Let $N_s$ a Poisson Process, is correct to say that the Lévy-Itô integral $\int_{[0,T]} e^{\beta N_s}\text{d}N_s = (1-e^{\beta N_s})/(1-e^{\beta})$?

65 Views Asked by At

For each measurable function $f$, a Poisson process defines a Lévy-Itô stochastic integral given by:

$$\int_{[0,T]} f(s)\ \text{d}N_s = \sum_{k=1}^{N_T} f(s_i)$$

where $s_1, s_2, \dots, s_{N_T}$ are the jump points of the process in the interval $[0,T]$. A possible problem appears when we apply this formula to a function $f(s) = \tilde{f}(N_s)$, because I am not sure how to check de measurability of the function.

If $f(s) = \tilde{f}(N_s) = e^{\beta N_s}$ is measurable, it seems clear that:

$$X_T = \int_{[0,T]} e^{\beta N_s}\ \text{d}N_s = e^{\beta N_{s_1}}+\dots + e^{\beta N_{s_{N_T}}}= e^{\beta}+\dots + e^{\beta(N_T-1)} = \frac{1-e^{\beta N_T}}{1-e^{\beta}}$$

For $\beta > 1$, we have an explosive stochastic process (it achieves infinite values in finite time). My question is: how can we prove if $f(s) = e^{\beta N_s}$ is an adapted measurable function?