# Why is the value of an adaptive stochastic process known at time t?

I am having a hard time to understand the concept of an adapted stochastic process. Using an analogy to finance, I have been told we can think of adaptiveness of a stock price process as having an access to a Bloomberg terminal and be able to check up the price of the stock at time $t$, i.e. at each point in time the price of the stock is known. I have also learned that a stochastic process is nothing but a collection of random variables and can thus be interpreted as function-valued random variable. Stochastic processes in general need not be adaptive, but as e.g. Shreve (Stochastic Calculus for Finance vol.2 page 53, 2004) notes it is often safe to assume for finance related stochastic processes to be adapted.

Now let us assume that we are dealing with an adapted stochastic process X and fix $t$. To me it seems that by doing this we will (at this arbitrary point in time) obtain a random variable $X(\omega; \text{t fixed})$ by the definition of a stochastic process. But wait a minute, the value of a random variable should not be known, right? On the contrary, it should be random!

How is this seeming puzzle reconciled? To me it is not clear how the definition of an adapted process implies that the value of $X(\omega; \text{t fixed})$ is known at time $t$. Rather, it just states that at the fixed $t$ $X(\omega; \text{t fixed})$ is $\mathcal{F}_{t}$-measurable, which is not enough. Just imagine a case of a single random variable (just one point in time) Y on $(\Omega, \mathcal{F})$ (i.e. Y is a $\mathcal{F}$-measurable function). Obviously the value of Y is not known but random.

I have found some earlier related questions (e.g. this) but these have not clarified the matter to me. Thank you in advance for the help!

• What do you mean $\mathcal {F}_t$-measurable is not enough? It is enough! – Quantuple Apr 18 '16 at 18:33
• To my knowledge random variables are always measurable functions where the measurable space we start from consists of the pair $(\Omega,\mathcal{F})$, i.e. sample space and set of events (see e.g. Def 8.1 in Jacod&Protter 2000). As such, to me it seems that measurability alone doesn't tell which realization the RV is going to take, it just allows us to measure the probabilities! In context of stochastic processes, the first answer in this thread is along my lines of thought. – vvv Apr 19 '16 at 6:44
• Okay I think I am starting to understand your question :) Bear with me, would things have changed for you if, say, the filtration $\mathcal {F}_t$ was the natural filtration of $X_t$ in your example? – Quantuple Apr 19 '16 at 7:22
• Another comment and I will shut up; In an adapted world, the past and present events and values are known and unchangeable, but the Future is random and unknown. That is all there is to it. Mathematical modeling must respect this real world fact. – noob2 Apr 19 '16 at 13:10
• Thank you for the comments! I think I started to get it now. Important insight for me was what I found here (1st page): we require that all RVs $X_t$ in the collection (stoch process) $(X_t)_{t \in T}$ are defined on the same prob space $(\Omega, \mathcal{F},P)$. This means every $X_t$ is $\mathcal{F}$-measurable by definition, but not all are $\mathcal{F}_t$-measurable. Namely, $X_s$ where $s < t$ are not $\mathcal{F}_t$-measurable, but when we get up to time t $X_t$ "becomes" $\mathcal{F}_t$-measurable and the uncertainty is resolved. Am I thinking this right? – vvv Apr 20 '16 at 7:09

## 1 Answer

I think you got it. Wrapping up:

Usually denoted by $(\mathcal {F}_t)_{t \geq 0}$, a filtration is a series of adaptive subsets of the $\sigma$-algebra $\mathcal{F}$ that keeps track of what really happened as time went by (i.e. fixed $\omega$).

Over the probability space $(\Omega, \mathcal{F}, \mathbb{P})$, a random variable $X_t$ is measurable iff $\mathbb {P}(X_t)$ can be defined in the usual sense.

If $X_t$ is in addition $\mathcal {F}_t$-measurable, over the filtered probability space $(\Omega, \mathcal {F}, (\mathcal {F}_t)_{t \geq 0}, \mathbb{P})$ we can further claim that $X_t$ is known almost surely given the information available at $t$:

$$\mathbb{P}(X_t=X(t) \vert \mathcal {F}_t)=1, \forall t \geq 0$$

where $X(s), \forall s \leq t$ figures the set of past values which the process $X_t$ has taken up to time $t$.

In financial mathematics, $\mathcal{F}_t$ usually corresponds to the natural filtration of a 'driving' process (e.g. Brownian motion $B_t$), which - as the name indicates - drives the target Markov process $X_t$ we would like to model.

One can then show that claiming that $X_t$ is $\mathcal {F}_t$-measurable is equivalent to saying that there exists a sufficiently well-behaved function $h$ such that $X_t = h (B_t)$ at time $t$.

• Got it, cheers! – vvv Apr 30 '16 at 6:13