Introduction to Markov Processes

Transcription

1 Introuction to Markov Processes Connexions moule m44014 Zzis law Gustav) Meglicki, Jr Office of the VP for Information Technology Iniana University RCS: Section-2.tex,v /12/21 18:03:08 gustav Exp Copyright c 2012 by Zzis law Meglicki December 21, 2012 Abstract We introuce the concept of stochastic an Markov processes an the Chapman-Kolmogorov equations that efine the latter. We show how Markov processes can be escribe in terms of the Markov propagator ensity function an the relate propagator moment functions. We introuce the Kramers-Moyal equations an use them to iscuss the evolution of the moments. We introuce two simple example processes that illustrate how Markov processes can be efine an characterize in practical terms. Finally, we introuce homogeneous Markov processes an show how the apparatus evelope so far simplifies for this class. This moule oes not iscuss more specific Markov processes continuous, jump, birth-eath, etc. Contents 1 Preliminaries 2 2 Stochastic an Markov Processes 2 3 The Chapman-Kolmogorov Equation 4 4 Moments of Markov State Density Function 5 5 The Markov Propagator 6 6 The Kramers-Moyal Equations 9 7 Evolution of the Moments Mean, Variance an Covariance Homogeneous Markov Processes 17 1

2 2 License to Connexions by Zzis law Meglicki, Jr 1 Preliminaries 1. Probability Distributions, Connexions moule m Stochastic an Markov Processes Stochastic processes A stochastic process is a process that evolves probabilistically through various states attaine at various times from a well efine initial state x 0 at time t 0. The probability ensity function of the process epens on the states an times at which they are reache, for example, P n 1 x n, t n), x n 1, t n 1),..., x 1, t 1) x 0, t 0)) 1) is the probability that the process will reach x 1 at time t 1, then will progress to x 2 at t 2, then will go through all the subsequent configurations liste, to reach x n at time t n, given that it has starte from x 0 at time t 0. Whether the process evolves between the points continuously or in jumps we on t enquire or care) at this stage. But we will evelop the means to specify this an it will let us say a lot about such processes, which is quite surprising given how general they seem to be at first sight. A stochastic process can be further characterize by other conitional probability ensities such as P n 1 2 x n, t n), x n 1, t n 1),..., x 2, t 2) x 1, t 1), x 0, t 0)), P n 2 3 x n, t n), x n 1, t n 1),..., x 3, t 3) x 2, t 2), x 1, t 1), x 0, t 0)) an so on... What is to the right of are spacetime points at which the system has been. What is to the left are spacetime points which the system may visit, an it is the probability of the system oing so that the function escribes. Let us consier function P 1 j x j, t j) x j 1, t j 1),..., x 1, t 1), x 0, t 0)) 3) 2) Markov processes A Markov process oes not remember its history Here we state that the probability of the system reaching x j at time t j is a function of where the system has been so far, that is, it epens on the system s entire history. We say that a stochastic process is Markovian if this is not the case, that is, if the probability of the system reaching x j at t j epens only on where it s been at t j 1, but not on the previous states. A Markov process is a process that remembers only the last state reache. We express it symbolically as follows P 1 j x j, t j) x j 1, t j 1),..., x 1, t 1), x 0, t 0)) = P x j, t j) x j 1, t j 1)). 4) This assumption simplifies the escription of the corresponing stochastic processes to the point of making them tractable, which is why we are so intereste in them. Let us consier P 2 1 x 2, t 2), x 1, t 1) x 0, t 0)). Clearly, this is equal the probability of the system reaching x 1 at t 1 times

