\otimes notation question

Question

what does this notation mean:

$f_t $ is $\mathscr B_t \otimes \mathscr F_t$-measurable for every $t\in[0,T]$ and $\Bbb E \left[ \int_0^T \mid f_t \mid^2 dt \right]$ and what alternatives may be used?

Answer 1

It seems to me that $f_t$ is defined on a product space, $\mathbb{R} \times \mathbb{\Omega}$. Usually $\mathbb{R}$ is equipped with the Borel $\sigma$-algebra $\mathcal{B}$, and usually $\mathcal{F}_t$ is a filtration on $\Omega$. I don't know why the Borel $\sigma$-algebra is subset by $t$. The $\otimes$ notation is the usual $\sigma$-algebra on the product space. Thus, if you have a product space, you usually equip it with the corresponding "product"'ish $\sigma$-algebra.

My guess is, that $f_t$ is a stochastic process in time. Thus it is a function of time $t$, and it is stochastic, thus it can be considered a function from $\Omega$ into $\mathbb{R}$, \begin{align*} (t,\omega) \mapsto f_t(\omega). \end{align*} That $f_t$ is stochastic corresponds to you not knowing what $\omega$ is input into the function.

Now, $\mathbb{E}\left[ \int_0^T |f_t|^2 dt \right]$ is the expectation of integral of the stochastic process $f_t$.

The filtration $\mathcal{F}_t$ can be thought of as "all information known up until time $t$". That $f_t$ is $\mathcal{F}_t$-measurable means that at time $t$, where we know the value of $f_t$ at time $t$. The filtration $\mathcal{F}_t$ grows in time, corresponding to additional information being added to our knowledge (e.g. as time goes, we know more and more of the path of $f_t$).

I hope it somewhat answers what your want to know. It is a bit unclear from your question. Also, as said, I don't know why the Borel $\sigma$-algebra is subset by $t$.