# Conditional Expectation with Indicator Functions for Poisson Process First Jump Time (Option Pricing PDE)

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!

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*}

• 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! – Slade Jan 30 '19 at 21:21
• Yes. That is the formal definition for the expectation conditional on $X_t=x$. – Gordon Jan 30 '19 at 21:28
• See Definition 4 on Page 220 of this book by Shiryaev. – Gordon Jan 30 '19 at 21:35
• It is the expectation by treating it as a function of $\tau$, while others are held fixed. That is, something like $E(F(\tau))$. – Gordon Jan 30 '19 at 21:40
• 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. – Gordon Jan 31 '19 at 18:35

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:

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

• 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 – Slade Jan 30 '19 at 3:23
• 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 $\delta(\tau - s)h(s,X_s)$ – Slade Jan 30 '19 at 5:28
• 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. – Ezy Jan 30 '19 at 10:58
• $X$ and $\tau$ are independent but isn't a function of $\tau$ ($h(\tau, X_\tau)$) not independent of $\tau$? – Slade Jan 30 '19 at 18:21
• @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. – Ezy Jan 30 '19 at 18:29