14 Random Sources and Sampling
Finite product measures determine the second moments of independent source slots. Diagonal coefficient moments then determine the second moments of rectangular matrix sums. Summing estimates over nonempty correction subsets gives an explicit sample count and a deterministic realization. These calculations occur in the proof of [ Ope26a , Theorem 5.2 ] .
All integrals are Bochner integrals, with value zero when the integrand is not integrable. Empty products have value one, and scalar inversion uses \(0^{-1}=0\). For a complex rectangular matrix \(A\), write \(\lVert A\rVert _2^2=\sum _{r,c}|A_{rc}|^2\) for its Hilbert–Schmidt square.
14.1 Independent source slots
Let \(S\) be a finite set. For each \(s\in S\), let \(\Omega _s\) be a measurable space and let \(I_s\) and \(L_s\) be two index sets. Given complex functions \(c_s(a,b;\cdot )\) on \(\Omega _s\), define the coefficient associated with multiindices \(i\in \prod _s I_s\) and \(l\in \prod _s L_s\) by
where \(\omega \in \prod _s\Omega _s\). The ket and bra index sets may have different cardinalities. This is the coefficient product in [ Ope26a , Equation (5.11) ] .
Let each \(\mu _s\) be a \(\sigma \)-finite measure on \(\Omega _s\), and set \(\mu =\bigotimes _{s\in S}\mu _s\). For complex functions \(f_s,g_s\),
If each \(f_s\overline{g_s}\) is integrable, the product integrand is integrable. In particular, pair-product integrability of the slot coefficients implies pair-product integrability of \(z_{il}\).
Conjugation commutes with the finite product, so the integrand is \(\prod _s(f_s(\omega _s)\overline{g_s(\omega _s)})\). The finite-product integral formula gives (2); its integrability statement applies to the same individual factors.
On the product of \(\sigma \)-finite measures, suppose the individual slot moments satisfy
Then the full multiindices have diagonal moments:
In particular, on probability spaces, let \(k\) be a positive integer and let \(p_s,q_s\) be nonnegative weights. Put \(p_i=\prod _s p_s(i_s)\) and \(\widetilde p_l=\prod _s q_s(l_s)\). If \(\nu _s(a,b)=k^{-1}\sqrt{p_s(a)q_s(b)}\), then
For finite \(I_s\), the equalities \(\sum _a p_s(a)=1\) imply \(\sum _i\prod _s p_s(i_s)=1\); this normalization identity requires no sign condition on the individual weights. The weighted moment formula is [ Ope26a , Equation (5.11) ] .
Apply (2) and substitute (3). A product of Kronecker deltas is one precisely when both complete multiindices agree, and is zero otherwise. The nonnegative-weight identity \(\prod _s\sqrt{p_s(i_s)q_s(l_s)}=\sqrt{p_i\widetilde p_l}\) gives (5). Finally, finite distributivity gives \(\sum _i\prod _s p_s(i_s)=\prod _s\sum _a p_s(a)=1\).
Let \(S\) be finite. At each slot \(s\), let \(A_s,C_s\) be finite endpoint supports with nonnegative weights \(\lambda _s,\mu _s\). Take independent standard circular complex Gaussian coordinates \(g^{(j,s)}_{ac}\), with \(\mathbb E|g^{(j,s)}_{ac}|^2=1\), and independent Gaussian factors across slots. For \(k{\gt}0\), define
The product law is a probability measure, and both \(z_{il}\) and every conjugate pair product are integrable. Set \(p_i=\prod _s\lambda _s(i_s^1)\lambda _s(i_s^2)\) and \(\widetilde p_l=\prod _s\mu _s(l_s^1)\mu _s(l_s^2)\). Their covariance is (5). If each endpoint weight list sums to one, both product weight lists also sum to one. For nonempty \(S\), \(\mathbb E z_{il}=0\). These are the actual independent coefficients in [ Ope26a , Equations (5.7), (5.8), and (5.11) ] .
The centered Gaussian pairing formula gives \(\mathbb E[(g_\alpha \overline{g_\beta }-\delta _{\alpha \beta }) \overline{(g_\gamma \overline{g_\delta }-\delta _{\gamma \delta })}] =\delta _{\alpha \gamma }\delta _{\beta \delta }\). Distinct sample coordinates remove the off-diagonal sample terms, so the slot covariance is \(k^{-1}\) times its quarter-weight square. Theorem 14.1.3 gives the full covariance. The Gaussian coefficients belong to \(L^2\), which gives the required product integrability. Finite-product factorization gives the zero mean for nonempty \(S\), and finite distributivity gives normalization.
