The expectation over an infinitesimal time interval

208 Views Asked by At

I have a deduction which I would like to formalize (I suppose with some additional measure theory):

Let $N(t)\sim Pois(\lambda_t)$, where $\lambda_t$ is stochastic (we are thus looking at a Cox process). Define $\mathcal{G}_t$ as the filtration taking into account all knowledge up til time $t$, then

\begin{equation} \mathbb{E}\left[dN(t)\bigg|\mathcal{G}_t\right]=1-e^{-\lambda_t dt}=(1-(1-\lambda_tdt))=\lambda_tdt. \end{equation}

The reasoning behind the deduction is that $dN(t)$ is the probability that an event occurs in the infinitesimal time interval $dt$. The probability of no default within the time interval $[a,b]\in\mathbb{R}$ is given by

\begin{equation} \mathbb{E}\left[e^{-\int_a^b\lambda_sds}\right] \end{equation}

and the intuition behind the deduction makes sense. However I think it is not formally correct. I suppose we cannot simply work like this with terms such as $dt$ and $dN(t)$. Would you have a suggestion on how to formalize this?

Thank you!!

1

There are 1 best solutions below

1
On BEST ANSWER

See e.g. slide 6 here for a precise martingale characterization. For example, if $M_t := N_t - \Lambda_t$ with $$ \Lambda_t:=\int_0^t \lambda_s\mathrm ds $$ is a martingale, then $$ \begin{align} \Bbb E\left[N_{t+h} - N_t|\mathcal G_t\right] &= \Bbb E\left[M_{t+h} - M_t|\mathcal G_t\right] + \Bbb E\left[\Lambda_{t+h} - \Lambda_t|\mathcal G_t\right] \\ &= \int_t^{t+h} \Bbb E\left[\lambda_s|\mathcal G_t\right]\mathrm ds \approx \lambda_t\cdot h \end{align} $$ for $h$ small enough, which justifies the idea behind this formal definition. I think any book on counting processes would have related material, however I am not really aware of particular references.