# Stochastic process and brownian motion

I just read the following and i am having some difficulty to interpret it:

We begin our analysis in the standard Black-Scholes world consisting of a bank account process of price denoted by $B_t$, and a risky stock process $S_t$, both defined on a filtered probability space $(0, \mathbb{F}, \mathfrak{F}_t, \mathbb{P})$, with the filtration $\mathfrak{F}_t$ generated by the standard Brownian motion $W_t$. Both asset processes are therefore adapted to the filtration $\mathfrak{F}_t$, with local dynamics shown below \begin{eqnarray} \mathrm{d}S_t & = & \mu S_t \mathrm{d}t + \sigma S_t \mathrm{d}W_t,\\ \mathrm{d}B_t & = & r B_t \mathrm{d}t. \end{eqnarray}

The problem is the last sentence in connection with the two equations?

Update: How is $B_t$ adapted to the same filtration $\mathfrak{F}_t$ as $S_t$ since $B_t$ is not driven by any Brownian motion?

• Could you please detail your question: which part of the last sentence? – lehalle Feb 15 '17 at 23:48
• "adapted to the filtration with local dynamics" – user10777 Feb 16 '17 at 11:12

I am not sure I understand your question. If not - then please clarify.

1. The process for $S$ follows from the Black-Scholes assumption of the risky asset price following a constant coefficient geometric Brownian motion.

2. The process for $B$ follows from the instantaneous interest rate being constant.

3. The process being adapted to the filtration $\left( \mathfrak{F}_t \right)_{t \in \mathbb{R}_+}$ simply means that for every $t \in \mathbb{R}_+$, the random variable $S_t$ is $\mathfrak{F}_t$-measurable. Think of this as the information in $\mathfrak{F}_t$ being sufficient to determine the value of $S_t$. Further think of the natural filtration $\mathfrak{F}_t$ of the Brownian motion $W$ as containing all the information of observing $W$ up until time $t$. Given that the process $S$ has only one driving source of uncertainty $W$, it follows that knowing the path of $W$ up to time $t$ is sufficient to determine the path of $S$ up to that time. Consequently, $S$ is adapted to the natural filtration generated by $W$.

As Quantuple remarked, the non-random process $B$ is adapted to any filtration. You don't need to know anything about the path of $W$ in order to know the value of $B_t$ for any $t \in \mathbb{R}_+$. I.e., even the information in $\mathfrak{F}_0$ (the trivial sigma-algebra) is sufficient to determine $B_t$. Since $\mathfrak{F}_0 \subseteq \mathfrak{F}_t$ it follows that $B$ is adapted.

• +1. As far as $B_t$ is concerned it is a bit of a tautology to say that it is adapted since it is a deterministic process: you can measure it even without the knowledge of $\mathfrak{F}_t$ at $t$ – Quantuple Feb 16 '17 at 8:18
• Good point @Quantuple. – LocalVolatility Feb 16 '17 at 8:24
• so Ft is a function of W and St is a function of Ft? – user10777 Feb 16 '17 at 11:08
• Ft includes all possible information about W for all times up to t, and St is a function only of that information, yes. – noob2 Feb 16 '17 at 14:07
• Extending on @noob2 comment: $\mathfrak{F}_t$ is not a function but a collection of subsets of $\mathbb{F}$. I suggest you to read a gentle introduction, such as Shreve's "Stochastic Calculus for Finance II". – LocalVolatility Feb 16 '17 at 14:17

Your problem seems to lie in the fundamentals of stochastic processes, so you should probalby refresh your knowledge in this field.

Every process, also $W_t$ comes with a "natural" filtration $\mathfrak{F}_t$. It's the minimal (in a certain sense) filtration for which the process is adapted. Adapted means for a process $X_t$ that for every $t$, $X_t$ is measurable on $\mathfrak{F_t}$.

Now to your sentence in question: With the Ito calculus, under certain conditions (basically the stochastic integrals have to be well defined and itos lemma has to hold), you can also "define" a process using its dynamics (also called Ito processes). Thats what the autor does for $S_t$ and $B_t$. Also, the result is an adapted process.