14.2 Rectangular trace-norm comparisons
For a complex rectangular matrix \(A\), let \(\lVert A\rVert _1\) be the sum of its singular values. Equivalently, \(\lVert A\rVert _1=\sum _j\sqrt{\lambda _j(A^\dagger A)}\); for square matrices this is the trace norm. Every rectangular contraction \(U\) of the same size satisfies \(|\operatorname{tr}(A^\dagger U)|\leq \lVert A\rVert _1\), and some such \(U\) satisfies \(\operatorname{tr}(A^\dagger U)=\lVert A\rVert _1\). This is the trace-norm duality used in [ Ope26a , Equation (5.13) ] .
Diagonalize \(A^\dagger A\) in an orthonormal basis \(v_j\). The trace is a sum of column inner products, each of absolute value at most \(\lVert Av_j\rVert \lVert Uv_j\rVert \leq \sqrt{\lambda _j(A^\dagger A)}\). Equality is attained by sending \(v_j\) to \(Av_j/\sqrt{\lambda _j(A^\dagger A)}\) for positive eigenvalues, and to zero on the kernel. The resulting columns are orthonormal on the positive eigenspaces, so this map is a contraction.
For complex matrices \(A\) of size \(m\times p\) and \(B\) of size \(n\times p\), \(\lVert AB^T\rVert _1\leq \lVert A\rVert _2\lVert B\rVert _2\). No dimension enters. This is the bound used in the proof of [ Ope26a , Lemma 6.2 ] .
Let \(v_j\) be an orthonormal eigenbasis of \((AB^T)^\dagger AB^T\), so \(\lVert AB^T\rVert _1=\sum _j\lVert AB^Tv_j\rVert \). For each \(j\) with \(AB^Tv_j\neq 0\), put \(u_j=AB^Tv_j/\lVert AB^Tv_j\rVert \); these vectors are orthonormal. Then \(\lVert AB^Tv_j\rVert =\langle A^\dagger u_j,B^Tv_j\rangle \leq \lVert A^\dagger u_j\rVert \lVert B^Tv_j\rVert \). The Cauchy–Schwarz inequality for the sum, Bessel’s inequality \(\sum _j\lVert A^\dagger u_j\rVert ^2\leq \lVert A\rVert _2^2\), and Parseval’s identity \(\sum _j\lVert B^Tv_j\rVert ^2=\lVert B\rVert _2^2\) give the bound.
Let \(M\) be a complex rectangular matrix and \(k\geq 0\). There is an orthonormal family \(f_1,\dots ,f_r\) with \(r\leq k\) whose orthogonal projector \(P=\sum _i f_if_i^\dagger \) satisfies \((k+1)\lVert M-MP\rVert _2^2\leq \lVert M\rVert _1^2\). If the domain of \(M\) has dimension at most \(k\), then \(M=MP\). No dimension enters the estimate. This is the singular-value tail estimate in the proof of [ Ope26a , Lemma 6.2 ] .
Let \(s_0\geq s_1\geq \cdots \) be the singular values of \(M\), with right singular vectors \(v_j\), and project onto the first \(\min (k,n)\) of them, where \(n\) is the dimension of the domain. By Parseval’s identity, \(\lVert M-MP\rVert _2^2=\sum _{j\geq k}s_j^2\). Each term is at most \(s_ks_j\), so the tail is at most \(s_k\lVert M\rVert _1\). Monotonicity gives \((k+1)s_k\leq s_0+\cdots +s_k\leq \lVert M\rVert _1\), and multiplying the two bounds gives the claim.
For nonnegative diagonal probability matrices \(\tau ,\widetilde\tau \) and an arbitrary rectangular matrix \(W\) of compatible size,
More generally, let \(R,\widetilde R\) be arbitrary positive semidefinite trace-one matrices and let \(O\) be a rectangular contraction between their spaces. Then
Zero eigenvalues and different input dimensions are allowed. These are the comparisons in [ Ope26a , Equations (5.15) and (5.17) ] .
A contraction’s squared column sums and squared row sums are at most one. Cauchy–Schwarz with the two probability weights therefore gives \(\lVert \tau ^{1/4}U\widetilde\tau ^{1/4}\rVert _2\leq 1\). Pull the half-weights into the trace pairing and apply Hilbert–Schmidt Cauchy–Schwarz against \(U\); attained duality gives (8). For general positive matrices, diagonalize each in its own orthonormal basis and conjugate \(O\) by the corresponding unitaries. Hilbert–Schmidt norm and the contraction bound are preserved, reducing (9) to the diagonal calculation.
Let \(Z\) be an arbitrary rectangular matrix. Let \(K\) and \(\widetilde K\) be contractions from its respective output and input spaces into a common physical space tensored with a discard space. Then
The same norm bound holds for left and right contractions alone, and partial trace alone cannot increase the trace norm. This is [ Ope26a , Equation (5.13) ] .
Pull a physical test contraction \(M\) back to \(\widetilde K^\dagger (M\otimes \mathbb {1})K\). Its operator norm is at most one, since tensoring with the identity preserves that norm bound. Trace-norm duality gives (10). Taking only the corresponding part of this calculation gives the separate multiplication and partial-trace bounds.