3 Creative Commons Attribution License CC-BY 3.0) 3 the probability of reaching x 2 at t 2 given that it s been at x 1 at t 1 an at x 0 at t 0, that is P 2 1 x 2, t 2), x 1, t 1) x 0, t 0)) = P x 1, t 1) x 0, t 0)) P 1 2 x 2, t 2) x 1, t 1), x 0, t 0)). 5) But if this is to be a Markov process then Consequently P 2 1 x 2, t 2) x 1, t 1), x 0, t 0)) = P x 2, t 2) x 1, t 1)). 6) P 2 1 x 2, t 2), x 1, t 1) x 0, t 0)) = P x 2, t 2) x 1, t 1)) P x 1, t 1) x 0, t 0)). 7) This extens naturally to an arbitrary number of transitions, so that P n 1 x n, t n),..., x 1, t 1) x 0, t 0)) = n P x i, t i) x i 1, t i 1)). 8) i=1 The magic function, P x i, t i) x i 1, t i 1)), is calle the Markov state ensity function. It is ifficult not to notice here similarity to quantum mechanical processes. If x i was to be a quantum particle position attaine at time t i, then the probability amplitue a complex number in general) of the particle progressing from x 0 at t 0 through x 1 at t 1, x 2 at t 2 an so on, until reaching x n at t n along this specific path woul be calculate similarly as n x i, t i) x i 1, t i 1). 9) i=1 The full probability amplitue of the particle starting from x 0 at t 0 an reaching x n at t n woul then be a sum of such proucts evaluate for all possible paths that the particle coul take: x n, t n) x 0, t 0) = n ) x i, t i) x i 1, t i 1). 10) paths The probability itself woul be evaluate by taking the square of the absolute value of x n, t n) x 0, t 0). But for a single, specific path 8) woul apply, because the square of the amplitue in this case woul be a prouct of squares of the single step amplitues, as liste by 9). We can therefore think of a progression of a quantum particle along a certain specific path as a typical Markovian process. We will also fin, eventually, that this is how the Brownian motion is escribe, another classic example of a Markov process. There is an intriguing brige between Brownian motion an quantum mechanics, pointe to by Ewar Nelson, a professor of Mathematics at Princeton University at the time, in 1966 a topic we inten to explore in further moules. i=1 Markov state ensity function Similarity to quantum mechanics A single Feynman path is a Markov process Brownian motion is a Markov process

4 4 License to Connexions by Zzis law Meglicki, Jr 3 The Chapman-Kolmogorov Equation Require properties of the Markov state ensity function The Markov state ensity function P x 2, t 2) x 1, t 1)) must satisfy certain obvious properties, namely P x 2, t 2) x 1, t 1)) 0 11) The Chapman-Kolmogorov equations an Ωx 2 ) P x 2, t 2) x 1, t 1)) x 2 = 1, 12) where Ωx 2) is the omain of x 2. These erive from the above integral representing probability. The fact that P relates to the Markov process is reflecte in the following property P 1 1 x 3, t 3) x 1, t 1)) = P 2 1 x 3, t 3), x 2, t 2) x 1, t 1)) x 2 = Ωx 2 ) Ωx 2 ) P x 3, t 3) x 2, t 2)) P x 2, t 2) x 1, t 1)) x 2. 13) This is the celebrate Chapman-Kolmogorov equation. We are going to rewrite the equation in two ways by making the following substitutions forwar x 1 x 0, t 1 t 0, x 2 x x, t 2 t, x 3 x, t 3 t + t, 14) which yiels the forwar Kolmogorov equation: backwar P x, t + t) x 0, t 0)) = P x, t + t) x x, t) ) P x x, t) x 0, t ) 0) x x 1 x 0, t 1 t 0, x 2 x 0 + x, t 2 t 0 + t, x 3 x, 15) t 3 t, 16)

5 Creative Commons Attribution License CC-BY 3.0) 5 which yiels the backwar Kolmogorov equation: P x, t) x 0, t 0)) = P x, t) x + x, t 0 + t) ) P x + x, t 0 + t) x 0, t ) 0) x Of course, we always assume that t 1 < t 2 < t 3 an that the same hols for the substitutions. Nothing stops us from inserting more intermeiate points into the progression of the observe system through the Kolmogorov steps, which leas to the compoune Chapman-Kolmogorov equation P x n, t n) x 0, t 0)) = n... P x i, t i) x i 1, t i 1)) x 1... x n 1. Ωx n 1 ) Ωx 1 ) i=1 18) 4 Moments of Markov State Density Function 17) The moments of the Markov state ensity function are compute as for any other probability ensity, namely x n = x n P x, t) x 0, t 0)) x, 19) Ωx) an similarly we o with variance an stanar eviation: varx) = σ 2 x) = x x ) 2. 20) We can also compute the first term in the covariance of the last two positions in the Markov chain: x 2x 1 = x 1x 2P x 2, t 2) x 1, t 1)) P x 1, t 1) x 0, t 0)) x 1 x 2. Ωx 2 ) Ωx 1 ) 21) Before we go any further then, let us first brush up on some properties of the moments, variances, covariances an stanar eviations: x x = 0, varx) = σ 2 x) = x x ) 2 = x 2 x 2, x 2 x 2, 22) where the equality hols for sure variable only. Some elementary properties of covariances an correlations are: covx, y) = x x )y y ) = xy x y, σx)σy) covx, y), covx, y) corrx, y) = σx)σy), 1 corrx, y) 23) Moments are compute as usual

