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Solution of the Kolmogorov forward equation for a linear process

  1. Apr 19, 2012 #1
    Consider the 1-D linear system governed by:

    "dx/dt = a*x(t) + n(t)"

    where "a" is a scalar and:
    x(t) = system state
    n(t) ~ N(0, sigma^2)

    ****************************

    We can write Ito's stochastic differential equation of the previous process as:

    "dx = a*x*dt + 1/2*sigma^2*dW_t"

    where:
    x = system state
    a*x = drift term
    1/2*sigma(t,x)^2 = diffusion term
    W_t = Wiener process

    ****************************

    The time evolution of the system state's probability density function " pdf(x(t)) = p(x,t) " is governed by the Kolmogorov forward equation (aka Fokker-Planck equation):

    "dp(x,t)/dt = -d/dx[a*x*p(x,t)] + 1/2*sigma^2*d^2p(x,t)/dx^2"


    which after expanding gives:

    "dp(x,t)/dt = -a*p(x,t) - a*x*dp(x,t)/dt + 1/2*sigma^2*d^2p(x,t)/dx^2"


    where "d/dt", "d/dx" and "d^2/dx^2" are partial derivatives.

    ****************************

    Because the process is linear and "n(t) ~ N(0, sigma^2)", if we further assume that "x(t=0) ~ N(x0, sigma_x^2)", then "pdf(x) = p(x,t)" is Gaussian for t >= 0. Which is basically why the Kalman filter works.

    ****************************

    Furthermore, in the case of "a = 0", i.e. pure diffusion, the solution of the KFE is:

    "p(x,t) = ((2*pi*sigma*t)^(-1/2))*exp(-((x-x0)^2)/(2*sigma*t))"

    which is indeed a Gaussian distribution thus supporting the previous statement.

    ****************************

    I therefore expected that a solution of the type (gaussian drift + diffusion):

    "p(x,t) = ((2*pi*sigma*t)^(-1/2))*exp(-((x-beta*t-x0)^2)/(2*sigma*t))"

    would solve the KFE. However, it is easy to verify that this is not a solution of the KFE above.

    ****************************

    Undoubtedly my reasoning is wrong and, most likely, in more than one place. Where?
     
  2. jcsd
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