Change of measure's impact on parameter value

Question

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:

1. There would be any impact on $\lambda$ when passing from $\mathbb{P}$ to $\mathbb{Q}$?
2. 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.

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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.

Answer 1

Theory

Define a single currency economy with $N+1$ tradable assets. Further assume that individually investing in each of these assets constitutes a self-financing strategy. Simply put, this amounts to considering that the assets of our model economy do not distribute capital in the form of dividends or coupons.

Let $S_1, \dots, S_N$ represent $N$ risky assets and $S_0$ a risk-free one. A common practice is to pick $S_0$ as the reference asset to define the pricing framework. We say that $S_0$ is chosen as the pricing "numéraire" (but we could pick any other traded asset price as numéraire without loss of generality).

Pricing theory then tells us that, in the absence of arbitrage opportunities, the processes $$ S_{1,t}/S_{0,t}, \dots, S_{N,t}/S_{0,t} $$ should all emerge as martingales under a measure $\Bbb{Q}$ which is equivalent to the physical measure $\Bbb{P}$ under which we observe the realisations of the prices of the assets of our model economy.

Intuitively, this means that the relative value of all the assets with respect to some reference asset (the numéraire) need to be martingales which is a model of a "fair evolution" of the market.

Because we picked $S_0$ (risk-free asset) as numéraire, the measure $\Bbb{Q}$ is known as the risk-neutral measure.

Your problem

Suppose a principal $N=1$ to keep notations uncluttered. You have 2 traded assets:

• Risk-free asset, $dP_t/P_t = r dt$
• Instrument to be priced, $V_t$, characterised by its payout, which is 1 unit of currency and which occurs either at maturity $T$ or when a certain event occurs at $\tau < T$. By absence of arbitrage this means that $V_\tau=1$ if early redemption occurs and $V_T=1$ otherwise.

Applying the pricing theory described earlier $V_t/P_t$ (or actually here the stopped process $V_t^\tau/P_t^\tau$ but this is a technicality) should be a martingale (see also optimal stopping theorem) i.e. \begin{align} V_0 &= \Bbb{E}_0^\Bbb{Q} \left[ V_\tau/P_\tau \mathbb{I}_{\{\tau \leq T\}} + V_T/P_T \mathbb{I}_{\{\tau > T\}} \right] \\ &= \Bbb{E}_0^\Bbb{Q} \left[ e^{-r\tau} \mathbb{I}_{\{\tau \leq T\}} + e^{-rT} \mathbb{I}_{\{\tau > T\}} \right] \end{align}

Remarks

As far as your side questions go, all depends on where the counting process $N_t$ intervenes. In your case, it is used to specify a stopping time. Hence it is exogenous to the specification of the model economy. This is why we reach the conclusion above that $\lambda$ is the same under $\Bbb{P}$ and $\Bbb{Q}$.

However, one could consider that your payoff is "degenerate", in the sense that we could build a more general version of it which would deliver a payoff proportional to the value of some other asset $S$ rather than just $1$ unit of currency. From there, let us further distinguish 2 situations:

• Situation 1: $dS_t/S_t = \mu dt + \sigma dW_t^\Bbb{P}$. Applying the theory described earlier gets to a very similar pricing equation namely: $$ V_0 = \Bbb{E}_0^\Bbb{Q} \left[ e^{-r\tau} S_{\tau} \mathbb{I}_{\{\tau \leq T\}} + e^{-rT} S_T \mathbb{I}_{\{\tau > T\}} \right] $$ and similarly, because now $S$ is also a traded asset, under $\Bbb{Q}$ you'll have: $$ dS_t/S_t = r dt + \sigma dW_t^\Bbb{Q} $$ which the basically the SDE form of the relationship (see martingale representation theorem + Itô's lemma) $$ S_t/P_t \text{ is a } \Bbb{Q} \text{ martingale } $$
• Situation 2: $dS_t/S_t = \mu dt + \sigma dW_t^\Bbb{P} + J dN_t^\Bbb{P}$. Now you see that the Poisson process becomes part of the specification of the asset price dynamics in the model economy (endogenous). In that case, because $S_t/P_t$ should be a $\Bbb{Q}$ martingale in our pricing framework, we have: $$ dS_t/S_t = (r-\lambda\Bbb{E}[J])dt + \sigma dW_t^\Bbb{Q} + J dN_t^\Bbb{Q} $$ so indeed the drift of the risky asset will have changed (Girsanov theorem -> both the drift parameter of the Brownian motion and the intensity parameter of the Poisson process are altered by the change of measure). And, still, $$ V_0 = \Bbb{E}_0^\Bbb{Q} \left[ e^{-r\tau} S_{\tau} \mathbb{I}_{\{\tau \leq T\}} + e^{-rT} S_T \mathbb{I}_{\{\tau > T\}} \right] $$ But now $S_t$ has a different dynamics than in situation 1.