14.3 Rectangular matrix second moments
Let \(J\) be finite, let \(\mu \) be a measure, and let \(z_j\) be complex functions with every product \(z_i\overline{z_j}\) integrable. Suppose \(\int z_i\overline{z_j}\, d\mu =\delta _{ij}v_i\), where \(v_i\in \mathbb {R}\). For complex numbers \(a_j\) and arbitrary rectangular matrices \(A_j\),
Both squared integrands are integrable. If \(\mu \) is a probability measure, then
These are the moment calculation and expectation estimate used in [ Ope26a , Equations (5.16) and (5.18) ] .
Expand \(|\sum _j z_j a_j|^2\) into the finite sum \(\sum _{i,j}z_i\overline{z_j}a_i\overline{a_j}\). Pair-product integrability permits integration term by term, and the diagonal moment identity removes every term with \(i\neq j\). Apply (11) to each matrix entry and sum over rows and columns to obtain (12). On a probability space, the nonnegative variance of \(f=\lVert \sum _j z_j A_j\rVert _2\) gives \((\mathbb E f)^2\leq \mathbb E(f^2)\), which implies (13).
14.4 Correction sums and deterministic realizations
Let \(T\) be a finite set of source positions, with \(N=|T|\). For every real \(a\),
For \(k{\gt}0\), the summand \(Q^{|U|}k^{-|U|/2}\) equals \((Q/\sqrt{k})^{|U|}\), so its sum is \((1+Q/\sqrt{k})^N-1\). The sum is zero when \(T\) is empty. If \(a\geq 0\), then \((1+a)^N-1\leq e^{Na}-1\). For \(0\leq \epsilon \leq 1\), one also has \(e^{\epsilon /8}-1\leq \epsilon /4\). These are the scalar estimates in [ Ope26a , Equation (5.19) and the following paragraph ] .
Finite distributivity gives \(\sum _{U\subseteq T}a^{|U|}=(1+a)^N\); remove the empty subset’s contribution one. The square-root identity \(k^{-n/2}=(\sqrt{k})^{-n}\) gives the sampling form. The inequality \(1+a\leq e^a\) gives the exponential estimate. Finally, \(|e^t-1|\leq 2|t|\) for \(|t|\leq 1\), applied to \(t=\epsilon /8\), gives the quarter-accuracy estimate.
For \(N\in \mathbb {N}\) and real \(Q,\epsilon \), put
Then \(k{\gt}0\) and \((8NQ/\epsilon )^2\leq k\leq (8NQ/\epsilon )^2+2\). If \(Q\geq 0\) and \(0{\lt}\epsilon \leq 1\), every positive integer \(k\) satisfying \((8NQ/\epsilon )^2\leq k\) obeys
For \(N{\gt}0\), the sample-count hypothesis gives \(Q/\sqrt{k}\leq \epsilon /(8N)\). This is an explicit integer choice for the sample count in [ Ope26a , Theorem 5.2, paragraph following Equation (5.19) ] .
Let \(T\) be finite and let \(F_U\) be integrable functions with values in a normed additive group, indexed by the nonempty subsets of \(T\). If \(\int \lVert F_U\rVert \, d\mu \leq a^{|U|}\) for each such \(U\), then
More generally, the same bound holds for an integrable real error \(e\) whenever \(e\leq \sum _U e_U\) pointwise and the integrable real correction estimates satisfy \(\int e_U\, d\mu \leq a^{|U|}\). In particular, for \(k{\gt}0\), per-subset bounds \(\int \lVert F_U\rVert \, d\mu \leq Q^{|U|}k^{-|U|/2}\) give the right side \((1+Q/\sqrt{k})^{|T|}-1\). This is the summation implication in [ Ope26a , Equation (5.19) ] .
Apply the triangle inequality pointwise, integrate the finite sum, and substitute the individual integral bounds. Formula (14) evaluates their sum. For a real error, integral monotonicity replaces the triangle inequality.
Let \(\mu \) be a probability measure, let \(Q\geq 0\) and \(0{\lt}\epsilon \leq 1\), and choose \(k\) by (15). Suppose \(e\) is integrable and \(\mathbb E e\leq (1+Q/\sqrt{k})^N-1\). If a condition \(V\) holds almost surely, there exists \(\omega \) satisfying \(V(\omega )\) and \(e(\omega )\leq \epsilon /2\). In particular, this conclusion applies with \(e=\lVert \sum _{\varnothing \neq U\subseteq T}F_U\rVert \) and \(N=|T|\), provided each \(F_U\) is integrable and \(\mathbb E\lVert F_U\rVert \leq Q^{|U|}k^{-|U|/2}\). This is the realization step after [ Ope26a , Equation (5.19) ] .