Hail to all, I am struggling to solve the following SDE for intensity:

$d\lambda_t = \kappa(\rho(t) - \lambda_t)dt + \delta dN_t $

I know to expect the solution in the form of

$\lambda_t = c(0)e^{-\kappa t} + \kappa \int_0^{t}e^{-\kappa (t-u)} \rho(u) du + \delta\int_0^{t}e^{-\kappa (t-u)}dN_u$

from here I am able to verify that this is indeed the solution but I am no able to reach this step rigorously.



Let us define the auxiliary process $\Lambda_t=e^{\kappa t}\lambda_t$. Note that:

$$ \Lambda_t = \kappa e^{\kappa t} \int_0^t(\rho_s-\lambda_s)ds+\delta e^{\kappa t}\int_0^tdN_t$$

Hence after a jump occurs at $t$:

$$ \Lambda_t=\Lambda_{t-}+\delta e^{\kappa t}$$

Therefore by Ito's lemma for jump-diffusion processes:

$$ \begin{align} d\Lambda_t & = \frac{\partial \Lambda_t}{\partial t}dt+\frac{\partial \Lambda_t}{\partial \lambda_t}\kappa(\rho_t-\lambda_t)dt+(\Lambda_t-\Lambda_{t-})dN_t \\[9pt] & = \kappa e^{\kappa t}\rho_tdt+\delta e^{\kappa t}dN_t \end{align}$$


$$ \Lambda_t=\Lambda_0+\kappa\int_0^te^{\kappa s}\rho_sds+\delta\int_0^te^{\kappa s}dN_s$$


$$ \lambda_t=\lambda_0+\kappa\int_0^te^{\kappa (s-t)}\rho_sds+\delta\int_0^te^{\kappa (s-t)}dN_s$$

You notice that in the original SDE the following factor is the "nuisance":

$$d\lambda_t = \cdots + \left(-\kappa \lambda_t dt\right) + \cdots$$

which corresponds to the differential of an exponential with constant $\kappa$:

$$ dx_t = \kappa x_tdt \quad \Leftrightarrow \quad x_t = Ce^{\kappa t}$$

Hence you need to try to get rid of it by making a $+\kappa \lambda_t dt$ appear somehow, which can be achieved by differentiating an exponential with constant $\kappa$ through Ito's lemma applied to $\Lambda_t$ as defined above.

  • $\begingroup$ excellent! the last paragraph was exactly what I was looking for. Thanks! $\endgroup$ – Michael Mark Oct 30 '17 at 14:05
  • $\begingroup$ You are welcome. $\endgroup$ – Daneel Olivaw Oct 30 '17 at 14:06

Set $\tilde{\lambda}_t = e^{\kappa t} \lambda_t$ and solve for $\tilde{\lambda}_t$

  • $\begingroup$ Neat, I love the fact it works but is there an intuition why I did such substitution in the first place? Cheers $\endgroup$ – Michael Mark Oct 30 '17 at 13:58
  • 1
    $\begingroup$ that's just a common trick for solving first order differential equations, stochastic or not :) $\endgroup$ – Antoine Conze Oct 30 '17 at 14:03

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