6 6 License to Connexions by Zzis law Meglicki, Jr Also, we observe that statistically inepenent variables x an y, that is, variables such that P xyx, y) = P xx)p yy) an uncorrelate, that is covx, y) = 0. But uncorrelate variables o not have to be statistically inepenent. 5 The Markov Propagator Markov propagator ensity function The observe similarity between Markov processes an quantum mechanics shoul have prepare us for what s coming now. A Markov propagator ensity function is the Markov state ensity function that yiels probability ensity at t +, where is an infinitesimal increment, given that the system has been at x at time t, that is P x + x, t + ) x, t) ), 24) where x, unlike, is not infinitesimal for example, the state may have jumpe in the time to somwhere quite far away from x. We consier it a function of x, parametrize by the initial state x, t) an labelle by an we employ the impressively looking capital pi, Π, to enote it: Π x x, t) ) = P x + x, t + ) x, t) ). 25) We can think of it as a notational shortcut for 24). The notation here reflects that of quantum mechanics. We can rea the construct as a evice that implements the infinitesimal time avance, applie to the initial state x, t). After the application of the evice, we ask about the probability of the system rifting from x by x. Require properties of the Being itself a probability ensity, Π must satisfy Markov propagator ensity function Π x x, t) ) 0, Π x x, t) ) x = 1. 26) For = 0 we must have Π x = 0 x, t) ) = δx ). 27) The Chapman-Kolmogorov equation In this case, the time avance evice oes not avance the time at all, so the system in question must remain at x. Being the Markov state ensity function, the propagator ensity function must also satisfy the Chapman-Kolmogorov equation, which in this case is usually written in the following form Π x x, t) ) = Π x x 1 α) x + x, t + α ) ) Ωx ) Π x α x, t) ) x, 28) where α ]0, 1[, which follows irectly from the evaluation of P x + x, t + ) x, t) )

7 Creative Commons Attribution License CC-BY 3.0) 7 through an intermeiate point x + x at t + α. The first step in 28) avances the state from its origin at x, t) by the infinitesimal time machine of α, an the state eflects by x. The secon step then commences with the state at x + x an the time avance to t + α. We apply again the time machine that avances the state by the remainer of, that is, by 1 α) an the state ens up eflecte by x from the original x. But this is not the starting point of this propagator. The starting point is x + x, so its en point of x + x must be recompute in reference to x + x, which is x + x x x = x x. Since Π x x, t)) is a probability ensity of x, the latter is its ranom variable. An it is this ranom variable, here enote by the orere pair that associates the probability ensity with it, x, Π x x, t) )) The Markov propagator is the ranom variable associate with Π that we call the Markov propagator. 25) remins us that we may think of it as x = xt + ) xt), which makes it a sort of a ifferential. But it is a ranom variable ifferential that is sensitive to the changes in the probability ensity across the, not just xt). So we shoul really write this more accurately as: x, P x x ) ) = x, P xx, t + )) x, P xx, t)). 29) The Ranom Variable Transformation theorem provies us with a formula for the probability ensity of variables that result from some functional operation on other ranom variables. The formula is The Ranom Variable Transformation theorem P y y 1,..., y m) = m... P x x 1,..., x n) δ y i f i x 1,..., x n)) x 1... x n, Ωx 1 ) Ωx n) i=1 where f i x 1,..., x n) are functions that transform x i into y j we o not insist on m = n), P x is a combine multivariate probability ensity of x 1,..., x n, an P y is the probability ensity of the y 1,..., y m ranom variables prouce by the operations f i. δ is the Dirac elta function. The Chapman-Kolmogorov equation for the propagator ensity function, 28), can be rewritten to reflect the Ranom Variable Transformation formula as follows Π x x, t) ) = Π x 2 1 α) x + x 1, t + α )) Ωx 1 ) Ωx 2 ) Π x 1 α x, t)) δ x x 1 x 2 ) x1 x 2, 31) which emonstrates that the ranom variable operation that is being performe here is x 1 + x 2. It is also in this sense that we shoul unerstan x = xt + ) xt). 30)

