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Consider an (inhomogeneous) Poisson process $N_t$ with intensity $\lambda_t$. Then I want to compute the following integral

$\mathbb{E} \left(\int f(t,N_{t-}) d\tilde{N}_t\right)^2$

for some smooth enough function $f$ and $d\tilde{N}_t=dN-\lambda_t dt$ denotes the compensated process.

Is it true that

$$ \mathbb{E} \left(\int_0^t f(t,N_{t-}) d\tilde{N}_t\right)^2$$

$$= \mathbb{E} \int_0^t |f(t,N_{t-}) |^2 dN_t$$

$$= \mathbb{E} \int_0^t |f(t,N_{t-}) |^2 \lambda_t dt?$$

If so, then how can I prove this fact?

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1 Answer 1

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Let \begin{align*} X_t=\left(\int_0^t f(s,N_{s-}) d\tilde{N}_s\right)^2 \end{align*} and \begin{align*} Y_t = \int_0^t f(s,N_{s-}) d\tilde{N}_s. \end{align*} By It$\hat{\text{o}}$'s product rule for jump processes (see Corollaries 11.4.10 and 11.5.5 of this book), \begin{align*} X_t &= 2\int_0^t Y_s dY_s + [Y, Y]_t\\ &=2\int_0^t Y_s dY_s + \sum_{0<s \le t}f(s,N_{s-})^2 (\Delta N_s)^2\\ &=2\int_0^t Y_s dY_s + \sum_{0<s \le t}f(s,N_{s-})^2 \Delta N_s\\ &=2\int_0^t Y_s dY_s + \int_0^t f(s,N_{s-})^2 dN_s\\ &=2\int_0^t Y_s dY_s + \int_0^t f(s,N_{s-})^2 d\tilde{N}_s + \int_0^t f(s,N_{s-})^2 \lambda_s ds. \end{align*} Given that both \begin{align*} \int_0^t Y_s dY_s \end{align*} and \begin{align*} \int_0^t f(s,N_{s-})^2 d\tilde{N}_s \end{align*} are martingale, we conclude that \begin{align*} E\left(\int_0^t f(s,N_{s-}) d\tilde{N}_s\right)^2 = E\left(\int_0^t f(s,N_{s-})^2 \lambda_s ds\right). \end{align*}

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