Math 289A: Lecture 7

In this lecture we will review what we have done so far from a slightly different perspective. Starting with the Fourier kernel

$F \colon \mathbb{R} \times \mathbb{R} \longrightarrow \mathbb{C},$

which is defined by $F(a,b) = e^{iab},$ we can define the characteristic function of a random variable $X$ with distribution $\mu$ by

$\varphi_X(a) = \mathbf{E}F(a,X) = \int\limits_{\mathbb{R}} F(a,x)\mu(\mathrm{d}x).$

This is an absolutely continuous function from $\mathbb{R}$ into the closed unit disc in $\mathbb{C}$ which uniquely determines the measure $\mu.$ To get some feel for which this might be, expand the Fourier kernel in a power series,

$F(a,b) = 1 + \sum_{d=1}^\infty \frac{i^d}{d!} (ab)^d.$

Then, assuming we can interchange expectation with the summation we get

$\mathbf{E}F(a,X) = 1 + \sum_{d=1}^\infty \frac{i^d}{d!} a^d\mathbf{E}[X^d],$

where

$\mathbf{E}[X^d] = \int\limits_{\mathbb{R}} x^d \mu(\mathrm{d}x)$ ,

is the degree $d$ moment of $X,$ or equivalently of $\mu.$ So at least heuristically, the characteristic function of $X$ is a generating function for the moments of its distribution. One situation in which this formal computation is quantitatively correct is when $\mu$ has compact support, so if the characteristic function does indeed characterize the distribution of $X$ then it must be the case that $\mu$ is characterized by its moments. It is not difficult to see why the why the moments of a probability measure with compact support determine it completely: if we know how to compute the expectation of polynomial functions of $X$ , then thanks to the Weierstrass approximation theorem we also know how to compute the expectation of continuous functions of $X,$ and this in turn tells us how $\mu$ measures intervals, by taking expectations of smooth bump functions. Of course, to obtain the fully general result that the characteristic function of $X$ completely determines its distribution a different argument is needed. For example, the Cauchy distribution has no moments whatsoever, but its characteristic function still exists and determines it uniquely.

The $N$ -dimensional Fourier kernel

$F_N \colon \mathbb{R^N} \times \mathbb{R^N} \longrightarrow \mathbb{C}$

is defined by

$F_N(A,B) = e^{i \langle A,B\rangle} = e^{i(a_1b_1+\dots+a_Nb_N)}, \quad A,B \in \mathbb{R}^N$

and the characteristic function of an $N$ -dimensional random vector $X_N$ is the continuous function $\mathbb{R}^N \to \mathbb{C}$ defined by

$\varphi_{X_N}(A) = \mathbf{E}F_N(A,X_N) = \int\limits_{\mathbb{R}^N} F_N(A,X) \mu_N(\mathrm{d}X),$

where $\mu_N$ is the distribution of $X_N$ in $\mathbb{R}^N.$ It is still true that $\varphi_{X_N}$ exists and uniquely determines $\mu_N$ , and we can tell a similar story about moments which supports this statement. To do so we first have to expand the $N$ -dimensional Fourier kernel in a power series. This is easy but let us first make some general observations about what this expansion must look like before doing the computation. Since $F_N(A,B) = F_N(B,A)$ and

$F_N(PA,PB) = e^{i\langle PA,PB\rangle} = e^{i \langle A,B \rangle} = F_N(A,B),$

for any $N \times N$ permutation matrix $P,$ we have

$F_N(A,B) =1 + \sum_{d=1}^\infty \frac{d^d}{d!} F_N^d(A,B)$

where $F_N^d(A,B)$ is a homogeneous polynomial of degree $d$ in $a_1b_1,\dots,a_Nb_N$ which is invariant under independent permutations of these variables. Therefore, we have

$F_N^d(A,B) = \sum\limits_{\alpha \in \mathsf{C}_N^d} (a_1b_1)^{\alpha_1}\dots (a_Nb_N)^{\alpha_N}F_N(\alpha)$

where $\mathsf{C}_N^d$ is the set of weak $N$ -part compositions $\alpha$ of $d \in \mathbb{N}$ and $F_N(\alpha)$ is a numerical coefficient. It is easy to determine this coefficient explicitly: since

$F_N(A,B) = e^{ia_1b_1} \dots e^{ia_Nb_N} = \left(\sum\limits_{d=0}^\infty \frac{i^d}{d!}(a_1b_1)^d \right)\dots \left(\sum\limits_{d=0}^\infty \frac{d^d}{d!}(a_Nb_N)^d \right),$

we have

$F_N(\alpha) = \frac{d!}{\alpha_1! \dots \alpha_N!}={d \choose \alpha_1,\dots,\alpha_N},$

the multinomial coefficient. Thus, the expected value of $F_N^d(A,X_N)$ is a generating polynomial for degree $d$ mixed moments of the entries of the random vector $X_N=(X_N(1),\dots,X_N(N))$ ,

$\mathbf{E}F_N^d(A,X_N) = \sum\limits_{\alpha \in \mathsf{C}_N^d} a_1^{\alpha_1} \dots a_N^{\alpha_N} \mathbf{E}[X^{\alpha_1}_N(1)\dots X^{\alpha_N}_N(N)]F_N(\alpha).$

