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In science, random processes are often described by a Langevin Equation:

where S is the variable of interest, a(S,t) and b(S,t) are certain known functions and is a Gaussian fluctuation with the following properties:

It can be shown by using the above properties that

We see that the integral in time of is X(t), the Wiener process, which itself is not differentiable. Therefore, mathematically speaking only the integral equation of the Langevin equation is defined. From this we use the following interpretation

Let G(t) be an arbitrary stochastic function of t and X(t).We shall consider the stochastic integral

where for all i up to The integral above is to be interpreted as a kind of Riemann-Stieltjes integral which can therefore be interpreted by the following limiting discretized sum

Since we can write it as with . Substituting this in the sum above we find that

This means that the outcome of our integration depends on where in each interval we evaluate G(t). This is not so if G(t) were a smooth function with no stochastic dependence. Indeed, if this were the case then taking (red) or (blue) would produce the same value for the integral as dt->0. Instead, for   we have

We will use the Ito stochastic integral where so that There is a good reason for this: in all natural cases unknown future events cannot affect the present. This means that the function G(t) is non-anticipating in that it cannot be used to predict the future increment in dX. This is of course equivalent in saying G(t) is a martingale since what we mean is exactly that

We only know what is the present value of G(t) which corresponds to that at the beginning integrating interval. For this reason it is more appropriate to use the Ito integral. It is important also to note that integrating a non-anticipative function with respect to dt or dX is itself non-anticipating.  So for G(t) =X(t) the Ito integral becomes:

Notice that   = and hence the result in Eq. [*] follows.

So what does this integral mean?

If we were to run many realizations of the Wiener processes in the interval [0,T] and compute the sum of X(s) dX(s) for each, the average sum would be zero. This is because vanish since   is statistically independent from the increment . Instead, the stochastic  has to be interpreted as how the Variance of the processes changes with time. It is therefore a mean square limit of sum of the discretized process.

We can use this to compute other integrals:

The integral of a polynomial in X(t):

(by the quadratic variation of X(t))

Note: the integral of G(t) with respect to dX and dt are two different things and, generally, have no connection.

Written by Raffaello Vardavas.

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