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

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

1 Answer 1

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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 $g$.

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  • $\begingroup$ 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. $\endgroup$ Commented Mar 12, 2019 at 10:43

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