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Gaussian orthogonality and the monomial expansion of the Hermite polynomials

Statement

Assume AC, and let Z∼N(0,1) (Standard normal and normal laws). For the monic Hermite polynomials Hm of The monic probabilists' Hermite polynomials:

(i) E[Hm(Z)]=0 for every m≥1, and E[Hm(Z)Hn(Z)]=m! δmn for all m,n≥0;

(ii) for every m≥0, xm=∑j=0⌊m/2⌋cm,jHm−2j(x) with cm,0=1 and cm,j=m!2j j! (m−2j)!=(m2j)(2j−1)!!∈Z(0≤j≤⌊m/2⌋); equivalently, the monomials and the Hermite polynomials are related by a unitriangular change of basis in each finite degree;

(iii) consequently, for any N≥1 and any m1,…,mN∈N, the mixed monomial ∏k=1Nxkmk is a Z-linear combination of products ∏kHmk′(xk) with mk′≤mk and mk′≡mk(mod2), the coefficient of ∏kHmk being 1.

Facts & Assumptions

Given: AC; a random variable Z∼N(0,1); the polynomials Hm defined by H0=1, H1(x)=x and xHm=Hm+1+mHm−1 (The monic probabilists' Hermite polynomials); φZ(t)=E[eitZ] is the characteristic function of Z.

[F1]

For Z∼N(0,1) and every real t, φZ(t)=e−t2/2 (Characteristic function of a normal law).

[F2]

If E∣Y∣k<∞ then φY∈Ck(R) with φY(j)(t)=E[(iY)jeitY] for 0≤j≤k; in particular φY(j)(0)=ijE[Yj] (Moments give derivatives of the characteristic function).

[F3]

For Z∼N(0,1), E∣Z∣2m=(2m−1)!! for every m≥1 (Gaussian even moments for Brownian increments); and E∣XY∣≤(EX2)1/2(EY2)1/2 for square-integrable X,Y (Cauchy-Schwarz for random variables).

[F6]

Expectations of integrable variables are linear, monotone for real variables, and satisfy ∣EU∣≤E∣U∣ (Linearity, monotonicity, and the modulus bound for expectation).

Proof

technique · direct
1.1givenF1F2F3F4algebra

Vanishing of odd moments: by [F1] the function φZ(t)=e−t2/2 is even, and by induction with [F4] each derivative φZ(j) has parity (−1)j (differentiating flips parity); hence φZ(2k+1) is odd and therefore φZ(2k+1)(0)=0 for every k≥0. All absolute moments of Z are finite, since by [F3] E∣Z∣2m=(2m−1)!! and [F3] gives E∣Z∣2m+1≤(E∣Z∣4m+2)1/2<∞; so [F2] applies to every order and gives E[Z2k+1]=i−(2k+1)φZ(2k+1)(0)=0.

1.2givenF4algebra

Derivative relation: Hm′=mHm−1 for every m≥1. This holds for m=1, since H1′=1=H0, and for m=2, since H2′=2x=2H1; for m≥2, if it holds for all indices up to m, then differentiating Hm+1=xHm−mHm−1 with [F4] gives Hm+1′=Hm+xHm′−mHm−1′=Hm+m xHm−1−m(m−1)Hm−2, and substituting xHm−1=Hm+(m−1)Hm−2 yields Hm+1′=(m+1)Hm.

1.3givenF5algebra

Monomial expansion: every m≥0 admits the expansion xm=∑j=0⌊m/2⌋cm,jHm−2j with cm,j=m!2jj!(m−2j)!. Indeed x0=H0 gives m=0; if the expansion holds for m−1≥0, then multiplying by x and using xHl=Hl+1+lHl−1 gives coefficient of Hm−2j equal to cm−1,j+(m+1−2j)cm−1,j−1 (with cr,l:=0 whenever l<0 or 2l>r, and xH0=H1 supplying the boundary case), which equals cm,j: for j≥1 the common denominator 2jj!(m−2j)! turns it into (m−1)!(m−2j)+2j (m−1)!2jj!(m−2j)!=m!2jj!(m−2j)! (using [F5]), and for j=0 it gives cm−1,0=1=cm,0. By [F5], cm,j=(m2j)(2j−1)!! is an integer and cm,0=1. Moreover the monicity and degree deg⁡Hl=l make the matrix of coefficients of x0,…,xm in the basis H0,…,Hm unitriangular with diagonal entries cm,0=1, so the expansion is the unique one and defines an invertible unitriangular change of basis in each finite degree.

