Unconditional expectation of a correlated Jump size

Question

I have a problem computing the expectation of this :

$J_{v, i} \sim \exp \left(\mu_{v}\right)$

$J_{S, i} \mid J_{v, i} \sim \mathcal{N}\left(\mu_{S}+\rho_{J} J_{v, i}, \sigma_{S}^2\right)$

How can we arrive at this results.When I try to compute it I find a non-Lebesgue integral, which can bursts on the infinite side(depending of the parameters) such as :

$\mathbb{E}\left[\mathrm{e}^{J_{S, i}}\right] = \mathrm{e}^{\mu_{S}+\frac{1}{2} \sigma_{S}^{2}}\int_{0}^{\infty}\mu_{v}\mathrm{e}^{J_{v, i}(\rho_{J} - \mu_{v} )} dJ_{v, i}$

Which is not the same as what is written by the author :

$\bar{\mu}=\mathbb{E}\left[\mathrm{e}^{J_{S, i}}\right]=\mathrm{e}^{\mu_{S}+\frac{1}{2} \sigma_{S}^{2}} /\left(1-\rho_{J} \mu_{v}\right)$

Could someone help me please?

Answer 1

I do not understand how you arrived at $\int_{0}^{\infty}\mu_{v}\mathrm{e}^{J_{v, i}(\rho_{J} - \mu_{v} )} dJ_{v, i}$. For one, $\mu_v$ is the scale parameter, not the rate parameter, so the pdf is $1/ \mu_v \exp(-x/\mu_v)$. The question is easier to solve with moment generating functions:

\begin{align}\mathbb{E}\left[\mathrm{e}^{J_{S, i}}\right] &=\mathbb{E}\left[\mathbb{E}\left[\mathrm{e}^{J_{S, i}} \mid J_{v,i} \right] \right] \\ &=\mathbb{E}\left[ \exp\left( \mu_s + \rho_J J_{v,i} + 0.5 \sigma_S^2 \right) \right] \\ &=\exp\left( \mu_s + 0.5 \sigma_S^2 \right) \mathbb{E}\left[ \exp\left( \rho_J J_{v,i} \right) \right] \\ &=\exp\left( \mu_s + 0.5 \sigma_S^2 \right) \frac{1}{1-\rho_J\mu_v} \end{align}