This is a follow-up question on Price of a prepayment-based claim.
Consider a zero-coupon bond of maturity $T$ with price $P_0$ for which the borrower can reimburse the principal $N$ at any time $\tau$ between $0$ excluded and $T$ included. Assuming a constant risk-free interest rate $r$, the price of this claim is given by the risk-neutral expectation of its payoff:
$$P_0 = \mathbb{E}^{\mathbb{Q}}\left[N\left(\mathbb{I}_{\{\tau \leq T\}}e^{-r\tau}+\mathbb{I}_{\{\tau>T\}}e^{-rT}\right)\right]$$
To model the stopping time $\tau$, we introduce a homogenous Poisson process $N(t)$ parameterised by $\lambda > 0$ such that for $t \in \mathbb{R}_+^*$ and $n \in \mathbb{N}$:
$$ \mathbb{P}(N(t) = n) = \frac{(\lambda t)^n}{n!}e^{-\lambda t}$$
Let $(\mathcal{F}_t)_{t \geq 0}$ be the natural filtration associated to the process $N(t)$. The stopping time $\tau$ with respect to filtration $(\mathcal{F}_t)_{t \geq 0}$ is then defined as:
$$ \tau = \min \{t>0 : N(t)>0 \}$$
We derive the stopping time distribution:
$$ \begin{align} & \mathbb{P}(\tau > t) = \mathbb{P}(N(t) = 0) = e^{-\lambda t} \\[8pt] & \mathbb{P}(\tau \leq t) = 1 - e^{-\lambda t} \end{align} $$
Hence as expected:
$$ \tau \sim \mathcal{E}(\lambda) $$
Now, my question concerns the effect of a change of measure $-$ from the real-world probability $\mathbb{P}$ to the risk-neutral measure $\mathbb{Q}$ $-$ on the parameter $\lambda$. For example, under the Black-Scholes model, the canonical change of measure modifies the asset $S_t$ drift from $\mu$ to $r$ but it does not have any impact on the diffusion coefficient $\sigma$.
I am familiar with the theory but not very familiar with the practicalities of the change of measure technique, so I do not really know how to tackle this issue here. Could someone please explain whether:
- There would be any impact on $\lambda$ when passing from $\mathbb{P}$ to $\mathbb{Q}$?
- Is my pricing problem sufficiently specified above or should there be any additional information to answer question 1?
Note: please if you think there is an impact do not post the derivation but a hint on how to proceed.
A first try:
The fundamental property of the risk-neutral measure is that (Brigo & Mercurio, 2007):
The price of any asset divided by a reference positive non dividend-paying asset (called numeraire) is a martingale (no drift) under the measure associated with that numeraire.
Now, in my setting I am not quite sure what would be considered the "asset": my guess is that it would be the Poisson process $N(t)$. Hence, letting $s<t$ we would have:
$$ \begin{align} \mathbb{E}^{\mathbb{Q}}\left[e^{-rt}N(t)|\mathcal{F}_s\right] & = \mathbb{E}^{\mathbb{Q}}\left[e^{-rt}N(s)|\mathcal{F}_s\right] + \mathbb{E}^{\mathbb{Q}}\left[e^{-rt}(N(t)-N(s))|\mathcal{F}_s\right] \\[10pt] & = e^{-rt}N(s) + \mathbb{E}^{\mathbb{Q}}\left[e^{-rt}(N(t)-N(s))\right] \\[10pt] & = e^{-rt}N(s) + e^{-rt}\lambda(t-s) \\[10pt] & = e^{-rt}\left(N(s) + \lambda (t-s)\right) \end{align} $$
Where the second step follows from the independent increments property of the Poisson process. Given that we want:
$$ \mathbb{E}^{\mathbb{Q}}\left[e^{-rt}N(t)|\mathcal{F}_s\right] = e^{-rs}N(s) $$
I have the impression that the implied - i.e. under the risk-neutral measure - lambda parameter $\tilde{\lambda}$ should be:
$$ \tilde{\lambda} \equiv \tilde{\lambda}(r,s,t) = N(s)\frac{e^{r(t-s)}-1}{t-s} $$
A few questions come to mind:
- Is my reasoning correct here?
- Does it make sense to define the parameter in terms of the process? My guess is no given that $\tilde{\lambda}(r,0,t) = 0$ because $N(0)=0$...
- Does it pose a problem to go from a constant $\lambda$ under the real-world measure $\mathbb{P}$ to a functional $\tilde{\lambda}(r,s,t)$ under the real-world measure $\mathbb{Q}$?
A second try:
Using the same logic, I now leverage the property than the compensated Poisson process is a martingale:
$$ \begin{align} & N_c(t) \equiv N(t) - \lambda t \\[10pt] & \mathbb{E}^{\mathbb{Q}}\left[N_c(t)|\mathcal{F}_s\right] = N_c(s) \end{align} $$
However when I discount I (obviously) get:
$$ \mathbb{E}^{\mathbb{Q}}\left[e^{-rt}N_c(t)|\mathcal{F}_s\right] = e^{-rt}N_c(s) $$
And hence I am stuck again.