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: enter image description here

  • $\begingroup$ Why not convert your handwriting to latex? $\endgroup$
    – Gordon
    Jun 27 '19 at 13:07
  • $\begingroup$ @Gordon Those aren't my notes but notes of the professor. $\endgroup$ Jun 27 '19 at 13:25
  • 1
    $\begingroup$ 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. $\endgroup$
    – Gordon
    Jun 27 '19 at 13:47
  • $\begingroup$ By the way, you can have a look of this question. $\endgroup$
    – Gordon
    Jun 27 '19 at 16:52

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