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

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 $$t = 1/4$$: $$\mathbb{E}[e^{t N}] = \frac{e^{\lambda /2}}{2^{d/2}}$$

• How far are you trying to get?
– user515599
Mar 12, 2019 at 8:35
• @orange is it hopeless to try to get a closed form formula in X ? Mar 12, 2019 at 8:36
• It turns into a product not asum, my bad
– user515599
Mar 12, 2019 at 9:53

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 $$g$$.
• Thaks, I had forgotten about the substitution formula to fix $x$. I have also been given a proof involving chi-square distributions, which I've included in an edit. Mar 12, 2019 at 10:43