8 = Π 2x, t) + O ) 2) Π 1x, t) + O ) 2)) 2. 38) 8 License to Connexions by Zzis law Meglicki, Jr Propagator moment functions Time erivative of the Markov process Π representing in some way a ifferential woul correspon to something like a erivative if we were to ivie it by Π x x, t)). 32) This is no longer a probability ensity, because unlike 26) it oes not integrate to 1 over the omain of x. But it is still everywhere positive an we may associate moments with it. Assuming x an x to be scalars, we efine Π nx, t) = lim x 0 Ωx ) n Π x x, t)) x. 33) ) Π nx, t) is calle the n-th propagator moment function of the Markov process escribe by 24). The above efinition oes not imply that Π nx, t) = x )n Π x x, t) ) x. 34) What it implies is that Π nx, t) = x )n Π x x, t) ) x + O ) 2). 35) Upon ivision of both sies by an upon taking the limit 0 the small term of the secon an higher orers in, O ) 2), isappears. We ve been careful to refer to Π x x, t)) / as something like a erivative. This is because a well efine time erivative for a real, genuinely stochastic Markov process oes not exist. A trajectory of Brownian motion, for example, is clearly not ifferentiable. This is just one such example. We can formalize this as follows. We begin with xt + ) xt) x ± σ x ). 36) Now we switch to 1-D an make use of the propagator moment functions. Clearly, from 35) an so σ 2 x ) = x 2 x 2 x = Π 1x, t) + O ) 2) 37) Therefore σ x ) = From this we get Π 2x, t) + O ) 2 ) Π 1x, t) + O ) 2 )) 2 = Π 2x, t) + O ) 2). 39) xt + ) xt) Π 1x, t) ± Π2x, t) ± O ) 2). 40)

9 Creative Commons Attribution License CC-BY 3.0) 9 We see now that the secon term, the one that contains Π 2, exploes as 0, unless Π 2 is zero as well. But Π 2, which is relate to the variance, is zero only if x is a sure variable, therefore not representing a genuinely stochastic Markov process. The compoune Chapman-Kolmogorov equation 18) can be rewritten in terms of propagator ensities assuming that the interval [t 0, t] is subivie into a large number n of infinitesimal segments. Then The compoune Chapman-Kolmogorov equation P x, t) x 0, t 0)) = n... Π x i x i 1, t i 1)) x 1... x n 1. Ωx 1 ) Ωx n 1 ) i=1 41) In principle, this equation lets us reconstruct the Markov process from the knowlege of the propagator ensity function. It can be thought of as a Markovian equivalent of the Schröinger equation, where the Hamiltonian plays a similar role. Equations 26), 27) an 28) supplemente by a small number of aitional requirements lea to tractable expressions for the propagator ensity functions, again in similarity to known expressions for Hamiltonians. The proceure then makes the resulting Markov processes tractable analytically an numerically. It is most surprising how much can be inferre about them starting from simple assumptions. 6 The Kramers-Moyal Equations The Kramers-Moyal Equations are partial ifferential equations for the Markov state ensity function expresse with the help of the propagator moment functions, efine by 33). They follow irectly from the forwar 15) an backwar 17) Kolmogorov equations. forwar Our starting point is the 1-imensional forwar Kolmogorov equation P x, t + t) x 0, t 0)) = P x, t + t) x x, t) ) P x x, t) x 0, t ) 0) x We introuce an auxiliary function 42) fx) = P x + x, t + t) x, t) ) P x, t) x 0, t 0)). 43) an observe that the expression uner the integral in the forwar Kolmogorov equation 42) is fx x ), which can be expane in the Taylor series aroun x, if f is analytic: fx x x ) n ) = fx) + n! n=1 n fx) x n. 44)

