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Working through an exercise in interest rate modelling and I have the following setup: $r_t = r_0 + \delta N_t$ where $\delta > 0$ and $\lambda > 0$ is the intensity of the Poisson pricess $N_t$. I want to calculate the bond price at time $0$, which is given by the formula $$ B(0,T) = \mathbb{E}_{\mathbb{P}^*}\left( e^{-\int_0^T r_u \,du}\right) = e^{-r_0T}\mathbb{E}_{\mathbb{P}^*}\left( e^{-\delta \int_0^T N_u\,du}\right). $$ where $\mathbb{P}^*$ is the risk-neutral probability measure. I haven't been dealing with Poisson processes lately and am not sure if there is a simple transformation or use of MGFs to get the second integral/expectation using properties of the Poisson process. I have tried to find some resources online but to no avail! Any help is appreciated. Thanks!

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  • $\begingroup$ Stochastic Fubini theorem might help? $\endgroup$ – Quantuple Jul 7 '16 at 7:44
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Let $$Y_t = \int_0^t N_u du$$

where $(N_t)_{t \geq 0}$ figures a Poisson process with intensity $\lambda$.

Using the stochastic Fubini theorem we have that: \begin{align} Y_T &= \int_0^T N_t dt \\ &= \int_0^T \int_0^t dN_u dt \\ &\color{lightgray}{= \int_0^T \int_0^T \mathbf{1}\{u \in [0,t]\} dN_u\ dt} \\ &\color{lightgray}{= \int_0^T \int_0^T \mathbf{1}\{u \in [0,t]\} dt\ dN_u} \\ &\color{lightgray}{= \int_0^T \int_0^T \mathbf{1}\{t \in [u,T]\} dt dN_u} \\ &= \int_0^T \int_u^T dt dN_u \\ &= \int_0^T (T-u) dN_u \\ &= \sum_{i=1}^{N_T} (T-T_i) \tag{0} \end{align} where $T_i$ figures the $i^{th}$ Poisson process jump time, which is Gamma distributed with integer parameter $i$ (also known as Erlang distribution see here Proposition 15.2 for instance). The jump times $T_i$ are therefore not unconditionally i.i.d. and $Y_t$ cannot be seen as a usual compound Poisson process.

But, conditioning helps! Indeed, conditionally on the number of jumps $N_T = n$ one can show that the $n$ jump times $\{T_i\}_{i=1}^n$ are actually independently and identically distributed uniformly over $[0,T]$ (see reference here top of page 443), that is \begin{align} T_i\ \vert\ N_T=n \ \ &\sim\ \ T U_i\\ \{U_i\}_{i=1}^n \ &\sim\ \ \mathcal{U}[0,1] \ \ \text{i.i.d.} \end{align}

We then have: \begin{align} \mathbb{E} \left[e^{-\delta Y_T} \vert \mathcal{F}_0 \right] &= \mathbb{E} \left[ \left. \exp\left(-\delta \sum_{i=1}^{N_T} (T-T_i)\right) \right\vert \mathcal{F}_0 \right] \tag{1} \\ &= \mathbb{E} \left[ \left. \mathbb{E} \left[ \left. \exp\left(-\delta \sum_{i=1}^{N_T} (T-T_i)\right) \right\vert \mathcal{F}_T \right] \right\vert \mathcal{F}_0 \right] \tag{2} \\ &= \mathbb{E} \left[ \left. \mathbb{E} \left[ \left. \exp\left(-\delta T \sum_{i=1}^{N_T} (1-U_i) \right) \right\vert \mathcal{F}_T \right] \right\vert \mathcal{F}_0 \right] \tag{3} \\ &= \sum_{n=0}^\infty \mathbb{E} \left[ \exp\left(-\delta T \sum_{i=1}^{n} \tilde{U}_i \right) \right] \mathbb{P}^*\left( N_T = n \right) \tag{4}\\ &= e^{-\lambda T} \sum_{n=0}^\infty \frac{(\lambda T)^n}{n!} \mathbb{E} \left[ \exp\left(-\delta T \sum_{i=1}^{n} \tilde{U}_i \right) \right] \tag{5} \\ &= e^{-\lambda T} \sum_{n=0}^\infty \frac{(\lambda T)^n}{n!} \prod_{i=1}^n \mathbb{E} \left[ \exp\left(-\delta T \tilde{U}_i \right) \right] \tag{6} \\ &= e^{-\lambda T} \sum_{n=0}^\infty \frac{(\lambda T)^n}{n!} \left(\mathbb{E} \left[ \exp\left(-\delta T \tilde{U}_1 \right) \right]\right)^n \tag{7} \\ &= \exp(-\lambda T) \exp\left(\lambda T \mathbb{E} \left[ \exp\left(-\delta T \tilde{U}_1 \right) \right] \right) \tag{8} \\ &= \exp\left(\lambda T \left(\mathbb{E} \left[ \exp\left(-\delta T \tilde{U}_1 \right) \right] - 1 \right) \right) \tag{9} \end{align} where $\tilde{U}_i = 1-U_i$ are also $\mathcal{U}[0,1]$ i.i.d. by construction and we used:

  1. Def. of $Y_T \vert \mathcal{F}_0 = \int_0^T N_u du = \sum_{i=1}^{N_T} (T-T_i)$ see $(0)$
  2. Tower property of conditional expectation
  3. Conditionally on $N_T$ the Poisson jump times are i.i.d. $\mathcal{U}[0,T]$
  4. Expectation over $N_T \vert \mathcal{F}_0$
  5. Def. of Poisson Process $(N_t)_{t \geq 0}$
  6. Properties of $\exp(.)$ + $\{\tilde{U}_i\}_{i=1}^n$ independent
  7. $\{\tilde{U}_i\}_{i=1}^n$ identically distributed
  8. Series expansion of $\exp(.)$
  9. Properties of $\exp(.)$

Now all you need to do is compute $Z = \mathbb{E}[\exp\left(-\delta T U \right)]$ with $U \sim \mathcal{U}[0,1]$:

\begin{align} Z &= \mathbb{E}[\exp\left(-\delta T U \right)] \\ &= \int_{0}^1 \exp\left(-\delta T u \right) \underbrace{p^*(u)}_{\mathbf{1}\{u \in [0,1]\}} du \\ &= \left[ \frac{\exp\left(-\delta T u \right)}{-\delta T} \right]_0^1 \\ &= \frac{1 - e^{-\delta T}}{\delta T} \end{align}

You will then be able to write: $$ e^{-r_0T}\mathbb{E}^{\mathbb{P}^*}\left[ e^{-\delta \int_0^T N_u\,du} \vert \mathcal{F}_0 \right] = \exp\left(-r_0T + \lambda T\left(\frac{1 - e^{-\delta T}}{\delta T}-1\right)\right) $$

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  • $\begingroup$ Wow thanks so much for that long type up! I suspected stochastic Fubini had to be used but wasn't sure of the details. Conditioning for Poisson (discrete) / Exponential (continuous) always seem to be a good approach for these type of things don't they. Thanks again!! $\endgroup$ – user89635 Jul 7 '16 at 12:01
  • $\begingroup$ You're welcome and yes indeed :) I am glad that it helped. I suspect there might be a less hand wavy way to do it though. $\endgroup$ – Quantuple Jul 7 '16 at 12:09
  • $\begingroup$ Just a quick one - the last expression does not need the expectation but all good apart from that. Cheers $\endgroup$ – user89635 Jul 8 '16 at 12:07
  • $\begingroup$ @user89635 You are absolutely right, thanks for proofreading. $\endgroup$ – Quantuple Jul 8 '16 at 12:17

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