3
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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 $\hat{N_t}$ $\endgroup$ Jun 25 at 5:17
1
$\begingroup$

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 $Y_t=N_t$):

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.