10 10 License to Connexions by Zzis law Meglicki, Jr We substitute this into 42) with the following effect P x, t + t) x 0, t 0)) = P x + x, t + t) x, t) ) P x, t) x 0, t 0)) x + n=1 1) n n! n x ) n P x + x, t + t) x, t) ) P x, t) x 0, t 0)) x. x n Let s have a look at the first integral. x appears only in the first P term. This, therefore, is a normalization integral which evaluates to 1 times the secon P term. The integrals in the sum similarly epen on x, which appears only in the first P in the integrate function. The secon P is therefore a coefficient that can be put in front of the integral. We subract P x, t) x 0, t 0)) from both sies an ivie both sie by t which yiels 45) P x, t + t) x 0, t 0)) P x, t) x 0, t 0)) = t 1) n n P x, t) x 0, t 0)) x ) n P x + x, t + t) x, t) ) ) x. n! x n t n=1 Now we take a limit t 0. In this limit the integral in the sum, upon its ivision by t becomes x ) n Π x x, t)) x = Π nx, t), 47) 46) The forwar Kramers-Moyal equation the efinition we have alreay introuce in 33). This, finally, leas to the forwar Kramers-Moyal equation t P x, t) x0, t0)) = n=1 1) n n! n Πnx, t)p x, t) x0, t0))). xn 48) backwar Our starting point is the 1-imensional backwar Kolmogorov equation P x, t) x 0, t 0)) = P x, t) x 0 + x, t 0 + t) ) P x 0 + x, t 0 + t) x 0, t ) 0) x We introuce an auxiliary function fx 0) = P x, t) x 0, t 0 + t 0)) 50) an observe that the first P uner the integral in the backwar Komogorov equation 49) is fx 0 + x ), which can be expane in 49)

11 Creative Commons Attribution License CC-BY 3.0) 11 the Taylor series aroun x 0, if f is analytic: fx 0 + x x ) n ) = fx 0) + n! n=1 We substitute this into 49) with the following effect P x, t) x 0, t 0)) = P x, t) x 0, t 0 + t 0)) + n=1 n fx 0). 51) x n 0 P x 0 + x, t 0 + t 0) x 0, t 0) ) x ) 1 n P x, t) x 0, t n! x n 0 + t 0)) 0 x ) n P x 0 + x, t 0 + t 0) x 0, t ) 0) x. 52) We observe that the integral in the first aen is a normalization integral for P an therefore equal to 1. This leaves P x, t) x 0, t 0 + t 0)) alone an we transfer it to the left sie of the equation an ivie both sies by t 0. In the limit of t 0 0, this yiels minus the erivative of P on the left sie. The right sie of the equation is left with the sum only an the integral turns into the moment integral of the Markov propagator ensity function, which, upon the ivision by we call Π nx 0, t 0), as per 33). The being use so can no longer be use again to play with t 0 in the ifferentiate P. Instea P x, t) x 0, t 0 + t 0)) t 0 P x, t) x0, t0)). 53) In summary we en up with the backwar Kramers-Moyal equation: 1 n P x, t) x 0, t 0)) = Πnx0, t0) P x, t) x 0, t 0)). t 0 n! x n n=1 0 54) The thing to observe is that the Π n coefficients are ifferentiate together with P in the forwar Kramers-Moyal equation, but not in the backwar one. There is also the 1) n factor in the forwar equation, but not in the backwar one, an the sign in front of the time erivative is negative in the backwar equation. Finally, looking at both Kramers-Moyal equations, it is easier to unerstan why one is calle forwar an the other backwar. This is not so clear when looking at the original Kolmogorov equations. In the forwar Kramers-Moyal equation, the x 0, t 0) pair is a fixe parameter an the ifferentiation is over t an x an procees forwar in time. The initial conition for the equation is P x, t = t 0) x 0, t 0)) = δx x 0). 55) In the backwar Kramers-Moyal equation, the x, t) pair is a fixe parameter an the ifferentiation is over t 0 an x 0 an procees backwar in time. The initial conition for the equation is P x, t) x 0, t 0 = t)) = δx x 0). 56) The backwar Kramers-Moyal equation

