How do I prove $$ p(w) = \int_{\mathbb{R}^D} e^{-jw^T \delta} k(\delta) d\delta $$ (i.e. $p(w)$ is the inverse Fourier transform of $k(\delta)$) where $p(w)$ is the multivariate Gaussian distribution $$ p(w) = (2 \pi /\sigma)^{-D/2} e^{-\sigma w^Tw/2} $$ and $k(\delta)$ is the Gaussian kernel $$ k(\delta) = e^{-\delta^T\delta/2\sigma}? $$ This paper guarantees it but I don't really see how this is proved by Bochner's Theorem.
2026-04-26 10:19:41.1777198781
Prove the inverse Fourier transform of Gaussian kernel is Gaussian distribution (by Bochner's)
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Assume that
Let $g(x) = e^{-\|x\|^2/2}$ and $S_\alpha f(x) = f(x/\alpha)$. Then $\mathcal Fg = g$ and $\mathcal F S_\alpha f(x) = \alpha^dS_{1/\alpha}\mathcal Ff(x)$. See Chapter 5.2 in [1].
Consider \begin{align*} \int_{\mathbb R^d}e^{-i\langle x,\omega\rangle}k(\omega)d\omega &= \int_{\mathbb R^d}e^{-i\langle x,\omega\rangle}e^{-\|\omega\|^2/(2\sigma)}d\omega = \left(2\pi\right)^{d/2} \mathcal Fe^{-\|x\|^2/(2\sigma)} = \left(2\pi\right)^{d/2} \mathcal FS_{\sqrt \sigma}g(x) \\ &= \left(2\pi\right)^{d/2} \sigma^{d/2}S_{1/{\sqrt \sigma}}\mathcal Fg(x) = \left(2\pi\right)^{d/2} \sigma^{d/2}S_{1/{\sqrt \sigma}}g(x) \\ &= \left(2\pi\right)^{d/2} \sigma^{d/2}e^{-\sigma\|x\|^2/2} = c_0 p(x), \end{align*} where $c_0$ is a normalizing constant (depending only on $\sigma$ and $d$).
Note that when we consider what is stated in their article as (2), we have (for $\mathbf z = \mathbf x -\mathbf y$) \begin{align*} \int_{\mathbb R^d}p(\omega)e^{-i\langle \omega,\mathbf z\rangle}d\omega = (2\pi)^{-d/2}\int_{\mathbb R^d} e^{-\|\omega\|^2/2}e^{-i\langle\omega,\mathbf z\rangle}d\omega = \mathcal Fg(\mathbf z) = g(\mathbf z) = k(\mathbf z), \end{align*} where we assume that $\sigma=1$ in the last step and use symmetry ($-\mathbf z$ corresponds to the definition of the Fourier transform in their article).
[1] Holger Wendland. Scattered data approximation. Cambridge University Press, 2005.