I have the following stochastic process:

$dX = (A-I)XdN$,

where $X$ is a $2\times1$ vector of random variables, $A$ is a constant, real, symmetric, $2\times2$ matrix, $I$ is the identity matrix and $dN$ is a simple Poisson process with constant intensity $\lambda$ and unit jump size.

I want to calculate the transition density of the process.

I am aware of three possible ways to do it:

  1. Calculate the probability distribution and then derive
  2. Find the characteristic function and then apply the inverse Fourier transform
  3. Solve the associated forward Kolmogorov equation

Here is what I have attempted:

  1. Nothing
  2. The characteristic function is defined as $C(z) = \mathrm{E}_{X_t}[\exp{iz^{\top}X_t}]$. I found

    $X_t = A^{N(t)}X(0)$,

    but I got stuck trying to calculate the expectation.


Let me show a method for 1-d case first: $dX = aX dN$

By Ito's formula ( In this special case, it is straightforward, I guess we don't even need Ito's formula)

$ln(X(t)) = ln(X(0)) + \sum_{s\in [0,t]}(ln(X_s)-ln(X_{s-}))$

Note that, whenever there is a jump at $s$, $ln(X_s)-ln(X_{s-})=ln(1+a)$, so we have

$ln(X(t)) = ln(X(0)) + N(t)ln(1+a)$

$X(t) = X(0) (1+a)^{N(t)}$, so you get all possible values for $X(t)$, and the associated probabilities, derived from $N(t)$ -- you have the transition probabilities.

For 2-d case, note that the coefficient matrix is symmetric, therefore diagonalizable , by linear transformation, you can reduce it to 1-d problems.

  • $\begingroup$ Thank you. This is precisely the procedure I had not attempted. $\endgroup$ – Marduk Feb 7 '16 at 13:49

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