12 12 License to Connexions by Zzis law Meglicki, Jr 7 Evolution of the Moments We re going to o it all in 1-D. Our starting point is xt + ) = xt) + x. 57) Hence x n t + ) = xt) + x )n = x n t) + ) n n x n k t)x k. 58) k k=1 Now we average both sies of the equation ) n x n t + ) = x n n t) + x n k t)x k. 59) k The probability ensity of x n t) is P x, t) x 0, t 0)) an the probability ensity of x is Π x x, t)). Therefore the combine probability ensity of x n k t)x k is k=1 Π x x, t) ) P x, t) x 0, t 0)) 60) an the integration to prouce x j t)x k for some j an k must run over x an x : x j t)x k = x j x k Π x x, t) ) P x, t) x 0, t 0)) x x = Ωx) Ωx) x j Π k x, t)p x, t) x 0, t 0)) x + O ) 2) = x j t)π k xt), t) + O ) 2). 61) Time evolution of the moments Let us go back to 59). We subtract x n t) from both sies, substitute 61) in place of x n k t)x k an ivie both sies by, which yiels xn t) = ) n n x n k t)π k xt), t). 62) k k=1 An this is our equation for the evolution of the moments of the Markov process xt) that oes not use explicitly the P s or the Πs. But, of course, the propagator ensity is hien insie the propagator moment functions Π n. The initial conition for the equation is x n t 0) = x n 0. 63) Time evolution of the mean 7.1 Mean, Variance an Covariance The equation for the evolution of the mean of the Markov process is trivial, xt) = Π1xt), t), 64)

13 Creative Commons Attribution License CC-BY 3.0) 13 an follows irectly from 62). The initial conition for this equation is For x 2 62) yiels x t = t 0) = x 0. 65) x 2 t) = 2 xt)π 1xt), t) + Π 2xt), t). 66) From this an from 64) we obtain Time evolution of the variance var xt)) = x 2 t) xt) 2) = 2 xt)π 1xt), t) + Π 2xt), t) 2 xt) Π 1xt), t). 67) The initial conition for this equation is var x t = t 0)) = 0. 68) Covariance is a function of two variables, for example, cov xt 1), xt 2)). Here we are going to evaluate its erivative with respect to t 2. cov xt 1), xt 2)) = xt 1)xt 2) xt 1) xt 2) ) 2 2 = 2 xt 1)xt 2) xt 1) 2 xt 2) = 2 xt 1)xt 2) xt 1) Π 1xt 2), t 2). 69) The first component of the sum on the right sie of the equation still requires some work. We procee as follows xt 1)xt 2 + 2) = xt 1) xt 2) + x t 2) ) = xt 1)xt 2) + xt 1)x t 2), 70) the average of which is where Therefore xt 1)xt 2) + xt 1)x t 2), 71) xt 1)xt 2) = x 1x 2P x 2, t 2) x 1, t 1)) P x 1, t 1) x 0, t 0)) x 1 x 2. Ωx 1 ) Ωx 2 ) 72) xt 1)xt 2 + 2) xt 1)xt 2) = xt 1)x t 2). 73)

14 14 License to Connexions by Zzis law Meglicki, Jr The right sie of this equation is somewhat tricky, because here we have to average x t 2). We o this as follows xt1)x t 2) = Ωx 1 ) Ωx 2 ) x 1x Π x 2 x 2, t 2) ) P x 2, t 2) x 1, t 1)) P x 1, t 1) x 0, t 0)) x x 2 x 1 = x 1 Π1x 2, t 2) 2 + O 2) 2)) Ωx 1 ) Ωx 2 ) P x 2, t 2) x 1, t 1)) P x 1, t 1) x 0, t 0)) x 2 x 1 = xt 1)Π 1xt 2), t 2) 2 + O 2) 2). 74) Substituting this result into 73) an iviing both sies by 2 yiels 2 xt 1)xt 2) = xt 1)Π 1xt 2), t 2). 75) Time evolution of the covariance Now we plug this result into 69) which yiels 2 cov xt 1), xt 2)) = xt 1)Π 1xt 2), t 2) xt 1) Π 1xt 2), t 2). 76) The initial conition for this equation is cov x t 1), x t 2 = t 1)) = var x t 1)). 77) Two examples of Markov processes It is useful to note that when Π 1x, t) oes not epen on x, then it falls out of the brackets on the right sie of 76) which makes the right sie zero. Thus cov xt 1), xt 2)) ens up inepenent of t 2 an so it must remain set to the initial conition, that is, var x t 1)). Looking at 64), 67) an 76) an the relate initial conitions 65), 68) an 77) we fin that these equations seem not only eminently tractable, but even relatively simple epening, that is, on the Π nx, t) functions. Thus, by persistent chipping at the problem, we have progresse from the initial view of stochastic processes that was intimiating, to say the least, to quite tractable equations that escribe the evolution of the mean, variance an covariance of the Markov process xt). To illustrate this point, we are going to look at two simple examples that happen to be applicable to some Markov processes of interest. The examples also illustrate how we woul use the propagator moment functions of the Markov process to specify it. Π 1x, t) = v, Π 2x, t) = γ where v an γ 0 are constants. In this case Π 1 = v an Π 2 = γ an the equations that escribe the evolution of the mean, the variance an the covariance plus their

