From the fundamental asset pricing theorem, we know that in the absence of arbitrage opportunities, the present value of an asset paying $\Psi(X)$ at maturity time $T$ is given by: \begin{equation} V_t = \beta(t)\mathbb{E}^{\mathbb{Q}}\Big[\frac{\Psi(X_T)}{\beta(T)}\Big |\mathcal{F}_t\Big] = \mathbb{E}^{\mathbb{Q}}\Big[D(t, T)\Psi(X_T) \Big| \mathcal{F}_t\Big] \end{equation} where the expectation is under the risk-neutral pricing measure $\mathbb{Q}$ that is equivalent to the "reference" measure $\mathbb{P}$, $\beta(.)$ stands for the saving account, $D(t, T)=\frac{\beta(t)}{\beta(T)}$ for the so-called discount factor, and $\mathcal{F}_t$ represents all infiomation available at time $t$ to an agent. According to this theory, the present price of a financial contact is the expectation of the future cash flow disocunted at risk-free rate $r(t)$. Here, $\mathbb{Q}$ is the risk-nuetral measure assocuited with the numerair process $\beta(t)$ under which any asset price disocunted by $\beta(t)$ is a Martinagle.The well-know Girsanove's Theorem tells us how the radon-Nikodym derivative process $\frac{d\mathbb{Q}}{d\mathbb{P}}$ looks like.

From the change of numeraire process technique, we learn how to change the measure. If we want to use another numeraire process such as a zero-coupon bond price, the price is computed under the T-forward measure $\mathbb{Q}^T$. Therefore, this measure is associated with the price of a zero-coupon bond with maturity time $T$. We have that \begin{equation} V_t = P(t, T)\mathbb{E}^{\mathbb{Q}^T}\Big[\frac{\Psi(X_T)}{P(T, T)}\Big |\mathcal{F}_t\Big] = P(t, T)\mathbb{E}^{\mathbb{Q}^T}\Big[ \Psi(X_T)\Big |\mathcal{F}_t\Big] \end{equation} where $P(t, T)$ is the zero-coupon bond price at time $t$ maturing at time $T$, for which $P(T, T)= 1$. We know that the Radon-Nikodym derivative $\frac{d\mathbb{Q}^T}{d\mathbb{Q}}|_{\mathcal{F}_T} = \frac{\beta(t)P(T, T)}{\beta(T)P(t, T)}$. In fact, this is how we get the above relation: \begin{equation} V_t = \mathbb{E}^{\mathbb{Q}}\Big[D(t, T)\Psi(X_T) \Big| \mathcal{F}_t\Big] = \mathbb{E}^{\mathbb{Q}^T}\Big[\Big(\frac{d\mathbb{Q}^T}{d\mathbb{Q}}|_{\mathcal{F}_T}\Big)^{-1}D(t, T)\Psi(X_T) \Big| \mathcal{F}_t\Big] \\ = \mathbb{E}^{\mathbb{Q}^T}\Bigg[ \Bigg(\frac{\beta(t)P(T, T)}{\beta(T)P(t, T)}\Bigg)^{-1}D(t, T)\Psi(X_T) \Big| \mathcal{F}_t\Bigg] \\ =P(t, T)\mathbb{E}^{\mathbb{Q}^T}\Big[ \Psi(X_T)\Big |\mathcal{F}_t\Big] \end{equation} Now, I amount to my main problem. Suppose that we can find a class of risk-neutral measure $\mathbb{Q}^{(B)}$ whose Radon-Nikodym derivative with respect to physical measure $\mathbb{P}$ is characterized as follows \begin{equation} \frac{d\mathbb{Q}^{(B)}}{d\mathbb{P}} = exp\Big\{ \sum_{k=1}^{N(t)}B(Y_k) - \lambda t\Big(\mathbb{E}^{\mathbb{P}}[e^{B(Y_1)}-1]\Big) \Big\} \end{equation}
where $B(.)$ is a Borel measurable function, $N(t)$ is a Poisson process with intensity $\lambda$, and $Y_k$ are positive random variables. In such a case, how can I define the price under measure $\mathbb{Q}^{(B)}$? Please consider two situations: a complete market and an incomplete market. For example, assume that the former refers to the situation where the payment is linked to a tradeable asset in the market while the latter refers to the situation where the cash flow payment is linked to an insurance risk, which is not tradeable. Please do not consider the form of Radon-Nikodym derivative $\frac{d\mathbb{Q}^{(B)}}{d\mathbb{P}}$. I aim to know if the pricing formula should look like the first relation but under measure $\mathbb{Q}^{(B)}$.

  • $\begingroup$ Risk neutral measure (as far as I can tell) usually implies that its numeraire is the money market account $\exp(\int_0^tr_s\,ds)\,.$ If you say that ${\mathbb Q}^{(B)}$ is 'a (class of) risk neutral measure' is that what you assume? If so the pricing formula for any payoff $H$ should simply be $V=\mathbb E[\exp(\int_0^Tr_s\,ds)H]$ as always. Regardless, how a risk neutral measure is related to the physical measure $\mathbb P$ is never a concern when one wants to know the pricing formula only. $\endgroup$
    – Kurt G.
    Sep 11, 2021 at 9:56
  • $\begingroup$ Yes, this is what I think. But in the case that the market is incomplete, i.e., there is an infinite number of pricing measures, so the final result can be impacted by the definition of our risk-neutral measure. Do you agree with me? $\endgroup$
    – user53249
    Sep 11, 2021 at 11:13
  • $\begingroup$ If we assume that $\mathbb{Q}^{(B)}$ is just a new measure equivalent to the real-world measure $\mathbb{P}$ (we say nothing about the fact that $\mathbb{Q}^{(B)}$ is a risk-neutral measure), then what will happen? $\endgroup$
    – user53249
    Sep 11, 2021 at 11:19
  • $\begingroup$ First, I agree that when you want to model an incomplete market you will not necessarily have a unique risk neutral measure. Secondly, if you assume ${\mathbb Q}^{(B)}$ is just a new measure I take from there that your numeraire is no longer the money market account. What is it ? In fact it doesn't matter for the form of the pricing formula: Let's call your numeraire $M(t)$. Then the pricing formula is $V={\mathbb E}_{{\mathbb Q}^{(B)}}[H/M(T)]$ as always. $\endgroup$
    – Kurt G.
    Sep 11, 2021 at 11:29
  • $\begingroup$ Ok, I got it. Thank you, bro. Just to complete your last equation (using Martingale property for the discounted price process): \begin{equation} V_t= M(t)\mathbb{E}^{\mathbb{Q}^{(B)}}[H/M(T)] \end{equation} $\endgroup$
    – user53249
    Sep 11, 2021 at 11:36


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