I am trying to solve the following problem.
Let $\mathcal{F}_t$ be the filtration generated by a Brownian motion $W$, and $\theta$ be a random variable on the same sample space $\Omega$, independent of $W$. Define an initially enlarged filtration $\mathcal{G}_t = \mathcal{F}_t \vee \sigma(\theta)$.
Now let $X$ be a $\mathcal{G}_s$-meaurable random variable. My question is, do I have $\mathbb{E}[X\vert \mathcal{F}_t] = \mathbb{E}[X\vert \mathcal{F}_s]$? I tried to use $\pi$-$\lambda$ theorem but failed.
Any hint is greatly appreciated!
I presume we have $0\leq s\leq t < \infty$.
Note that trivially, $$ \mathcal{F}_s \, \perp \!\!\! \perp \, \mathcal{F}_t \, \mid \, \mathcal{F}_s. $$ Since $\theta \perp \!\!\! \perp W$ and $\mathcal{F}_s \subset \mathcal{F}_t$, it also holds that $$ \sigma(\theta) \, \perp \!\!\! \perp \, \mathcal{F}_t \, \mid \, \mathcal{F}_s. $$ Collecting results, $$ \mathcal{G}_s \, \perp \!\!\! \perp \, \mathcal{F}_t \, \mid \, \mathcal{F}_s. $$ In other words, for any $A\in\mathcal{G}_s$ it holds that $$ \mathbb{P}(A \, | \, \mathcal{F}_t) = \mathbb{P}(A \, | \, \mathcal{F}_s). $$ By classic extension (monotone class argument), $$ \mathbb{E}[X \, | \, \mathcal{F}_t] = \mathbb{E}[X \, | \, \mathcal{F}_s] $$ for any $\mathcal{G}_s$-measurable random variable $X$ with finite expectation.
If you're not too familiar with conditional independence, please let me know.