Applying Itô’s Formula to Compensated and Uncompensated Poisson Processes
Summary
The document explains why the drift correction in Itô’s formula depends on whether a process is driven by a compensated or ordinary Poisson process. For an integral against the compensated process, the process has a negative compensator drift, producing a derivative term alongside the jump contribution. For a process driven by the ordinary Poisson process, there is no such drift term; rewriting its jump integral using the compensated process instead places the compensator in the time integral.
The answer compares these cases and applies the ordinary-process formula to the Poisson count itself, whose jumps have size one. This accounts for the generator term involving the rate and the change in the function at a jump. The discussion is a conceptual clarification rather than a full derivation: it assumes the relevant processes and functions satisfy conditions for the formulas, and does not address more general jump sizes or state-dependent intensities in detail.
Key ideas
- Compensation adds a continuous drift to a Poisson process.
- The derivative correction appears when the process includes that compensator drift.
- An ordinary Poisson count has unit jumps and no compensator drift.
- The count-process generator weights the function’s jump change by the event intensity.
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# Ito Lemma for Poisson Process
# Ito Lemma for Poisson Process
I'm new to stochastic calculus on jump processes and encountered a difficulty. I would appreciate some clarification from the community on the following question.
Let $g_t$ be a $\mathcal{F_t}$-adapted process where $\mathcal{F_t}$ is the natural filtration generated by a Poisson process $N_t$.
Define stochastic integral $Y_t$ as $$Y_t = \int_0^t g_{s^{-}} d\hat{N_s}$$ where $\hat{N_t} = N_t - \lambda t$ is the compensated Poisson process.
Suppose $Z_t = f(t,Y_t)$ is once differentiable in $t$, then the Ito formula for Poisson Process is \begin{equation}dZ_t = \bigg\{\partial_t f(t,Y_t) + \lambda \Big(\big[f(t, Y_{t^{-}} + g_{t^{-}}) - f(t,Y_{t^{-}})\big] - g_t \partial_y f(t,Y_t) \Big) \bigg\}dt + \bigg[ f(t,Y_{t^{-}} + g_{t^{-}}) - f(t, Y_{t^{-}}) \bigg]d\hat{N_t} \end{equation}
so my question is why then do we have, by Ito formula, \begin{equation} H(\tau, N_\tau) = H(t,N_t) + \int_t^\tau (\partial_t + \mathcal{L}_s)H(s, N_s)ds + \int_t^\tau \left[ H(s, N_{s^{-}} + 1) - H(s,N_{s^{-}})\right]d\hat{N_s} \end{equation} where $\mathcal{L_t}{H(t,n)} = \lambda(t,n,u) \left[ H(t,n+1) - H(t,n)\right]$? Take $N_t = n$.
Why did the $-g_t \partial_y{H(t, N_t)}$ term disappear? I believe $Y_t = \int_0^t 1 d\hat{N_s} = \hat{N_t}$ so $-g_t \partial_y{H(t, N_t)} = -\partial_{\hat{N_t}} H(t,N_t) = 0$? Wouldn't $\partial_\hat{N_t} H(t,N_t) = \partial_{N_t} H(t,N_t)$ instead?
Thanks
## Answer by ir7 (score 1, accepted)
https://quant.stackexchange.com/a/65741
The basic Ito formula for a Poisson process is $$ dY_t = \mu_t dt + g_t dN_t $$
$$ df(Y_t) = \mu_t f'(Y_t) dt + (f(Y_{t-}+g_t) - f(Y_{t-}))dN_t $$
(dropped $f$'s direct dependence on the time variable to avoid the partial derivative clutter).
Case $\mu_t = -\lambda g_t$ (this is your original case):
$$ df(Y_t) = -\lambda g_t f'(Y_t) dt+ (f(Y_{t-}+g_t) - f(Y_{t-}))dN_t $$
$$ = \lambda [(f(Y_{t-}+g_t) - f(Y_{t-})) - g_t f'(Y_t)]dt + (f(Y_{t-}+g_t) - f(Y_{t-}))d\hat{N}_t $$
Case $\mu_t = 0$ (here $gf'$ vanishes):
$$ df(Y_t) = (f(Y_{t-}+g_t) - f(Y_{t-}))dN_t $$
$$ = \lambda (f(Y_{t-}+g_t) - f(Y_{t-})) dt + (f(Y_{t-}+g_t) - f(Y_{t-}))d\hat{N}_t $$
Case $\mu_t = 0, g_t =1$ (this is the mysterious case, where $Y_t=N_t$):
$$ df(N_t) = (f(N_{t-}+1) - f(N_{t-}))dN_t $$
$$ = \lambda (f(N_{t-}+1) - f(N_{t-})) dt + (f(N_{t-}+1) - f(N_{t-}))d\hat{N}_t $$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.