Solving a Mean-Reverting Hawkes Intensity with an Integrating Factor
Summary
The document derives the solution to an intensity process that mean-reverts toward a potentially time-varying baseline and jumps upward when events occur. It uses an integrating factor: multiply the intensity by an exponential in time so the mean-reversion term cancels under the product rule for jump processes. Integrating the resulting expression yields the initial intensity, a discounted integral of the baseline, and a discounted sum of past jumps.
The answer outlines the derivation and provides the resulting representation, making the method useful for solving linear jump-driven stochastic differential equations. It assumes a constant mean-reversion rate and jump size, and it does not discuss parameter estimation, stability conditions, or how the process is used in a trading model. The displayed derivation has some notation inconsistencies around the initial value and jump integral, so the integral form is clearer than its intermediate expressions.
Key ideas
- Multiplying the intensity by an exponential integrating factor removes its linear mean-reversion term.
- The intensity solution combines its initial value, the baseline path, and the history of event jumps.
- Past baseline values and jumps are weighted by exponential decay determined by the mean-reversion rate.
- The derivation assumes constant decay and jump parameters and does not address model calibration.
Tags
Full text
# Hawkes process intensity solution
# Hawkes process intensity solution
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.
Thanks!
## Answer by Daneel Olivaw (score 4, accepted)
https://quant.stackexchange.com/a/36657
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:
$$ \begin{align} d\Lambda_t & = \frac{\partial \Lambda_t}{\partial t}dt+\frac{\partial \Lambda_t}{\partial \lambda_t}\kappa(\rho_t-\lambda_t)dt+(\Lambda_t-\Lambda_{t-})dN_t \\[9pt] & = \kappa e^{\kappa t}\rho_tdt+\delta e^{\kappa t}dN_t \end{align}$$
Integrating:
$$ \Lambda_t=\Lambda_0+\kappa\int_0^te^{\kappa s}\rho_sds+\delta\int_0^te^{\kappa s}dN_s$$
Finally:
$$ \lambda_t=\lambda_0+\kappa\int_0^te^{\kappa (s-t)}\rho_sds+\delta\int_0^te^{\kappa (s-t)}dN_s$$
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.
## Answer by Antoine Conze (score 1)
https://quant.stackexchange.com/a/36656
Set $\tilde{\lambda}_t = e^{\kappa t} \lambda_t$ and solve for $\tilde{\lambda}_t$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.