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This is supposed to be for the derivation of a PDE for pricing a specific type of option, from the book 'Nonlinear Option Pricing' (Guyon).

The option delivers $g(\tau, X_{\tau})$ at time $\tau$ if $\tau < T$, or it delivers $g(T,X_T)$ at time $T$ if $\tau \geq T$. $\tau$ is the first time of jump for a Poisson process with intensity $\beta(t)$ (which is independent of information up to time $t$). So the current time is $t$ and the option maturity is $T$ (unless jump occurs earlier).

The $r(s,X_s)$ values below is just the rate used to discount the payoffs, so I'm not sure it's relevant for the question I have.

So the option price at time $t$ is $$\mathbb{E}{\large[}1_{\tau \geq T} * e^{-\int_{t}^{T}r(s,X_s)ds}g(T,X_T) + 1_{\tau < T} * e^{-\int_{t}^{\tau}r(s,X_s)ds}g(\tau, X_{\tau}) | X_t = x {\large]}$$

I understand up to that point. After that this equation is set equal to the following: $$\mathbb{E}{\large[}e^{-\int_{t}^{T}r(s,X_s) + \beta(s)ds}g(T,X_T) + \int_{t}^{T}\beta(s)g(s,X_s)e^{-\int_{t}^{s}r(u,X_u) + \beta(u)du}ds) | X_t = x {\large]}$$

I have no idea how the second term in the sum comes about. The first term I can see comes from the following (I think): $$\mathbb{E}{\large[}1_{\tau \geq T} * e^{-\int_{t}^{T}r(s,X_s)ds}g(T,X_T)| X_t = x {\large]} = \\ \mathbb{E}{\large[}1_{\tau \geq T} | X_t = x {\large]} * \mathbb{E}{\large[}e^{-\int_{t}^{T}r(s,X_s)ds}g(T,X_T)| X_t = x {\large]} = \\ \mathbb{E}{\large[}1_{\tau \geq T} {\large]} * \mathbb{E}{\large[}e^{-\int_{t}^{T}r(s,X_s)ds}g(T,X_T)| X_t = x {\large]} = \\ e^{-\int_{t}^{T}\beta(s)ds} * \mathbb{E}{\large[}e^{-\int_{t}^{T}r(s,X_s)ds}g(T,X_T)| X_t = x {\large]} = \\ \mathbb{E}{\large[}e^{-\int_{t}^{T}r(s,X_s) + \beta(s)ds}g(T,X_T) | X_t = x {\large]} $$

So basically I am wondering how to get from $$\mathbb{E}{\large[}1_{\tau < T} * e^{-\int_{t}^{\tau}r(s,X_s)ds}g(\tau, X_{\tau}) | X_t = x {\large]}$$ to $$\mathbb{E}{\large[}\int_{t}^{T}\beta(s)g(s,X_s)e^{-\int_{t}^{s}r(u,X_u) + \beta(u)du}ds | X_t = x {\large]}$$ assuming that I calculated the other part properly.

Thanks a lot for the help!

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In the book, it is assumed that $\tau$ is the first time of jump of the Poisson process $N_t$ with deterministic intensity $\beta(t) >0$, independent of the filtration $(\mathcal{F}_t)$. Then, for any $ u > t \ge 0$, \begin{align*} \mathbb{P}(\tau > u \mid \tau > t) &= e^{-\int_t^u \beta(s) ds}. \end{align*} That is, the density of $\tau$, conditional on $\tau > t$, is given by $\beta(u) e^{-\int_t^u \beta(s) ds}$, for $u > t$.

