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Let $R := (R_1, R_2)$ be a two dimensional diffusion process defined by the following SDE:

$$\mathrm{d}R_{1,t} = -\lambda_1 R_{1,t}\mathrm{d}t + \lambda_1 \sigma(R_{1,t}, R_{2,t}) \mathrm{d}W_t$$ $$\mathrm{d}R_{2,t} = \lambda_2\left(\sigma^2(R_{1,t}, R_{2,t}) - R_{2,t}\right)\mathrm{d}t$$

where $W$ is a standard brownian motion, $\lambda_1, \lambda_2 >0$, $\sigma: (R_1, R_2) \mapsto \beta_0 + \beta_1 R_1 + \beta_2 \sqrt{R_2}$ ; $\beta_0 > 0, \beta_1 < 0, \beta_2 \in (0, 1)$

for $T >0$ and $t\in (0, T]$, i know that the law of $R_{1,t}$ is : $$R_{1,t} \sim \mathcal{N}\left(R_{1,0}e^{-\lambda_1 t}, \lambda_1^2 \int_0^t e^{-2\lambda_1 (t - s)}\mathbb{E}\left(\sigma^2(R_{1,s}, R_{2,s})\right)\mathrm{d} s \right)$$

I'm interested in knowing the law of $R_{2,t}$, or at least its laplace transform : $$\mathbb{E}\left(e^{-\lambda R_{2,t}}\right)$$

The goal is to compute $\mathbb{E}\sqrt{R_{2,t}}$, using the identity : $$\mathbb{E}\sqrt{R_{2,t}} = \frac{\sqrt{\pi}}{2} \int_0^{+\infty} \frac{1 - \mathbb{E}\left(e^{-\lambda R_{2,t}}\right)}{\sqrt{\lambda^3}} \mathrm{d} \lambda$$

Greyearl
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  • Actually, it is not immediate that $R_{1,t}$ is Gaussian. https://math.stackexchange.com/questions/2480786/is-the-ito-integral-of-a-predictable-process-gaussian-distributed – Thomas Kojar Aug 24 '23 at 20:09
  • good to include both links in your post https://mathoverflow.net/questions/453374/laplace-transform-of-a-stochastic-process – Thomas Kojar Aug 24 '23 at 20:17
  • @ThomasKojar Thanks for adding the link. Well, for the law of $R_{1,t}$, one can start by finding its law conditional on the filtration generated by the BM, i.e $R_{1,t} \ | \ \mathcal{F}t$ and notice that the process defined by : $\sigma_t = \sigma(R{1,t}, R_{2,t})$ is actually adapted to the filtration. Why do you think otherwise ? – Greyearl Aug 25 '23 at 13:19
  • as the counterexample in the linked post shows, predictable/adapted is not enough. In fact, if that was the case then by the martingale-representation theorem, every martingale adapted to Brownian filtration would have been Gaussian. – Thomas Kojar Aug 25 '23 at 15:08
  • @ThomasKojar Thanks, you're right, it's not that immediate, but still, the expressions of the expectation and the variance of $R_{1,t}$ are true! – Greyearl Aug 26 '23 at 13:39

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