# 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 $$$$dZ_t = \bigg\{\partial_t f(t,Y_t) + \lambda \Big(\big[f(t, Y_{t^{-}} + g_{t^{-}}) - f(t,Y_{t^{-}})\big] - g_t \partial_y f(t,Y_t) \Big) \bigg\}dt + \bigg[ f(t,Y_{t^{-}} + g_{t^{-}}) - f(t, Y_{t^{-}}) \bigg]d\hat{N_t}$$$$

so my question is why then do we have, by Ito formula, $$$$H(\tau, N_\tau) = H(t,N_t) + \int_t^\tau (\partial_t + \mathcal{L}_s)H(s, N_s)ds + \int_t^\tau \left[ H(s, N_{s^{-}} + 1) - H(s,N_{s^{-}})\right]d\hat{N_s}$$$$ where $$\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

• I am referring to the wrong example. Should instead be using Ito formula for $N_t$ then for a stochastic integral of a compensated Poisson process $\hat{N_t}$ Jun 25, 2021 at 5:17

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 $$Y_t=N_t$$):

$$df(N_t) = (f(N_{t-}+1) - f(N_{t-}))dN_t$$

$$= \lambda (f(N_{t-}+1) - f(N_{t-})) dt + (f(N_{t-}+1) - f(N_{t-}))d\hat{N}_t$$