I'm reading Baxter's "Introduction to Financial Calculus" and he defines a stochastic process as a continuous process $X_t $ that can be written as

$X_t = X_0 + \int_0^t \sigma_s dW_s + \int_0^t \mu_s ds $, where $\sigma, \mu $ are are visible processes that satisfy $\int_0^t (\sigma_s^2 +|\mu_s|) ds $ is finite for all $t$. My question is where does this last condition come from and why must it be finite?

Secondly, once we have the above definition, he writes that two stochastic processes $X_t, \tilde{X}_t$ that have the same volatility $\sigma_t$ and drift $\mu_t$, and that agree at time zero ($X_0=\tilde{X}_0$) are identical i.e. $X_t = \tilde{X}_t$ for all $t$. I'm confused about this since, by the above definition, both processes involve a random number (one from $W_t$ and one from $\tilde{W}_t$ and unless these two Brownian motions are equal, I cannot convince myself that even a single random draw from each of the $N (0,t)$ distribution would yield the same result - the probability is infinitesimally small. If we then demand this to be true for all $t#, it will be even less likely. Why then do these two stochastic processes agree? If it's because the Brownian motion parts are the same, why is this the case - I'd have thought that for two processes, this would not generally be the case.

  • $\begingroup$ Btw, the author must be assuming that the driving Brownian motion is the same, $W = \tilde{W}$. $\endgroup$ – Michael Jul 31 '17 at 9:17

The so-called "stochastic processes" are actually Ito processes---an Ito integral plus an $\omega$-by-$\omega$ Lebesgue integral. By "visible", you probably mean previsible. This is not general.

The integrability condition is just there to ensure that the two summand integrals exist, in the appropriate sense. The proper statement should be, for all $t$, $\int_0^t \sigma^2 + |\mu| ds < \infty$ almost surely.

$\int_0^t \sigma^2 ds < \infty$ almost surely ensures that the Ito integral is well-defined (therefore, as a consequence of the construction, a local martingale). $\int_0^t |\mu_s| ds$ almost surely ensures that the $\omega$-by-$\omega$ Lebesgue integral is well-defined.

"Identical" here means indistinguishable. In this context, equivalence of processes is defined up to modification (equality a.s. for all $t$) or indistinguishability (equal sample paths a.s.).


A stochastic process is an infinite collection of random variables, where each random variable is indexed by t (usually time but not necessarily). A random variable is a (deterministic) function of the experiment outcome ω (ω can be one-dimensional, finite-dimensional, or infinite-dimensional which it usually is if a stochastic process is to be defined on it). The randomness in the term "random variable" does not come from the variable itself but from the underlying random experiment. Two random variables are equal if they are equal for each outcome ω of the random experiment. For example: The random experiment of tossing 3 coins has 8 possible outcomes (ω's) - HHH,HHT,HTH.HTT,THH,THT,TTH,TTT. in this example each ω is 3-dimensional. Define one random variable X as the number of heads in ω (note that X is deterministic once ω is given). Define a second random variable Y as 3 minus the number of tails in ω. Clearly, once ω is known, the values of X and Y are equal and this can be written as X=Y. This is just an equality of functions and has nothing to do with probability. The same can be said about a stochastic process (which is just a collection of functions). Two stochastic processes X_t and Y_t are equal if, for every given outcome ω, X_t(ω)=Y_t(w) for every t. This has nothing to do with probability. To answer your question more specifically - you don't draw twice. You only draw once and define two processes on the result of this one draw (this is why they can be equal).

  • $\begingroup$ I suspect the sense of "identical" in the question is not the same as your sense of "equal" in this answer. In particular, two processes on completely different probability spaces can be "identical" but they obviously wouldn't be equal as functions. $\endgroup$ – whuber Jul 12 '17 at 18:15
  • $\begingroup$ Then how would you define their being "identical"? $\endgroup$ – Zahava Kor Jul 12 '17 at 18:31
  • $\begingroup$ One way to characterize "identical" would be that for all finite subsets of times $\{t_1,\ldots, t_n\}$, the variables $(X_{t_1},\ldots,X_{t_n})$ and $(\tilde{X}_{t_1},\ldots,\tilde{X}_{t_n})$ are identically distributed. $\endgroup$ – whuber Jul 12 '17 at 18:47
  • $\begingroup$ This becomes a philosophical question - is "Identically distributed" the same as "identical"? In my opinion, "Identical" resembles "Equal" more than it does "Identically distributed". $\endgroup$ – Zahava Kor Jul 12 '17 at 21:44
  • $\begingroup$ Perhaps. But please observe that your definition of "equal" is mathematical, not philosophical, and that it is too restrictive to be useful for this particular question. I agree that this issue might be a distraction, though, because the interesting and harder question in this thread is about why the finiteness of the stochastic integral is needed as an assumption. $\endgroup$ – whuber Jul 12 '17 at 22:10

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