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Hermite expansion of a correlated Gaussian feature kernel

Statement

Assume AC. Let γ be standard Gaussian measure, let hm=Hm/m! be the normalized probabilists’ Hermite basis, and let f,g∈L2(γ) have coefficients am=∫fhm dγ and bm=∫ghm dγ. For ρ∈[−1,1], let (X,Y) have the centered bivariate normal law with unit variances and correlation ρ. Then E[f(X)g(Y)]=∑m=0∞ambmρm. The series is absolutely convergent, including at ρ=±1. In particular, the Gaussian dual kernel of f is ∑m≥0am2ρm, and ∑m≥0am2=∥f∥L2(γ)2.

Facts & Assumptions

Given: AC; f,g∈L2(γ); their normalized Hermite coefficients; and a centered unit-variance Gaussian pair (X,Y) of correlation ρ∈[−1,1].

[A1]

AC supplies the countable and dependent choice required by the Gaussian and Hermite suppliers (The Axiom of Choice, AC supplies countable selections and prescribed serial paths).

[F1]

The functions hm form a complete orthonormal basis of real L2(γ), so the Hermite expansions converge in norm and satisfy Parseval (Completeness of normalized probabilists’ Hermite polynomials in Gaussian L²).

[F2]

The monic Hermite polynomials satisfy H0=1, H1(x)=x, xHm=Hm+1+mHm−1, and Hm(x)=m!∑j=0⌊m/2⌋(−1/2)jxm−2j/(j!(m−2j)!) (The monic probabilists' Hermite polynomials).

[F3]

The standard normal has density (2π)−1/2e−x2/2, and each linear projection of a centered Gaussian vector is a scalar normal with the corresponding covariance quadratic form as its variance, including variance zero (Standard normal and normal laws, Multivariate normal law, including singular covariance).

[F4]

Finite linear combinations commute with expectation for integrable random variables (Linearity, monotonicity, and the modulus bound for expectation).

[F5]

Dominated convergence applies under a single integrable majorant (Dominated convergence).

[F6]

The L2 Cauchy–Schwarz inequality bounds product expectations (Cauchy-Schwarz for random variables).

[F7]

The real exponential is defined by its power series, has derivative equal to itself, and obeys exp⁡(x+y)=exp⁡(x)exp⁡(y) (The real exponential function and the number e by a power series, The exponential function is smooth and (exp⁡)′=exp⁡, The exponential addition formula exp⁡(x+y)=exp⁡(x)exp⁡(y)).

[F9]

A difference quotient of a differentiable real function on a compact interval is bounded in absolute value by a bound for its derivative on that interval (The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a)).

Proof

technique · direct
1.1F2F7F8algebra

For real s and x, set Gx(s)=exp⁡(sx−s2/2). By [F7]–[F8], Gx′(s)=(x−s)Gx(s). Put Pm(x)=Gx(m)(0). Then P0=1 and P1=x; differentiating the identity Gx′=(x−s)Gx m times and setting s=0 gives Pm+1=xPm−mPm−1 for m≥1. The Hermite recurrence in [F2] therefore gives Pm=Hm for every m.

2.1A1F3F7step 1.1algebra

For real s,t, [F3] gives sX+tY∼N(0,v) with v=s2+2ρst+t2≥0. Completing the square in the standard-normal density gives EeaZ=ea2/2 for real a and Z∼N(0,1), so EesX+tY=ev/2 by the scalar normal's affine definition; when v=0, its law is a point mass and the same formula holds. Thus [F7] gives E[GX(s)GY(t)]=e−(s2+t2)/2EesX+tY=eρst. This uses the given pair directly and includes ρ=±1.

3.1F3F5F6F7F8F9step 1.1step 2.1algebra

For any fixed finite m,n, differentiating the expectation identity in step 2.1 m times in s and n times in t at (0,0) is justified as follows. On a bounded parameter rectangle, each required derivative of the integrand is bounded by K(1+∣X∣)q(1+∣Y∣)qeR(∣X∣+∣Y∣) for finite K,q,R. This is integrable by [F6], since completing the square gives finite second moments for each marginal factor (1+∣Z∣)qeR∣Z∣ under Z∼N(0,1). On a slightly larger rectangle, [F9] gives the same majorant for each difference quotient, so [F5] passes each of the finitely many derivatives through expectation. By step 1.1 the derivative on the left is E[Hm(X)Hn(Y)]; differentiating eρst on the right gives m!ρm if m=n and 0 otherwise. Hence E[Hm(X)Hn(Y)]=m!δmnρm.

4.1F1F4step 3.1algebra

For finite partial sums fM=∑m=0Mamhm and gM=∑m=0Mbmhm, step 3.1, finite expectation linearity and orthonormality give E[fM(X)gM(Y)]=∑m=0Mambmρm.

5.1F1F6step 4.1∎

By [F1], fM→f and gM→g in L2(γ). Since both marginals of (X,Y) are γ, [F6] bounds the product-expectation error by ∥fM−f∥2∥gM∥2+∥f∥2∥gM−g∥2, which tends to zero. Also ∑m∣ambmρm∣≤(∑mam2)1/2(∑mbm2)1/2 by [F1] and Cauchy–Schwarz for sequences. Taking M→∞ proves the stated identity and endpoint absolute convergence; the case f=g gives the dual-kernel formula.

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