15 Creative Commons Attribution License CC-BY 3.0) 15 initial conitions are xt) = v x t = t 0) = x 0 var xt)) = γ var x t = t 0)) = 0 cov x t 1), x t 2)) 2 = 0. cov x t 1), x t 2 = t 1)) = var x t 1)). 78) The erivative of the covariance is zero, because Π 1 is a constant, which means that cov xt 1), xt 2)) = var x t 1)). 79) The solution to this problem is therefore xt) = x 0 + v t t 0) var x t)) = γ t t 0) cov xt 1), xt 2)) = γ t 1 t 0). 80) For γ = 0 the variance an the covariance remain zero an the process becomes eterministic. Π 1x, t) = λx, Π 2x, t) = γ where λ > 0 an γ 0 are constants. This example is a little more complicate. The equations that govern it are as follows xt) = λ xt) x t = t 0) = x 0 var xt)) = 2λ var xt)) + γ var x t = t 0)) = 0 cov x t 1), x t 2)) 2 = λ cov x t 1), x t 2)) cov x t 1), x t 2 = t 1)) = var x t 1)). 81) Before we go any further, we re going to explain how these equations come about. The first one is obvious. The equation for the variance is obtaine by substituting our specific expressions for Π 1 an Π 2 in 67) which yiels var xt)) = 2 xt) λxt)) + γ 2 xt) λxt) = 2λ x 2 t) xt) 2) + γ = 2λ var xt)) + γ. 82)

16 16 License to Connexions by Zzis law Meglicki, Jr The equation for the covariance is obtaine by substituting Π 1 an Π 2 as efine above in 76) which yiels 2 cov x t 1), x t 2)) = xt 1) λxt 2)) xt 1) λxt 2) = λ xt 1)xt 2) xt 1) xt 2) ) = λ cov xt 1), xt 2)). 83) The solution of the first equation in 81) is of the form e λt. The initial conition forces the following choice of constants xt) = x 0e λt t 0). 84) The variance equation woul be like the mean equation were it not for the non-homogeneous term γ. We eal with this by postulating a solution of the form var xt)) = Ae 2λt t 0) ft), 85) where A is a constant an ) Ae 2λt t0) ft) = Ae 2λt t 0) ) ft) 2λft) 86) Subsituting this solution into the equation for the variance yiels ) Ae 2λt t 0) ft) 2λft) = 2λAe 2λt t0) ft) + γ 87) We a 2λAe 2λt t 0) ft) to both sies of the equation, which kills this term an we are left with ft) = γ A e2λt t 0), 88) which solves to ft) = γ 2λA e2λt t 0) + B, 89) where B is another constant. We substitute this into 85) an obtain var xt)) = Ae 2λt t 0) γ ) 2λA e2λt t 0) + B = γ 2λ + ABe 2λt t 0), 90) leaving us with just one constant AB as shoul be expecte. For t = t 0 the exp function is 1 an we obtain var xt = t 0)) = γ + AB = 0 91) 2λ which implies that AB = γ/2λ). Therefore var xt)) = γ ) 1 e 2λt t 0). 92) 2λ

