If I have a two norm distributed RV, i.e. $L \sim N(m,s^2)$ and $M \sim N(0,1)$, with coo coefficient $ \operatorname{Corr}(L,M) =\rho$, what is then:
$$ E[e^L M^2] = \mathop{?} $$
If I have a two norm distributed RV, i.e. $L \sim N(m,s^2)$ and $M \sim N(0,1)$, with coo coefficient $ \operatorname{Corr}(L,M) =\rho$, what is then:
$$ E[e^L M^2] = \mathop{?} $$
On
I don't think there's enough information to know this expectation. For example, suppose that $m = 0$, $s = 1$, and $\rho = 0$. This would be equally consistent with either i) $A$ and $B$ being independent or ii) $B = AW$ where $W$ is the Rademacher distribution (see Wikipedia). These result in different values for the sought expectation. Specifically in case ii) the expectation will be larger because whenever $A$ is large $B^2$ is also large.
Assuming that $A$ and $B$ are jointly normal, then $X=(A,B)$ has a normal distribution with the mean $\mu$ and the covariance matrix $\Sigma$ given by
$$ \mu = (m, 0) \qquad\text{and}\qquad \Sigma = \begin{pmatrix} s^2 & \rho s \\ \rho s & 1 \end{pmatrix}. $$
Then we know that the moment generating function of $X$ is given by
$$ \mathbb{E}[e^{\xi \cdot X}] = e^{\xi\cdot\mu + \frac{1}{2}\xi\cdot\Sigma \xi}, \qquad \xi \in \mathbb{R}^2. $$
Now plugging $\xi = (1, t)$, we have
$$ \mathbb{E}[e^{A}e^{tB}] = e^{m + \frac{1}{2}(s^2 + 2\rho s t + t^2)}. $$
Differentiating both sides with respect to $t$ twice and plugging $t = 0$, we get
$$ \mathbb{E}[e^{A}B^2] = e^{m + \frac{1}{2}s^2}(1 + s^2\rho^2). $$