2.1givenF1F3step 1.1algebraF6

Stein identity: for every real polynomial p, E[Zp(Z)]=E[p′(Z)]. Write p(x)=∑j=0dajxj; then E[Zp(Z)]=∑j=0dajE[Zj+1] and E[p′(Z)]=∑j=1djajE[Zj−1], both finite sums. The constant term contributes a0E[Z]=0 on the left and nothing on the right, and for every j≥1 one has E[Zj+1]=j E[Zj−1]: when j is even both sides vanish by step 1.1; when j is odd, [F3] gives E[Zj+1]=(j)!! and jE[Zj−1]=j(j−2)!!=j!!; here (−1)!!:=1 covers j=1, where both sides are E[Z2]=1. Hence the two sums are equal.

2.2givenstep 1.3algebra

Multivariate expansion: let N≥1 and m1,…,mN≥0. Expanding each factor by step 1.3 and multiplying out, ∏kxkmk=∑j1,…,jN(∏kcmk,jk)∏kHmk−2jk(xk); each index mk′:=mk−2jk satisfies mk′≤mk and mk′≡mk(mod2), the coefficients are integers by step 1.3, and the single tuple (j1,…,jN)=(0,…,0) contributes ∏kHmk with coefficient 1.

3.1givenstep 1.1step 1.2step 2.1algebraF6

Zero means: E[H0(Z)]=1 and E[Hm(Z)]=0 for every m≥1, by induction on m: the case m=1 is E[Z]=0 from step 1.1, and for m≥1 the recurrence and step 2.1 give E[Hm+1(Z)]=E[ZHm(Z)]−mE[Hm−1(Z)]=E[Hm′(Z)]−mE[Hm−1(Z)]=mE[Hm−1(Z)]−mE[Hm−1(Z)]=0, where step 1.2 identifies Hm′=mHm−1 and the induction hypothesis handles E[Hm−1].

4.1givenstep 1.2step 2.1step 3.1algebraF6

Orthogonality: put a(M,n):=E[HM(Z)Hn(Z)] for M,n≥0. We show a(M,n)=n! δMn. First a(0,n)=E[Hn(Z)]=δ0n by step 3.1. For M=1 and n≥1, step 2.1 gives a(1,n)=E[ZHn]=E[Hn′]=na(0,n−1). For M≥2 and n≥1, the recurrence HM=xHM−1−(M−1)HM−2, the Stein identity of step 2.1 applied to p=HM−1Hn and the derivative relation of step 1.2 give a(M,n)=E[ZHM−1Hn]−(M−1)a(M−2,n)=E[(HM−1Hn)′]−(M−1)a(M−2,n)=(M−1)a(M−2,n)+n a(M−1,n−1)−(M−1)a(M−2,n)=n a(M−1,n−1), while for n=0 and M≥1 one has a(M,0)=E[HM(Z)]=0 by step 3.1. If k≤M and k≤n, iteration gives a(M,n)=n(n−1)⋯(n−k+1) a(M−k,n−k); taking k=n when n≤M yields a(M,n)=n! a(M−n,0), which is 0 for M>n and is n! a(0,0)=n! for M=n by step 3.1, while taking k=M when M<n yields a(M,n)=n(n−1)⋯(n−M+1)a(0,n−M)=0 because n−M≥1. Hence a(M,n)=n! δMn, which is claim (i) together with step 3.1.

5.1givenstep 1.3step 2.2step 3.1step 4.1∎

Conclusion: step 3.1 and step 4.1 prove (i); step 1.3 proves (ii) (including the unitriangularity clause); step 2.2 proves (iii). No step used anything beyond the published derivative, moment and characteristic-function facts listed above.

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