I am trying to calculate the covariance between a Johnson SU distribution and a Normal distribution.
Let $z \sim\mathcal{N}(0,1)$, we have $\operatorname{cov}(\lambda \sinh(\frac{z-\gamma}{\delta})+\xi, \sigma z+\mu)$, where $\lambda, \gamma, \delta, \xi$ are the parameters of my JSU distribution and $\sigma, \mu$ are the parameters of the Normal. Calculation yields (EDIT: added more details and noticed a mistake):
$\operatorname{cov}(X, Y) = \operatorname{E}{\big[(X - \operatorname{E}[X])(Y - \operatorname{E}[Y])\big]}$ so
$\operatorname{cov}(\lambda\sinh(\frac{z-\gamma}{\delta})+\xi, \sigma z+\mu)=\operatorname{E}(\lambda\sinh(\frac{z-\gamma}{\delta})+\xi-\operatorname{E}(\lambda\sinh(\frac{z-\gamma}{\delta})+\xi))(\sigma z+\mu-\operatorname{E}(\sigma z+\mu))$
According to Wikipedia the mean of the Johnson's SU distribution is $\xi - \lambda \exp \frac{\delta^{-2}}{2} \sinh\left(\frac{\gamma}{\delta}\right)$
$\operatorname{cov}(\lambda\sinh(\frac{z-\gamma}{\delta})+\xi, \sigma z+\mu)=\operatorname{E}((\lambda\sinh(\frac{z-\gamma}{\delta})+\xi - \xi + \lambda \exp \frac{\delta^{-2}}{2} \sinh\left(\frac{\gamma}{\delta}\right))(\sigma z-\sigma))$
$=\lambda(\operatorname{E}( \sigma z\sinh(\frac{z-\gamma}{\delta}) + \sigma z\exp \frac{\delta^{-2}}{2} \sinh\left(\frac{\gamma}{\delta}\right)-\sigma \sinh(\frac{z-\gamma}{\delta}) -\sigma \exp \frac{\delta^{-2}}{2} \sinh\left(\frac{\gamma}{\delta}\right)))$
$=\lambda(\operatorname{E}( \sigma z\sinh(\frac{z-\gamma}{\delta})) + \operatorname{E}(z)\sigma \exp \frac{\delta^{-2}}{2} \sinh\left(\frac{\gamma}{\delta}\right)-\sigma \operatorname{E}(\sinh(\frac{z-\gamma}{\delta})) -\sigma \exp \frac{\delta^{-2}}{2} \sinh\left(\frac{\gamma}{\delta}\right))$
$=\lambda(\operatorname{E}( \sigma z\sinh(\frac{z-\gamma}{\delta})) -\frac{\sigma}{2} (\operatorname{E}(\exp\frac{z-\gamma}{\delta})-\operatorname{E}(\exp-\frac{z-\gamma}{\delta})) -\sigma \exp \frac{\delta^{-2}}{2} \sinh\left(\frac{\gamma}{\delta}\right))$
$=\lambda(\operatorname{E}( \sigma z\sinh(\frac{z-\gamma}{\delta})) -\frac{\sigma}{2} (\exp-\frac{\gamma}{\delta}\operatorname{E}(\exp\frac{z}{\delta})-\exp\frac{\gamma}{\delta}\operatorname{E}(\exp-\frac{z}{\delta})) -\sigma \exp \frac{\delta^{-2}}{2} \sinh\left(\frac{\gamma}{\delta}\right))$
$=\lambda(\operatorname{E}( \sigma z\sinh(\frac{z-\gamma}{\delta})) -\frac{\sigma}{2} ((\exp-\frac{\gamma}{\delta}\exp\frac{1}{2\delta^2})-(\exp\frac{\gamma}{\delta}\exp-\frac{1}{2\delta^2})) -\sigma \exp \frac{\delta^{-2}}{2} \sinh\left(\frac{\gamma}{\delta}\right))$
$=\lambda(\operatorname{E}( \sigma z\sinh(\frac{z-\gamma}{\delta})) -\sigma \sinh(\frac{1}{2\delta^2}-\frac{\gamma}{\delta}) -\sigma \exp \frac{\delta^{-2}}{2} \sinh\left(\frac{\gamma}{\delta}\right))$
$\operatorname{cov}(\lambda\sinh(\frac{z-\gamma}{\delta})+\xi, \sigma z+\mu)=\lambda\sigma(\operatorname{E}( z\sinh(\frac{z-\gamma}{\delta})) -\sinh(\frac{1}{2\delta^2}-\frac{\gamma}{\delta}) -\exp \frac{\delta^{-2}}{2} \sinh\left(\frac{\gamma}{\delta}\right))$
I am unable to calculate $\operatorname{E}(z\sinh(\frac{z-\gamma}{\delta}))$.
I have tried going back to to the definition of the expected value $\operatorname{E}[X] = \int_{-\infty}^\infty x f(x)\,dx$, with no results. I have also attempted a numerical approach since I'm using the distributions to model data, but it defeats the point of modelling in the first place and I would want to avoid it if possible.
Any help would be appreciated.
Let $z=\delta w$ where $\delta>0$ and $\alpha=\gamma/\delta$ so $w\sim{\sf N}(0,1/\delta^2)$ and \begin{align}\Bbb E\left[z\sinh\frac{z-\gamma}\delta\right]&=\delta\Bbb E[w\sinh(w-\alpha)]\\&=\delta\int_{-\infty}^\infty w\frac{e^{w-\alpha}-e^{-w+\alpha}}2\frac\delta{\sqrt{2\pi}}e^{-\delta^2w^2/2}\,dw\\&=\frac{\delta^2}{2\sqrt{2\pi}}\left[e^{-\alpha}\int_{-\infty}^\infty we^{w-\delta^2w^2/2}\,dw-e^{\alpha}\int_{-\infty}^\infty we^{-w-\delta^2w^2/2}\,dw\right].\end{align} Now \begin{align}\int_{-\infty}^\infty we^{\pm w-\delta^2w^2/2}\,dw&=e^{1/(2\delta^2)}\int_{-\infty}^\infty we^{-\delta^2(w\mp 1/\delta^2)^2/2}\,dw=\pm e^{1/(2\delta^2)}\frac{\sqrt{2\pi}}{\delta^3}\end{align} using standard Gaussian integral identities so \begin{align}\Bbb E\left[z\sinh\frac{z-\gamma}\delta\right]&=e^{1/(2\delta^2)}\frac{e^{-\alpha}+e^\alpha}{2\delta}=\frac{e^{1/(2\delta^2)}}\delta\cosh\frac\gamma\delta.\end{align}