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Poisson Counting Processes and Default-Time Densities

Article Quant Q&A · Author: Marco Pittella
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Summary

The document asks how to express the discounted probability of a default occurring between two times as an integral. It proposes integrating the discount factor against a default-time density and asks why that density has the form of an intensity multiplied by an exponential survival term, rather than a Poisson probability mass function.

The key distinction is between the count of events by a time and the time of the first event. For a Poisson process with time-varying intensity, the first-event density at time s is the intensity at s times the probability of surviving without an event up to s. The document poses this distinction but does not provide a complete derivation or discuss assumptions, such as whether intensity is deterministic or how discount rates relate to default risk.

Key ideas

  • A Poisson probability mass function describes the number of events in an interval, not the time of the first event.
  • For a process with deterministic intensity, the first-event density combines the instantaneous intensity with survival up to that time.
  • The discounted default expectation can be expressed by integrating the discount factor against the default-time density.
  • The document raises the distinction between event counts and default timing but does not resolve it fully.

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StatisticsRisk management
Full text
# Poisson distribution and counting process


# 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:

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.