Dynamic Hedging of Portfolio Credit Risk in a Markov Copula Model

Abstract

We devise a bottom-up dynamic model of portfolio credit risk where instantaneous contagion is represented by the possibility of simultaneous defaults. Due to a Markovian copula nature of the model, calibration of marginals and dependence parameters can be performed separately using a two-step procedure, much like in a standard static copula setup. In this sense this solves the bottom-up top-down puzzle which the CDO industry had been trying to do for a long time. This model can be used for any dynamic portfolio credit risk issue, such as dynamic hedging of CDOs by CDSs, or CVA computations on credit portfolios.

Appendix

Here is the proof of Proposition 2.1. For I⊆N, we define the filtration $\mathbb{X}^{I}=(\mathcal{X}_{t}^{I})_{t\geq0}$ as the initial enlargement of $\mathbb{X}$ by the τ _i for i∈I, i.e. for every t:

$$ \mathcal{X}_t^{I}= \mathcal{X}_{t} \vee \bigvee_{i\in I} \sigma(\tau_i ). $$

By an application of Lemma 2.5 in [14], writing J=N∖I for every I⊆N, we obtain:

(8)

Now, in the common shocks model of this paper, writing

we have on {I _t =I} (and therefore $\{\tau_{i} = \tau^{J}_{i}, i\in I\}$):

where $\mathbb{P} (E_{Y}> \varLambda_{{ \bar{t}_{Y} }} , Y\in \mathcal{Y}_{J} \mid\bar{\mathcal{X}}^{I}_{ \bar{t} } )= \exp (-\sum_{Y\in{ \mathcal{Y}_{J} }} \varLambda^{Y}_{{ \bar{t}_{Y} }} )$ in the last identity holds by independence of $(E_{Y})_{Y\in{ \mathcal{Y}_{J} }}$ from $\bar{\mathcal{X}}^{I}_{ \bar{t} }$ and by $\bar{\mathcal{X}}^{I}_{ \bar{t} }$- (in fact, $\mathcal{X}_{ \bar{t} }$-) measurability of $(\varLambda^{Y}_{ \bar{t}_{Y} })_{Y\in{ \mathcal{Y}_{J} }}$.

Note moreover that we have on {I _t =I}:

$$\mathcal{X}_t \subseteq \bar{\mathcal{X}}^I_t \subseteq \mathcal{X}_{t} \vee \bigvee_{Y\in\bar{\mathcal{Y}}_J} \sigma(E_Y), $$

where the $E_{Y}, Y\in\bar{\mathcal{Y}}_{J}$ are independent of $\mathbb{X}$. We thus have on {I _t =I}:

The Markov property of X _t finally yields that

Plugging this into (8) yields that

which is (3). Setting all t _j but t _i equal to 0, we deduce (4) from (3), for t _i ≥t.