expectation of some variable, $x$, at time $t$. I would like to take the Ito differential and equate coefficients.
So, how could one think about this ito differential:
$ d\mathbb{E}^{G_t}[x] = d\left(\int x dG_t(x) \right) $
I hope this is right! If anyone happens along later, please let me know if I'm on the right track.
References:
Da Prato, et. al., "Stochastic Equations in Infinite Dimensions" Carmona, et. al., "Interest Rate Models: an Infinite Dimensional Stochastic Analysis Perspective"
The above expression can be rewritten (assuming $G_t(x)$ admits a density) as $$ \begin{align*} d\mathbb{E}^{G_t}[x] &= d\left(\int x dG_t(x) \right) \\ &= d\left(\int x g_t(x)d(x) \right) \\ &= dF(g_t)\\ \end{align*} \\ $$ where $F(g_t)$ is a functional from the space of probability measures to $\mathbb{R}$.
Suppose that the dynamics of the stochastic distribution $g_t(x)$ are known, such that $$ \begin{align*} g_t(x) = g_0(x) + \int_0^t \phi_s(x)ds + \int_0^t \Phi_s(x) dW_s(x) \end{align*} $$
The following definition of an Ito's lemma on such functionals can be used to calculate the differential (see theorem 4.32 from Da Prato. Also, I maintain their notation here because it is more concise.): $$ \begin{align*} F(t, X(t)) &= F(0, X(0)) + \int_0^t \langle F_x(s, X(s), \Phi(s)dW(s) \rangle +\\ &\int_0^t \left \lbrace F_t(s, X(s)) + \langle F_x(s, X(s), \phi(s) + \frac{1}{2}\text{Tr}[F_{xx}(s, X(s))(\Phi(s)Q^{1/2})(\Phi(s)Q^{1/2})^*]\rangle \right \rbrace ds \end{align*} $$ What's left to do in my answer is define the derivative of $F(g_t)$, but this should simply be a functional derivative.