conditional distribution of $X_t$ in a jump-diffusion model

Question

I'm working with a variant of the Ornstein Uhlenbeck model that includes Lévy-type jumps, (i.e. a jump-diffusion model with mean reversion).

I want the distribution of $X_t$ (given the parameters of the model and a value for $X_0$), but I'm not sure how to calculate it and I haven't been able to find an explicit solution in the literature.

The stochastic differential equation would be something like:

$$dX_t = \theta(\mu - X_t)dt + \sigma_W d\mathbf{W}_t + \sigma_j d\mathbf{P}_\lambda,$$

where the first term controls the mean-reversion, the second term is a standard Wiener process, and and $\mathbf{P}_\lambda$ is a compound Poisson process that has intensity $\lambda$ and zero-mean jumps with variance $\sigma^2_j$.

If I didn't have any mean-reversion, $X_t$ would be Gaussian with mean $X_0$ and variance $t(\sigma_W^2 + \lambda\sigma^2_j)$, i.e. the sum of the variances from the two processes.

Alternatively, if I didn't have any jumps, $X_t$ would also be normally distributed, with mean $X_0 e^{-\theta t} + \mu (1 - e^{-\theta t})$ and variance $\sigma_W^2 (1 - e^{-2 \theta t})/(2 \theta)$ (i.e. exponential decay toward the stationary distribution of the OU process).

Since I have jumps and mean-reversion, $X_t$ is no longer Gaussian; its tails would be too fat for that. However, it does seem like $P(X_t=x|N_j)$ (where $N_j$ is the number of jumps) would be Gaussian, because it would just be a sum of Gaussian random variables.

My problem is that I don't know what the mean and variance of this conditional Gaussian would be. Depending on how long it had been since the last jump, the variance could be anywhere between the OU variance and that amount plus the jump variance, and I'm not sure how to integrate over the possibilities.

Could someone please point me to a solution or help me calculate it?

Answer 1

I'll preface this response by saying I've worked with OU processes quite a bit and point processes quite a bit, but never the combination. So I could be way off and there's a much easier way...

I gave this a quick look and I think the best solution is to condition on the times (and hence number, $N(t)$ as well) of jumps in the Poisson process, say $\tau_0, \dots, \tau_t$, where $\tau_0 = 0$ and $\tau_{N(t)+1}=t$. Then if you look at the joint distribution, all you have to do is integrate over all the $X_{\tau_j}$, i.e., $$ [X_t|X_0, \boldsymbol{\tau}] = \int\prod_{j=1}^{N(t)+1}[X_{\tau_j}|X_{\tau_{j-1}}]dX_{\tau_1}\dots dX_{\tau_{N(t)}}. $$ Good thing is that the expectation is still $E(X_t) = X_0 e^{-\theta t} + \mu(1-e^{-\theta t})$ because of the zero mean shocks. You can also iterate the conditional expectations. To get the variance of $X_t|X_0,\boldsymbol{\tau}$ you have to iterate through finding the variance of $X_2|X_0$, $X_3|X_0$, etc. till you get to $X_t|X_0$ I tried a few iterations and it's possible but maybe a little messy. Starting with $$ var(X_{\tau_1}|X_0) = \sigma^2_j + \frac{\sigma^2}{2\theta}(1-e^{-2\theta\tau_1}), $$ Then, $$ \begin{aligned} var(X_{\tau_2}|X_0) &= var(E(X_{\tau_2}|X_{\tau_1},X_0)) +E(var(X_{\tau_2}|X_{\tau_1},X_0))\\ &= \left[\left\{\sigma^2_j + \frac{\sigma^2}{2\theta}(1-e^{-2\theta\tau_1})\right\} e^{-2\theta(\tau_2-\tau_1)}\right] + \left[\sigma_j^2 + \frac{\sigma^2}{2\theta}(1-e^{-2\theta(\tau_2-\tau_1)})\right]\\ &= \sigma_j^2(1+e^{-2\theta(\tau_2-\tau_1)}) + \frac{\sigma^2}{2\theta}(1-e^{-2\theta\tau_2}) \end{aligned} $$ (If I did all the algebra correctly, which I may not have. It's late ;-)) Now, you follow this procedure for finding the variance of $X_{\tau_3}|X_0$, etc.

After all that, you'll have $[X_t|X_0, \boldsymbol{\tau}]$. So, to finish you'll have to integrate overall the times $$ [X_t|X_0] = \int [X_t|X_0, \boldsymbol{\tau}] [\boldsymbol{\tau}] d\boldsymbol{\tau}, $$ where $[\boldsymbol{\tau}]$ is the pdf of a time homogenous Poisson process. I think you might be stuck for an analytical solution there, but you might be able to break it up into times given $N(t)$ and $N(t)$ which is Poisson($\lambda$). You might make a little more analytical headway.