any guidance on the following problem would be greatly appreciated!
Denote $M_t = N_t - \lambda t$ as the compensated Poisson process and $N_t$ as the Poisson process then my question is determine $$\textrm{var}\left[ \int_{a}^{b} M_t \, dM_t\right]$$.
I have the following properties:
- $\int_{a}^{b} M_{t-} dM_t = \frac{1}{2}({M_b}^2-{M_a}^2) - \frac{1}{2}(N_b-N_a)$ where $M_{t-}$ is the left-hand limit of $M_t$
- $\int_{a}^{b} c \,dM_t = c(M_b - M_a)$ for $c$ is a constant
- $E\bigl[\int_{a}^{b} M_{t-} \, dM_t\bigr] = 0$
Edit: What I thought is: \begin{align} \textrm{var}\left[ \int_{a}^{b} M_{t-} \, dM_t\right] &= E\left[\left(\int_{a}^{b} M_{t-} \, dM_t\right)^2\right] \\&= E\left[\left(\frac{1}{2}({M_b}^2-{M_a}^2) - \frac{1}{2}(N_b-N_a)\right)^2\right] \\ &= \frac{1}{4}E\left[({M_b}^2 - {M_a}^2)^2\right] - \frac{1}{2}E\left[\left({M_b}^2 - {M_a}^2 \right)\left(N_b - N_a\right)\right] + \frac{1}{4}E\left[(N_b - N_a)^2\right] \end{align}
However, I have troubles determining the former 2 terms.
For instance, $E\left[({M_b}^2 - {M_a}^2)^2\right] = E[{M_b}^4] + E[{M_a}^4] - 2E[{M_a}^2{M_b}^2]$
but what is $E[{M_a}^2{M_b}^2] \iff$ what is $E[{N_a}^2{N_b}^2]$?
Edit: I believe the trick is to use the independent increments of the Poisson point process.
For $a < b$, observe that ${M_b}^2 - {M_a}^2 = (M_{b-a} + M_a)^2 - {M_a}^2 = {M_{b-a}}^2 + 2M_{b-a}M_a$ so $E\left[({M_b}^2 - {M_a}^2)^2\right] = E[M_{b-a}^4] - 4E[M_{b-a}^3]E[M_a] + 4E[M_{b-a}^2]E[M_a^2] = \lambda(b-a) + 3\lambda^2(b-a)^2 + 4\lambda^2a(b-a)$
This procedure can be similarly repeated to obtain $E\left[\left({M_b}^2 - {M_a}^2 \right)\left(N_b - N_a\right)\right] = \lambda^2(b-a)^2 + \lambda(b-a)$ and $E[(N_b - N_a)^2] = E[N_{b-a}^2] = \lambda^2(b-a)^2 + \lambda(b-a)$
Thus $\textrm{var}\left[ \int_{a}^{b} M_{t-} \, dM_t\right] = \frac{1}{4}\lambda^2(b-a)^2 + \lambda^2a(b-a)$
Thanks!