17 Creative Commons Attribution License CC-BY 3.0) 17 The covariance equation is just like the mean equation. But here the initial conition at t 2 = t 1 sets the covariance to variance of xt 1). Consequently the solution is cov x t 1), x t 2)) = var xt 1)) e λt 2 t 1 ) 93) But we have the expression for the variance in the form of 92). Plugging it into the above solution yiels cov x t 1), x t 2)) = γ ) 1 e 2λt 1 t 0 ) e λt 2 t 1 ). 94) 2λ In summary xt) = x 0e λt t 0) var xt)) = γ 1 ) e 2λt t 0) 2λ cov x t 1), x t 2)) = γ 2λ 1 e 2λt 1 t 0 ) ) e λt 2 t 1 ). 95) 8 Homogeneous Markov Processes Markov processes are sai to be homogeneous if the corresponing Markov process ensity function oes not epen on time that is Π x x, t) ) = Π x x ), 96) in which case they are calle temporally homogeneous; space that is Π x x, t) ) = Π x t ), 97) in which case they are calle spatially homogeneous; time an space that is Π x x, t) ) = Π x ), 98) in which case they are calle completely homogeneous. Brownian motion is an example of a completely homogeneous process. A great eal can be sai about such processes, because the equations that escribe them are simpler an their certain properties can be seen right away. One such important property of temporally homogeneous Markov processes, an by extension also of completely homogeneous ones, is that the probability ensity P x, t) x 0, t 0)) epens on time through t t 0 only, in other wors Temporally homogeneous Markov processes P x, t) x 0, t 0)) = P x, t t 0) x 0, 0)). 99) From this it follows immeiately that P x, t) x0, t0)) = P x, t) x 0, t 0)) 100) t t 0

18 18 License to Connexions by Zzis law Meglicki, Jr Because the propagator ensity oes not epen on time, the propagator moment functions Π n o not epen on time either Π nx) = 1 x n Πx x) x. 101) The Kramers-Moyal equations Consequently, the Kramers-Moyal equations simplify too, with the left sie of the equations fully interchangeable on account of 100) t P x, t) x0, t0)) = t 0 P x, t) x 0, t 0)) = n=1 n=1 1) n n! n Πnx)P x, t) x0, t0))), xn 1 n Πnx0) P x, t) x 0, t 0)). n! x n 0 102) Completely homogeneous Markov processes If the process is completely homogeneous, the above equations apply, but we can simplify even more. Let us ivie [t, t 0] into n infinitesimal intervals, each of length, then we can write the compoune Chapman- Kolmogorov equation as follows P x, t) x 0, t 0)) =... Ωx 1 ) Ωx n 1 ) j=1 n Π x j ), 103) where n 1 x n = x x 0 x i x x 0 = i=1 n i=1 = t t0 n. 104) The integrals run over x 1,..., x n 1, but not over x n. We can a the latter by making use of the above expression for x x 0 an inserting the corresponing Dirac elta in 103), then we make use of the fact that Π x i ) epen on their own x i only an obtain P x, t) x 0, t 0)) = n j=1 Ωx j ) Π x i ) δ x x 0 n i=1 x i ) x j. 105) The next step in gaining further insights is to express the Dirac elta in the form of the Fourier transform: which in this case becomes δx) = 1 2π 1 2π e ikx x 0) e ikx k, 106) n e ikx j k. 107) j=1

19 Creative Commons Attribution License CC-BY 3.0) 19 We substitute this into 105), also assume that for each j, Ωx j) = [, ], which yiels e ikx x 0) P x, t) x 0, t 0)) = 1 2π where = 1 2π ˆΠ k ) = Π x ) ) n e ikx x k e ikx x 0) ˆΠ k ) ) n k, 108) Π x ) e ikx x 109) is the Fourier transform of Π. Looking at 108) we see that not only oes P x, t) x 0, t 0)) epen on t t 0, as per 104). It epens on x through x x 0 too. Hence P x, t) x 0, t 0)) = P x x 0, t t 0) 0, 0)), 110) wherefrom P t P x = P t 0 = P x ) For the completely homogeneous Markov processes the propagator moment function Π nx, t) no longer epen on x or t. They are therefore constants, Π n. In combination with 111), this reuces the two Kramers- Moyal equations to just one t P x, t) x0, t0)) = n=1 1) n n! n Π n P x, t) x0, t0)). 112) xn The Kramers-Moyal equations The equation for the evolution of the moments, 62), similarly simplifies to ) n n xn t) = Π k x n k t). 113) k k=1