Consider the linear SDE
\[ dX_t = F X_t dt + L dB_t \]
where \(F, L\) are known constant matrices. This SDE can be considered as a multivariate version of the Ornstein-Uhlenbeck process. Below, we will solve this SDE and derive its exact transition density.
Itô Lemma
A key result we will be using is the Itô lemma, which we state below. Consider an SDE
\[ dX_t = \mu(t, X_t) dt + \sigma(t,X_t) dB_t \]
with drift function \(\mu\) and volatility function \(\sigma\). For a function \(f(t, X_t)\) that depends (potentially) both \(t\) and \(X_t\), the Itô lemma states that
\[ df = \left[ \frac{\partial f}{\partial t} + (\nabla_x f)^T \mu + \frac{1}{2} \text{tr} [\sigma^T (\nabla_{xx} f) \sigma] \right] dt + (\nabla_x f)^T \sigma dB_t \]
which is obtained by a Taylor expansion of \(f\) and substitution of \(dX_t\).
Conditional Distribution
Returning to solving the SDE, we notice first that if we consider a change-of-variable \(Y_t := e^{-Ft} X_t\) with matrix exponential, applying Itô to \(Y_t\) yields
\[ dY_t = [- F e^{- Ft} X_t + e^{-F t} F X_t ] dt + e^{-F t} L dB_t = e^{-F t} L dB_t \] where the drift components cancel out as \(F e^{-Ft} = e^{-Ft} F\).
So, integrating the above equation from \(n\) to \(n+h\) and substituting back \(Y_t = e^{-Ft} X_t\) gives us
\[ e^{-F(n+h)} X_{n+h} - e^{-Fn} X_n = \int_{n}^{n+h} e^{-Ft} L dB_t, \]
which can be converted to the following form by multiplying \(e^{F(n+h)}\) both sides
\[ X_{n+h} = e^{Fh} X_n + \int_{n}^{n+h} e^{F(n+h-t)} L dB_t. \]
Therefore, we have the following exact transition density
\[ \begin{split} X_{n+h} | X_n = x_n &\sim N(A_h x_n, Q_h) \\ A_h &:= e^{Fh} \\ Q_h &:= \int_{n}^{n+h} e^{F(n+h-t)} L L^T e^{F^T(n+h-t)} dt \end{split} \]
where \(Q_h\) is obtained using the Itô isometry. While it is obvious how \(A_h\) can be found, the calculation of \(Q_h\) does not appear immediate. A nice way to compute \(Q_h\) is presented below.
Conditional Distribution Covariance Calculation
Consider the auxiliary block matrix
\[ K = \begin{bmatrix} F & LL^T \\ 0 & -F^T \end{bmatrix} \]
and we are interested in its matrix exponential \(\Phi(t) = e^{Kt}\) with entries
\[ e^{Kt} =: \Phi(t) = \begin{bmatrix} \Phi_{11}(t) & \Phi_{12}(t) \\ 0 & \Phi_{22}(t) \end{bmatrix}. \]
Since \(\Phi\) is a matrix exponential, we know \(\Phi'(t) = K \Phi(t)\), so
\[ \begin{bmatrix} \Phi_{11}'(t) & \Phi_{12}'(t) \\ 0 & \Phi_{22}'(t) \end{bmatrix} = \begin{bmatrix} F & LL^T \\ 0 & -F^T \end{bmatrix} \begin{bmatrix} \Phi_{11}(t) & \Phi_{12}(t) \\ 0 & \Phi_{22}(t) \end{bmatrix}. \]
Expanding out the matrix multiplication and matching the entries yield
\[ \begin{split} \Phi_{11}' &= F \Phi_{11} \implies \Phi_{11} = e^{Ft} \\ \Phi_{22}' &= -F^T \Phi_{22} \implies \Phi_{22} = e^{-F^Tt}\\ \Phi_{12}' &= F \Phi_{12} + LL^T \Phi_{22} \end{split} \]
where the first two lines are established by realising \(\Phi_{11}(0) = \Phi_{22}(0) = I\). For the final line, we apply an integrating factor \(e^{-Ft}\) so
\[ \begin{split} \Phi_{12}' &= F \Phi_{12} + LL^T e^{-F^Tt} \\ e^{-Ft} \Phi_{12}' - e^{-Ft} F \Phi_{12} &= e^{-Ft} LL^T e^{-F^Tt} \\ \frac{d}{dt} [e^{-Ft} \Phi_{12}] &= e^{-Ft} LL^T e^{-F^Tt} \\ \end{split} \]
and integrating from \(0\) to \(h\) as well as using the initial condition \(\Phi_{12}(0) = 0\) yields
\[ \begin{split} e^{-Fh} \Phi_{12}(h) &= \int_{0}^{h} e^{-Ft} LL^T e^{-F^Tt} dt \\ \Phi_{12}(h) &= e^{Fh} \int_{n}^{n+h} e^{-F(t-n)} LL^T e^{-F^T(t-n)} dt \\ \Phi_{12}(h) &= \int_{n}^{n+h} e^{F(n+h-t)} LL^T e^{F^T(n+h-t)} dt \cdot e^{-F^Th} \\ \Phi_{12}(h) &= Q_h e^{-F^Th} = Q_h \Phi_{22}(h). \\ \end{split} \]
Therefore, we have \(Q_h = \Phi_{12}(h) \Phi^{-1}_{22}(h)\). The derivations above is often known as the Van Loan identity.