# Poisson distribution and counting process

Let $$\begin{Bmatrix} N_t \end{Bmatrix}_{(t\in[0,T])}:=\mathbb{I}_{(\tau \leq T)}:=k, \forall t \in [\tau_{k}\leq \tau_{k+1})\sim \mathrm{Po}(\lambda_{t}:=\int_{0}^{t}\lambda_{s}ds<+\infty)$$ a counting process with $$\tau$$ generical default time. I have to express $$\mathbb{E}^{\mathbb{Q}}[\mathbb{I}_{(t\leq\tau\leq T)}e^{-\int_{t}^{\tau}r_sds}]$$ in an integral form.

Keeping in mind that for continuous random variables we have $$\mathbb{E}[X]:=\int_{\mathbb{R}}xf(x)dx$$, i can rewrite the expected value under neutrality measure $$\mathbb{Q}$$ in an integral form defined over the entire space (which in my case is $$(t,T)$$, that is the range of indicator function). The random variable which will form the integral will be right the indicator function, that indicates the probability a default may occur in the considered range. But at the same time, $$\mathbb{I}$$ describes the process that counts the numbers of potential defaults that may occur in $$(t,T)$$, ergo a counting process $$N_t$$. And because PDF of a counting process is a Poisson distribution, professor says, that we can write:

$$\mathbb{E}^{\mathbb{Q}}[\mathbb{I}_{(t\leq\tau\leq T)}e^{-\int_{t}^{\tau}r_sds}]\Rightarrow =\int_{t}^{T}e^{-\int_{t}^{s}r_udu}f(s)ds=[\int_{t}^{T}e^{-\int_{t}^{s}r_udu}\gamma(s)e^{-\int_{0}^{s}\gamma(u)du}ds]$$

1) Since PDF of a Poisson Distribution is $$\frac{e^{-\lambda} \lambda^{k}}{k!}$$, where'd that $$\gamma(s)e^{-\int_{0}^{s}\gamma(u)du}ds$$ come from?

2) Isn't $$\lambda$$ the parameter of a $$\mathrm{Po}$$?

Thanks who's going to help me!

N.B.: For a greater clearness look below:

• Why not convert your handwriting to latex? – Gordon Jun 27 '19 at 13:07
• @Gordon Those aren't my notes but notes of the professor. – Marco Pittella Jun 27 '19 at 13:25
• You can still use latex to type here to make it more readable -- you should have the closest idea for your prof's notations than anyone else. – Gordon Jun 27 '19 at 13:47
• By the way, you can have a look of this question. – Gordon Jun 27 '19 at 16:52