Let $\mathcal{F}_{\infty} = \cup_{t\ge 0} \mathcal{F}_t$. Then, for any Borel set $A$, based on the independence condition of $\tau$ and $\mathcal{F}_{\infty}$, \begin{align*} &\ \mathbb{E}\left(\left(\mathbb{I}_{\tau \geq T} \, e^{-\int_{t}^{T}r(s,X_s)ds}g(T,X_T) + \mathbb{I}_{\tau < T}\, e^{-\int_{t}^{\tau}r(s,X_s)ds}g(\tau, X_{\tau})\right) \mathbb{I}_{X_t \in A}\, \big|\, \tau > t \right)\\ =&\ \mathbb{E}\bigg(\bigg(e^{-\int_t^T \beta(s) ds} e^{-\int_{t}^{T}r(s,X_s)ds}g(T,X_T) \\ &\qquad\quad + \int_t^T e^{-\int_{t}^{u}r(s,X_s)ds}g(u, X_{u})\, \beta(u)\, e^{-\int_t^u \beta(s) ds} du\bigg) \mathbb{I}_{X_t \in A}\bigg)\\ =&\ \mathbb{E}\left(\left(e^{-\int_t^T r(s,X_s) + \beta(s) ds} g(T,X_T) + \int_t^T e^{-\int_{t}^{u}\beta(s) + r(s,X_s)ds}g(u, X_{u})\, \beta(u)\, du\right) \mathbb{I}_{X_t \in A}\right). \end{align*} Therefore, \begin{align*} &\ \mathbb{E}\left( 1_{\tau \geq T} * e^{-\int_{t}^{T}r(s,X_s)ds}g(T,X_T) + 1_{\tau < T} e^{-\int_{t}^{\tau}r(s,X_s)ds}g(\tau, X_{\tau})\, \big|\, \tau > t, X_t = x \right)\\ =&\ \mathbb{E}\left(e^{-\int_t^T r(s,X_s) + \beta(s) ds} g(T,X_T) + \int_t^T \beta(u)\, e^{-\int_{t}^{u}\beta(s) + r(s,X_s)ds}g(u, X_{u})\, du \, \big|\, X_t = x \right). \end{align*}

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  • $\begingroup$ Where does the indicator function for $X_t \in A$ come from? It looks like it's related to the condition $X_t = x$ but I've never seen something done like that before. Thanks! $\endgroup$ – Slade Jan 30 at 21:21
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    $\begingroup$ Yes. That is the formal definition for the expectation conditional on $X_t=x$. $\endgroup$ – Gordon Jan 30 at 21:28
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    $\begingroup$ See Definition 4 on Page 220 of this book by Shiryaev. $\endgroup$ – Gordon Jan 30 at 21:35
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    $\begingroup$ It is the expectation by treating it as a function of $\tau$, while others are held fixed. That is, something like $E(F(\tau))$. $\endgroup$ – Gordon Jan 30 at 21:40
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    $\begingroup$ Note that $X_t=X(\omega, t)$, where $\omega$ is a random sample, That is, $X_{\tau} = X(\omega, \tau)$, where $\omega$ is independent of $\tau$. Then you can use the independence as in Lemma 2.3.4 of this book by Shreve. $\endgroup$ – Gordon Jan 31 at 18:35
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The density of the random variable $\tau$ is like you pointed out;

$$\phi(s):=E[\delta(\tau-s)|\tau \geq t] = e^{-\int_t^s\beta(u)du}\beta(s)$$

where we called $\delta$ the Dirac density function ($P(X=x):=E[\delta(X-x)]$ for any random variable eg)

So you just need to plug this explicitly in the expectation to get the result (exactly same way as what you did to show that $E[1_{\tau>T}]=e^{-\int_t^T\beta(u)du}$ )

In general you can write for any function $h$

$$h(\tau,X_\tau) = \int ds \delta(\tau -s)h(s,X_s)$$

so taking expectation one has (taking into account the independence property of $\tau$ from previous information:

$$E[h(\tau,X_\tau)|X_t=x] = \int ds E[\delta(\tau -s)h(s,X_s)|X_t=x]$$ $$E[h(\tau,X_\tau)|X_t=x] = \int ds E[\delta(\tau -s)]E[h(s,X_s)|X_t=x] = \int ds \phi(s)E[h(s,X_s)|X_t=x]$$

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  • $\begingroup$ I still can't see how the function $g(\tau, X_{\tau})$ becomes a part of the integral. Would you mind showing some of the steps? Thanks $\endgroup$ – Slade Jan 30 at 3:23
  • $\begingroup$ How come $\delta(\tau - s)$ is conditionally independent of $h(s,X_s)$? Isn't information about the value of $\tau$ affecting the value of $h(s,X_s)$. In the part that I was able to do before, $g(T,X_T)$ was set regardless of the value taken on by $\tau$, but this doesn't seem to be the case with $\delta(\tau - s)h(s,X_s)$ $\endgroup$ – Slade Jan 30 at 5:28
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    $\begingroup$ Pleasd specify clearly the assumptions made on $\tau$ as done in your book then. From your question it seemed you assumed $X$ and $\tau$ are independent. $\endgroup$ – Ezy Jan 30 at 10:58
  • $\begingroup$ $X$ and $\tau$ are independent but isn't a function of $\tau$ ($h(\tau, X_\tau)$) not independent of $\tau$? $\endgroup$ – Slade Jan 30 at 18:21
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    $\begingroup$ @Slade i showed in my answer how to extract the dependency in $\tau$ by using the dirac function. I believe my argument is complete with the assumption of independence. $\endgroup$ – Ezy Jan 30 at 18:29

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