How to take expectation of $x*e^{-\bf{a}-\bf{b}*x}$ when $x$ is standard multivariate-normal Let $\varepsilon \sim N(\vec{0},\mathbb{I}_{N})$.
Let $\bf{b}$ be a non-stochastic $(N \times 1)$ vector.
How would I go about taking the following expectation?
$\mathbb{E}\big[ \exp \{-\bf{b}^\prime \bf{b} - \bf{b}^\prime \varepsilon    \}  \varepsilon \big]$ 
Note the term inside the exponential is a scalar, and the term outside the exponential is a $(N \times 1)$ vector. 
Here is my work so far:
$\mathbb{E}\big[ \exp \{-\bf{b}^\prime \bf{b} - \bf{b}^\prime \varepsilon     \}  \varepsilon\big] = \int_{-\infty}^{\infty} \exp \{-\bf{b}^\prime \bf{b} - \bf{b}^\prime x\}  x | 2\pi|^{-0.5}\exp\{ -0.5 x^{\prime} x \} dx = |2\pi|^{-0.5} \exp\{-b^\prime b \}\int_{-\infty}^{\infty} \exp \{- \bf{b}^\prime x - 0.5x^{\prime}x\}  x dx  $
I'm not sure how the scalar interacts with the vector in the integration.
 A: As there is another proposed solution, I see value in writing my hint explicitly. This is a very common method of solving such integrals.
Note that $ b^T \varepsilon = \varepsilon^T b$ as the result product is scalar. Then, by completing $x^Tx - 2b^Tx$ to the square, we get
$$
\begin{align*}
\mathbb{E}_{\varepsilon \sim N\left(0, I\right)}\left[\varepsilon e^{-b^Tb - b^T \varepsilon}\right] &= \int_{-\infty}^{\infty} \frac{1}{(2\pi)^{N/2}} x e^{-\frac{2b^Tb + 2b^T x + x^T  x}{2}} dx \\
&= \int_{-\infty}^{\infty} \frac{1}{(2\pi)^{N/2}} x e^{-\frac{b^T b}{2}}e^{-\frac{\left(x + b\right)^T \left(x + b\right)}{2}} dx \\
&= \mathbb{E}_{\varepsilon{'} \sim N\left(-b, I\right)}\left[\varepsilon{'} e^{-\frac{b^Tb}{2}}\right] 
\end{align*}
$$
Which we know already to calculate explicitly.
Also note that with adequate changes, we can calucate the term for any covariance matrix (by replacing $b = \Sigma^{-1}c$ and completing to $\left(x+c\right)^T\Sigma^{-1}\left(x+c\right)$).
A: The crucial property here is that the variables in the random vector are independent. Consider the typical element, say the 1st, of the vector holding the expected values:
$$\mathbb{E}\big[ \exp \{-\bf{b}^\prime \bf{b} - \bf{b}^\prime \varepsilon    \}  \varepsilon_1 \big] = \exp \{-\bf{b}^\prime \bf{b}\}\cdot \mathbb{E}\left[\varepsilon_1 \cdot \exp\left\{-\sum b_j\varepsilon_j\right\}\right] $$
$$=\exp \{-\bf{b}^\prime \bf{b}\}\cdot \mathbb{E}\left[\varepsilon_1 \cdot \exp\left\{b_1\varepsilon_1\right\}\right]\cdot \mathbb{E}\left[\exp\left\{-b_2\varepsilon_2\right\}\right]\cdot ... \cdot \mathbb{E}\left[\exp\left\{-b_n\varepsilon_n\right\}\right]$$
Except from the first expected value, all the others are MGFs of the univariate standard normal each evaluated at $-b_j, j=2,...,n$. The first expected value is
$$\mathbb{E}\left[\varepsilon_1 \cdot \exp\left\{-b_1\varepsilon_1\right\}\right]=\int_{-\infty}^{\infty} \varepsilon_1 \exp\{-b_1 \varepsilon_1\} \phi(\varepsilon_1) d\varepsilon_1$$
and has a solution expressed in terms of the standard normal CDF $\Phi$ and the standard  normal pdf $\phi$.
Then do the lot for every element of the vector.
