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'''Theorem:''' <math>\operatorname E[\langle \varphi(x), \varphi(y)\rangle] = e^{\frac{\|x-y\|^2}{2\sigma^2}}. </math>
'''Proof:''' It suffices to prove the case of <math>D=1</math>. Use the trigonometric identity <math>\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)</math>, the spherical symmetry of gaussian distribution, then evaluate the integral
: <math>\int_{-\infty}^\infty \frac{\cos (k x) e^{-x^2 / 2}}{\sqrt{2 \pi}} d x=e^{-k^2 / 2}. </math> '''Theorem:''' <math>\operatorname{Var}[\langle \varphi(x), \varphi(y)\rangle] = O(D^{-1})</math>. (Appendix A.2<ref>{{Cite arXiv |last1=Peng |first1=Hao |last2=Pappas |first2=Nikolaos |last3=Yogatama |first3=Dani |last4=Schwartz |first4=Roy |last5=Smith |first5=Noah A. |last6=Kong |first6=Lingpeng |date=2021-03-19 |title=Random Feature Attention |class=cs.CL |eprint=2103.02143 }}</ref>).
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