Why is the value of an adaptive stochastic process known at time t? - Quantitative Finance Stack Exchange most recent 30 from quant.stackexchange.com 2019-08-23T11:35:10Z https://quant.stackexchange.com/feeds/question/25481 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://quant.stackexchange.com/q/25481 3 Why is the value of an adaptive stochastic process known at time t? vvv https://quant.stackexchange.com/users/17650 2016-04-18T16:16:19Z 2016-04-30T12:18:01Z <p>I am having a hard time to understand the concept of an <a href="https://en.wikipedia.org/wiki/Adapted_process#Definition" rel="nofollow noreferrer">adapted stochastic process</a>. 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 <a href="https://en.wikipedia.org/wiki/Stochastic_process#Definition" rel="nofollow noreferrer">stochastic process</a> 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.</p> <p>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!</p> <p>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.</p> <p>I have found some earlier related questions (e.g. <a href="https://math.stackexchange.com/questions/690531/intuition-for-random-variable-being-sigma-algebra-measurable">this</a>) but these have not clarified the matter to me. Thank you in advance for the help!</p> https://quant.stackexchange.com/questions/25481/-/25600#25600 1 Answer by Quantuple for Why is the value of an adaptive stochastic process known at time t? Quantuple https://quant.stackexchange.com/users/19887 2016-04-24T12:28:47Z 2016-04-30T12:18:01Z <p>I think you got it. Wrapping up:</p> <p>Usually denoted by $(\mathcal {F}_t)_{t \geq 0}$, a filtration is a series of <em>adaptive subsets</em> of the $\sigma$-algebra $\mathcal{F}$ that keeps track of <em>what really happened as time went by</em> (i.e. fixed $\omega$).</p> <p>Over the <em>probability space</em> $(\Omega, \mathcal{F}, \mathbb{P})$, a random variable $X_t$ is measurable iff $\mathbb {P}(X_t)$ can be defined in the usual sense. </p> <p>If $X_t$ is in addition $\mathcal {F}_t$-measurable, over the <em>filtered probability space</em> $(\Omega, \mathcal {F}, (\mathcal {F}_t)_{t \geq 0}, \mathbb{P})$ we can further claim that $X_t$ is known almost surely given the information available at $t$:</p> <p>$$\mathbb{P}(X_t=X(t) \vert \mathcal {F}_t)=1, \forall t \geq 0$$</p> <p>where $X(s), \forall s \leq t$ figures the set of past values which the process $X_t$ has taken up to time $t$.</p> <hr> <p>In financial mathematics, $\mathcal{F}_t$ usually corresponds to the <em>natural filtration</em> 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. </p> <p>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 $t$. </p>