Expected Value of Mean-Reverting Jump Process

Question

I cant see the link between my method of calculation and the method done in the book Cartea and Jaimungal (Algorithmic and High Frequency Trading, page 220.) We have a mean-reverting process $$d\mu_t=-k\mu_tdt+\eta_{1+N_{t^-}}dN_t$$ which has solution $$\begin{align}\mu_t&=e^{-kt}\mu_0+\int^t_0e^{-k(t-u)}\eta_{1+N_{u^-}}dN_u \\&=e^{-kt}\mu_0+\sum_{m=1}^{N_t} e^{-k(t-\tau_m)}\eta_m. \end{align}$$ where $N_t$ is a Poisson process with intensity $\lambda$ and $\eta_i$ are i.i.d. and independent of $N_t$, $k>0$ and $\tau_m$ denote the Poisson arrival times.

Now they calculate the expected value via the integral to achieve $$\mathbb{E}[\mu_t]=e^{-kt}\mu_0+\frac{\lambda}{k}\mathbb{E}[\eta_1](1-e^{-kt}).$$

I try to do the same via the summation definition which isn't quite working.

The arrival times are Gamma($m,\lambda$) distributed and the MGF of a Gamma r.v. $X$ gives $\mathbb{E}[e^{kX}]=(\frac{\lambda}{\lambda-k})^m$.

Hence, $$\begin{align} e^{kt}\mathbb{E}[\mu_t] -\mu_0 &= \mathbb{E}\bigg[\sum_{m=1}^{N_t} e^{k\tau_m}\eta_m\bigg]\\ &= \mathbb{E}\bigg[\mathbb{E}\bigg[\sum_{m=1}^{N_t} e^{k\tau_m}\eta_m\bigg]|N_t\bigg]\\ &= \mathbb{E}\bigg[\sum_{m=1}^{N_t} \mathbb{E}\Big[e^{k\tau_m}\Big]\mathbb{E}\Big[\eta_m\Big]|N_t\bigg]\\ &= \mathbb{E}\Big[\eta_1\Big] \cdot \mathbb{E}\bigg[\sum_{m=1}^{N_t} \mathbb{E}\Big[e^{k\tau_m}\Big]|N_t\bigg] \end{align}$$

Now if I substitute the result for the MGF of a gamma function into this and work out the geometric sum, I find that I don't get the correct answer. I am certain the the geometric summation and MGF steps are correct and the final step after this would be to work out the unconditional expectation of the Poisson process, but I believe there must be an error somewhere here, as the other steps seem fine.

For completeness, the geometric sum becomes $$\frac{\lambda}{k}\bigg(1-\Big(\frac{\lambda}{\lambda-k}\Big)^{N_t}\bigg)$$ and so the terms from the Poisson process do not match up nicely. Strangely, if this product were to be $\frac{\lambda}{k}\bigg(1-\Big(\frac{\lambda}{\lambda+k}\Big)^{-N_t}\bigg)$ I think it works, but I cannot find any reason to support where this comes from. Any help on this would be fantastic.

Answer 1

First, we need to be careful about putting the condition at the right place:

\begin{align} e^{kt}\mathbb{E}[\mu_t] -\mu_0 &= \mathbb{E}\bigg[\sum_{m=1}^{N_t} e^{k\tau_m}\eta_m\bigg]\\ &= \mathbb{E}\bigg[\sum_{m=1}^{N_t} \mathbb{E}\bigg[e^{k\tau_m}\eta_m|N_t\bigg]\bigg]\\ &=\mathbb{E}\Big[\eta_1\Big] \cdot \mathbb{E}\bigg[\sum_{m=1}^{N_t} \mathbb{E}\Big[e^{k\tau_m}|N_t\Big]\bigg]. \end{align} Now the distribution of $\tau_m$ given $N_t$ is not Gamma, but uniform between $0$ and $t$. It cannot be Gamma because this would have a finite probability that $\tau_m>t$. Because of this, and by noting that the MGF of a uniform random variable between $0$ and $t$ is $\frac{e^{kt}-1}{kt}$, we obtain the desired result.