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Using the Compensated Poisson Isometry for Squared Integrals

Article Quant Q&A · Author: Student
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Summary

The document shows how to compute the second moment of a stochastic integral against a compensated Poisson process. Its method defines the integral as a martingale, applies the product rule for jump processes to its square, and separates the jump quadratic variation from the martingale terms. Since a Poisson process has unit jumps, the quadratic variation becomes an integral against the counting process.

Replacing that counting-process integral with its compensator plus a compensated martingale yields the expected integrated squared integrand weighted by the intensity. The result is the Poisson-process version of the stochastic integral isometry. The argument requires suitable integrability so the stochastic integrals are true martingales and their expectations vanish. The question uses an inhomogeneous intensity, and the answer allows the intensity to vary over time; it does not spell out detailed regularity conditions or handle cases where those conditions fail.

Key ideas

  • The square of a compensated Poisson integral can be analyzed with the jump-process product rule.
  • Unit jumps make the quadratic variation equal to an integral of the squared integrand against the counting process.
  • The expected quadratic variation is expressed using the intensity compensator.
  • The identity relies on integrability conditions that ensure the martingale terms have zero expectation.

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# Stochastic integral involving Poisson Process


# Stochastic integral involving Poisson Process












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?

## Answer by Gordon (score 4, accepted)

https://quant.stackexchange.com/a/74855

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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