Conditional expectation involving exponentiated square norm and independent Gaussian r.v.

Question

Given a $d$-dimensional standard Gaussian variable $\varepsilon$, a variable $X$ independent of $\varepsilon$ and a continuous function $F : \mathbb{R}^d \to \mathbb{R}^d$ , I am trying to compute $$\mathbb{E}[ \mathrm{exp} \, (\langle F(X), \epsilon \rangle + \tfrac{1}{4} \Vert \varepsilon \Vert^2) | X]$$

which "amounts" to computing

$$\mathbb{E}[ \mathrm{exp} \,(\Vert F(X) + \tfrac{1}{2} \varepsilon \Vert^2) | X]$$ as the first term if we develop the square is $X$-measurable. But I don't really know where to go further, as the exponetial prevents me from splitting the sum of the scalar product.

Edit: After discussion, it is related to the MGF of a non central chi-square distribution: for a fixed $x$, $N = \Vert 2 F(x) + \varepsilon \Vert^2$ follows a chi-square distribution with $d$ degrees of freedom and $\lambda = 4 \Vert F(x) \Vert^2 $. It remains to apply the formula for the MGF of a non central chi-square, with $t = 1/4$: $$\mathbb{E}[e^{t N}] = \frac{e^{\lambda /2}}{2^{d/2}}$$

Answer 1

First we can use the formula $$ \mathbb E\left[h\left(X,Y\right)\mid X\right]=g(X) $$ where $X$ is a vector independent of $Y$ and $g(x)=\mathbb E\left[h\left(x,Y\right)\right]$. The wanted conditional expectation is thus $g(X)$, where $$ g(x)=\mathbb E\left[\exp\left(\sum_{k=1}^d\left(F_k(x)\varepsilon_k+\frac 14\varepsilon_k^2\right)\right)\right]. $$ Using independence of the sequence $\left(\varepsilon_k\right)$ gives $$ g(x)=\prod_{k=1}^d\mathbb E\left[\exp\left( F_k(x)N+\frac 14N^2 \right)\right], $$ where $N$ has a standard normal distribution. Rewriting the later expectation as an integral and reducing the square leads to a nice formula for $g$.