# Expected Value of Mean-Reverting Jump Process

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.

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.

• I think I assumed the inter-arrival times were i.i.d. exponential and so assumed the arrival times were gamma as a sum of exponential random variables... I thought this was always assumed for a Poisson process. Jun 8 '20 at 0:50
• Yes this is still true unconditional on any other data. But as soon as you know how many events you have in some interval, the actual timings are just uniform (without ordering them). Jun 8 '20 at 6:14