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Let $u = \epsilon e^{\frac{1}{2} \eta}$ where \begin{equation*} \left( \begin{array}{c} \epsilon \\ \eta \\ \end{array} \right) \sim N\left( \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right), \left( \begin{array}{cc} 1 & \rho \sigma \\ \rho \sigma & \sigma^2 \\ \end{array} \right) \right), \quad -1 < \rho < 1. \end{equation*} What is $\mathbb{E}[u \eta]$?
Thank you.
I think I may have solved this myself, although my method is rather long-winded. Please correct me if I have made a mistake anywhere.
I believe the answer is $\frac{5}{4} \rho \sigma e^{\frac{1}{4} \sigma^2}$.
Here is the proof.
By definition, \begin{equation*} E[\vartheta U] = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \vartheta \chi e^{\frac{1}{2} \vartheta} \, \text{pdf}_{\chi, \vartheta}(\chi, \vartheta) \, d\chi \, d\vartheta, \end{equation*} where $\text{pdf}_{\chi, \vartheta}$ is the joint pdf of $\chi$ and $\vartheta$, which is well known for the bivariate normal distribution. By Fubini's theorem, we can rewrite this as \begin{equation*} E[\vartheta U] = \int_{-\infty}^{+\infty} \vartheta e^{\frac{1}{2} \vartheta} I_1 \, d\vartheta, \end{equation*} where \begin{align*} I_1 &:= \int_{-\infty}^{+\infty} \chi \, \text{pdf}_{\chi, \vartheta}(\chi, \vartheta) \, d\chi \\ &= \int_{\chi=-\infty}^{\chi=+\infty} \frac{1}{2\pi \sigma \sqrt{1 - \rho^2}} \chi \exp\left( -\frac{z}{2 (1 - \rho^2)} \right) \, d\chi \end{align*} with $z := \chi^2 - \frac{2\rho}{\sigma} \chi \vartheta + \frac{\vartheta^2}{\sigma^2}$, whence $\chi = \frac{1}{2} \frac{dz}{d\chi} + \frac{\rho}{\sigma} \vartheta$. We substitute this for $\chi$ in the integral above to obtain \begin{align*} I_1 &= \int_{\chi=-\infty}^{\chi=+\infty} \frac{1}{2\pi \sigma \sqrt{1 - \rho^2}} \left( \frac{1}{2} \frac{dz}{d\chi} + \frac{\rho}{\sigma} \vartheta \right) \exp\left( -\frac{z}{2 (1 - \rho^2)} \right) \, d\chi \\ &= \int_{\chi=-\infty}^{\chi=+\infty} \frac{1}{2\pi \sigma \sqrt{1 - \rho^2}} \frac{1}{2} \frac{dz}{d\chi} \exp\left( -\frac{z}{2 (1 - \rho^2)} \right) \, d\chi + \int_{\chi=-\infty}^{\chi=+\infty} \frac{1}{2\pi \sigma \sqrt{1 - \rho^2}} \frac{\rho}{\sigma} \vartheta \exp\left( -\frac{z}{2 (1 - \rho^2)} \right) \, d\chi \\ &= \frac{1}{4\pi \sigma \sqrt{1 - \rho^2}} I_2 + \frac{\rho}{\sigma} \vartheta \int_{\chi=-\infty}^{\chi=+\infty} \text{pdf}_{\chi, \vartheta}(\chi, \vartheta) \, d\chi \end{align*} where \begin{equation*} I_2 := \int_{\chi=-\infty}^{\chi=+\infty} \exp\left( -\frac{z}{2 (1 - \rho^2)} \right) \, dz. \end{equation*} Since $z(\chi)$ is a parabola with a minimum attained at $(\chi, z) = \left( \frac{\rho}{\sigma} \vartheta, \frac{1 - \rho^2}{\sigma^2} \vartheta^2 \right)$, \begin{align*} I_2 &= \int_{\chi=-\infty}^{\chi=\frac{\rho}{\sigma} \vartheta} \exp\left( -\frac{z}{2 (1 - \rho^2)} \right) \, dz + \int_{\chi=\frac{\rho}{\sigma} \vartheta}^{\chi=+\infty} \exp\left( -\frac{z}{2 (1 - \rho^2)} \right) \, dz \\ &= -\int_{z=\frac{1 - \rho^2}{\sigma^2} \vartheta^2}^{z=\frac{\rho}{\sigma} \vartheta} \exp\left( -\frac{z}{2 (1 - \rho^2)} \right) \, dz + \int_{z=\frac{1 - \rho^2}{\sigma^2} \vartheta^2}^{\chi=+\infty} \exp\left( -\frac{z}{2 (1 - \rho^2)} \right) \, dz \\ &= 0. \end{align*} Thus \begin{equation*} I_1 = \frac{\rho}{\sigma} \vartheta \int_{\chi=-\infty}^{\chi=+\infty} \text{pdf}_{\chi, \vartheta}(\chi, \vartheta) \, d\chi = \frac{\rho}{\sigma} \vartheta \, \text{pdf}_{\vartheta}(\vartheta) \end{equation*} and \begin{align*} E[\vartheta U] &= \frac{\rho}{\sigma} \int_{-\infty}^{+\infty} \vartheta^2 e^{\frac{1}{2} \vartheta} \, \text{pdf}_{\vartheta}(\vartheta) \, d\vartheta \\ &= \frac{\rho}{\sigma} \int_{-\infty}^{+\infty} \vartheta^2 e^{\frac{1}{2} \vartheta} \, \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-\vartheta^2}{2\sigma^2}}\, d\vartheta \\ &= \frac{\rho}{\sigma} \int_{-\infty}^{+\infty} \vartheta^2 \, \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-\vartheta^2 + \vartheta \sigma}{2\sigma^2}}\, d\vartheta \\ &= \frac{\rho}{\sigma} \int_{-\infty}^{+\infty} \vartheta^2 \, \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(\vartheta - \sigma/2)^2 + \sigma^2/4}{2\sigma^2}}\, d\vartheta \\ &= \frac{\rho}{\sigma} e^{\frac{1}{4}\sigma^2} \int_{-\infty}^{+\infty} \vartheta^2 \, \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(\vartheta - \sigma/2)^2}{2\sigma^2}}\, d\vartheta, \end{align*} where we recognise the integral as the second moment of a $N(\sigma/2, \sigma^2)$ random variable, which is $\frac{5}{4} \sigma^2$.
For simplicity, let $X = \eta/2 \sim N(0, \sigma^2/4)$. You want to compute $E\eta\epsilon e^{\eta/2} = 2EX\epsilon e^X$. Using properties of the conditional distributions in the multivariate normal one gets:
\begin{align} E\epsilon Xe^X &= E E\left(\epsilon Xe^X\mid X\right) \\ &= E Xe^X E\left(\epsilon\mid X\right)\\ &= EXe^X \frac{\mathrm{cov}(X, \epsilon)}{\mathrm{var}(X)}X \\ &= \frac{\rho \sigma/2}{\sigma^2/4} EX^2e^X. \end{align}
Then by using Stein's lemma twice: \begin{align} EXXe^X &= \frac{\sigma^2}{4}E\left(Xe^X + e^X \right) \\ &= \frac{\sigma^2}{4}E\left(\frac{\sigma^2}{4}e^X + e^X \right) \\ &= \frac{\sigma^2}{4}\left(\frac{\sigma^2}{4} + 1 \right)Ee^X \\ &= \frac{\sigma^2}{4}\left(\frac{\sigma^2}{4} + 1 \right)e^{\sigma^2/8}. \end{align}
Thus,
\begin{align} 2EX\epsilon e^X &= 2 \times \frac{\rho \sigma/2}{\sigma^2/4}\times \frac{\sigma^2}{4}\left(\frac{\sigma^2}{4} + 1 \right)e^{\sigma^2/8} \\ &= {\rho \sigma}\left(\frac{\sigma^2}{4} + 1 \right)e^{\sigma^2/8}, \end{align}
where we have used that the moment generating function of $X$ is $m(t) = Ee^{tX} = e^{t^2\sigma^2/8}$ to get the expectation of the exponential term.
