User:Egm6321.f10.team4.Yoon/FPE

=Markov Process= Probability density of the process at any future instant is independent from the past values. Those ensemble have given past and present values,

Conditional probability density fuction at the future instant t1 is as follows and since ,in Markov process, the past arguments are independent from the future yields

For general case, x(t) is vector Markov process with n component. To derive Fokker Plank Equation, the Chapman-Kolmogorov Equation is required

=Chapman Kolomogorov Eqn= Consider a stochastic process x(t) that is composed with scalar x. The members have fixed values at each time,

constitute a sub-ensenble and each fractions of sub-ensemble

Thus the fraction

Fraction $$\displaystyle P(x_2,t_2|x_1,t_1)\delta x_2 $$ of the sub-ensemble have values in an element dx2 at x2 at time t2

For the above holds for any process, restricting to Markov process becomes, simplifying,

The above is still applicapable in Vector process. i.e) in two dimensional vector process($$\rho, \sigma$$)

=Derivation of FPE simplest case= Simplest case of FPE is Scalar process x(t) obtained by integrating a white noise $$\xi$$(t) White noise here will appear from various postulated properties in the ensuing development First, Langevin Eqn of the process is

Taking integral on the both side of the equation,

Since $$\xi(t)$$ consist of infinitesimal impulses, it must be interpreted w/ care

Substituting the Eq in to

note 1.$$\Delta i$$ i is statiscally independent of x(t) for $$t<\tau i$$ 2.x(t) is called " a process with independent increaments"

Recall Chapman-Kolmogov eqn, with omiting $$x_1<\tau i$$ and $$t_1<\tau i$$(Since the arguments are not essential)

let $$\displaystyle t_3=t+\delta t,t_2=t$$ Substituting and rewriting,

let $$\displaystyle z=x_3-x_2 $$

Note x3 determines Z when x2 is fixed. The probability that $$\displaystyle x(t+\delta t) $$ is in an element $$\displaystyle \delta x_3 $$ at $$\displaystyle x_3 $$ (given that x(t)=x2) is the same thing as the probability that the change during the interval $$\displaystyle \delta t $$ is in an element $$\displaystyle \delta z $$ at $$\displaystyle z$$ (given that x(t)=x2). if we designate the latter probability $$\displaystyle q(z,\delta|x_2,t)\delta z$$. we have

Since x2 si fixed,

Then, we can obtain

Confining the above to the simplest case, transition probability density is independent of the value x2, t becomes,

Returning to the main eqn, and taking the subscript off yields

Which implies, the probability that the system will be in a differential element at x at time t+dt is equal to how the prob changes from differential element at x-z at time t multiplied by transition probability of the change z, summed WRT all change values Z. When dt is small the prob that a large change z occurs is small, so that transition probability density q(z,dt) has appreciable magnitude only when z is near zero. Let's expend the both side of equation with Tayler series,

if (dt)^2 is neglected,

note 1. the first integral of a probability density is unity - canceled with the first term of LHS 2. the second integral is the mean of the transition probability density 3. the third integral represents the mean-square value of the change z 4. the integral after fourth integral are small to neglect compared to the third moment.

rewriting and simplifying,

When dt->0, the ratio $$\displaystyle \frac{\bar z^2 (\delta t)}{\delta t} $$ goes to a constant value b(parameter of the white noise)

if dt is increased by the addition of a sub-interval we expect the variance of z to increase linearly. Applying the above eqn, we obtain