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Suppose I have $dP_t^i = (r^i + h_i^{\mathbb{P}})P_t^i dt - P_{t-}^i dH_t^i$ where $H_i(t) = \mathbb{1}_{\tau_t \leq t}$ denotes a default indicator process of i. $\tau_i$ is the default time and $h_i$ is some constant.

Let $w_t^{i, \mathbb{P}} := H_t^i - \int_0^t (1-H_u^i) h_i^{\mathbb{P}}du$ a jump martingale.

Then the dynamics under the new measure $\mathbb{Q}$ is given by $dP_t^i = r_D P_t^i dt - P_{t-}^i dw_t^{i,Q}$ with $h_i^{\mathbb{Q}} = r^i - r_D + h_i^{\mathbb{P}}$

The Radon-Nikodym density is not relevant here. They say that they apply Ito in the first formula for $dP_t^i$ and get the new dynamics with the new measure.

I don't get the new dynamics. Can somebody help?

SinusK
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  • Perhaps I'm just not getting the point, but this question seems to be missing some essential information about how the new measure is related to the first. Also what is $r_D$? – spaceisdarkgreen Jun 30 '17 at 00:08
  • I think what's going on is you need the processes to have drift $r_D$ in the risk neutral measure so the default rates are adjusted to compensate. – spaceisdarkgreen Jun 30 '17 at 00:18
  • Thank you for your answer. $r_D$ is the risk free rate. here is the link to the paper: http://users.wpi.edu/~ssturm/Papers/BCS1_rev_SSRN.pdf
    On page 8 and 9 there are the equations. The Radon Nikodym density is not relevant for the calculation of the dynamics.     – SinusK Jun 30 '17 at 00:30
  • I think one must apply Ito's formula to the jump process and replace $dH_t^i$ with the new process $dw_t$. – SinusK Jun 30 '17 at 00:34
  • how does one "apply ito's lemma to the jump process"? Equivalent measures to the physical one for the drift-jump process are found by changing the default intensity. For the pricing measure they will select the intensity so that the return is the discount rate. This means at the end of the day it's that return plus a martingale (the suitably compensated jump process). I think that's all there is to it but could easily be wrong. – spaceisdarkgreen Jun 30 '17 at 01:04
  • There is Ito's Lemma for Levy Jump processes. But I don't know if this is the correct way. Your idea is right but how to get the new dynamics? – SinusK Jun 30 '17 at 10:31

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