Why we have this equality ?
$$\mathbb{E}[\ \mathbb{\hat{E}}(X(.)|\mathcal{F}_t)_G K(G) |\mathcal{F}_t] = \int_{\mathbb{R}}\mathbb{\hat{E}}(X(.)|\hat{\mathcal{F}}_t)_u K(u) dP_t^G(u)$$
For all $t\in[0,T]$, on a probability space $(\hat{\Omega},\hat{\mathbb{P}}, \hat{\mathcal{F}})$, where :
- $\hat{\mathcal{F}}_t=\cap_{s>t}\mathcal{F}_s\otimes\mathcal{B}(\mathbb{R})$
- $K(.)$ is $\mathcal{F}_t\otimes\mathcal{B}(\mathbb{R})$-measurable
- $\hat{E}$ the expectation w.r.t $\hat{\mathbb{P}}=\mathbb{P}\otimes\eta$ (with $\eta$ probability measure on ($\mathbb{R},\mathcal{B}(\mathbb{R})$)
- $G:\Omega\rightarrow\mathbb{R}$ a random variable
- $P_t^G$ probability of $G$ given $\mathcal{F}_t$
- X could be a process like that : $X:\Omega \times \mathbb{R}\rightarrow \mathbb{R}$
My question is notably on the expectation with $\hat{\mathcal{F}}$ and why it appears (right side of the equality) ?