In the previous lectures we defined a strange new kernel

$K_N \colon \mathbb{R}^N \times \mathbb{R}^N \longrightarrow \mathbb{C}$

$K_N(A,B) = \int\limits_{\mathbb{U}_N} F_N(A,|U|^2B)\mathrm{d}U,$

where the integration is against Haar probability measure on the unitary group $\mathbb{U}_N,$ and for any $U \in \mathbb{U}_N$ the matrix

$|U|^2=\begin{bmatrix} {} & \vdots & {} \\ \dots & |U_{xy}|^2 & \dots \\ {} & \vdots & {} \end{bmatrix}$

is obtained from $U=[U_{xy}]$ by taking the squared modulus of each entry. Thus for any $B \in \mathbb{R}^N$ the vector

$|U|^2B=\begin{bmatrix} |U_{11}|^2b_1 + \dots + |U_{1N}|^2b_N \\ |U_{21}|^2b_1 + \dots + |U_{2N}|^2b_N \\ \vdots \\ |U_{N1}|^2b_1 + \dots + |U_{NN}|^2b_N\end{bmatrix} \in \mathbb{R}^N$

has coordinates which are superpositions of the coordinates of $B,$ and because of this we named $K_N$ the $N$ -dimensional “quantum Fourier kernel.”

Problem 7.1. Prove that $K_N$ is a bounded symmetric kernel on $late \mathbb{R}^N$, i.e. $|K_N(A,B)|\leq 1$ and $K_N(A,B)=K_N(B,A)$ for all $A,B \in \mathbb{R}^N.$

Then, given a random vector $B_N$ with distribution $\sigma_N$ in $\mathbb{R}^N,$ we defined its quantum characteristic function by

$\psi_{B_N}(A)=\mathbf{E}K_N(A,B_N) = \int\limits_{\mathbb{R}^N} K_N(A,B)\sigma(\mathrm{d}B).$

This is a continuous function from $\mathbb{R}^N$ into the closed unit disc in $\mathbb{C}$ , and we claimed that $\psi_{B_N}$ uniquely determines the distribution $\sigma_N$ of $B_N.$ Since the quantum characteristic function of $B_N$ is clearly distinct from its classical characteristic function

$\varphi_{B_N}(A) = \mathbf{E}F_N(A,B_N)=\int\limits_{\mathbb{R}^N} F_N(A,B)\sigma(\mathrm{d}B),$

this must be true for some different reason. To understand what this reason might be, at least heuristically, we could proceed along the same lines we did with $F_N$ and try to determine if $\varphi_{B_N}(A)$ is some sort of generating function encoding natural statistics of $B_N$ .

Let’s give this a try. The first step is to expand the quantum Fourier kernel $K_N$ as a power series. Before doing the computation, we can look for some easy symmetries of $K_N$ which will tell us in advance what a series expansion of $K_N$ ought to look like. In fact, the quantum Fourier kernel is much more symmetric than the classical Fourier kernel – it is invariant under independent (as opposed to simultaneous) permutations of the coordinates of its arguments.

Problem 7.2. Prove that $K_N(PA,QB)=K_N(A,B)$ for any $A,B \in \mathbb{R}^N$ and any $N\times N$ permutation matrices $P,Q.$

From Problems 7.1 and 7.2, we can indeed infer the general form of a series expansion of the kernel $K_N.$ Namely, we have

$K_N(A,B) = 1 + \sum\limits_{d=1}^\infty \frac{i^d}{d!} K_N^d(A,B),$

where $K_N^d(A,B)$ is a homogeneously degree $d$ polynomial function of the coordinates $a_1,\dots,a_N,b_1,\dots,b_N$ which is invariant under independent permutations of $a_1,\dots,a_N$ and $b_1,\dots,b_N$ and stable under swapping these two sets of variables. What does such a polynomial look like?

A classical theorem of Newton asserts that a symmetric homogeneous degree $d$ polynomial in $N$ variables is a polynomial function of the power-sum polynomials

$p_1(x_1,\dots,x_N) = x_1+\dots+x_N \\ p_2(x_1,\dots,x_N)=x_1^2+\dots+x_N^2\\ \vdots \\ p_N(x_1,\dots,x_N)=x_1^N+\dots+x_N^N.$

Equivalently, the vector space $\Lambda_N^d$ of symmetric homogeneous degree $d$ polynomials in $N$ variables has as a linear basis the polynomials

$p_\alpha(x_1,\dots,x_N)=\prod\limits_{k=1}^{\ell(\alpha)} p_{\alpha_1k}(x_1,\dots,x_N),$

where $\alpha=(\alpha_1,\dots,\alpha_r)$ ranges over the set $\mathsf{P}_N^d$ of partitions of $d$ with largest part at most $N,$ and $\ell(\alpha)$ denoting the number of parts.

Problem 7.3. Prove that for any $\alpha \in \mathsf{P}_N^d$ and any $A \in \mathbb{R}^N,$ the evolution $p_\alpha(A)=p_\alpha(a_1,\dots,a_N)$ satisfies $|p_\alpha(A)| \leq \|A\|_\infty N^{\ell(\alpha)},$ and that this bound is sharp.

Thus, a power series expansion of the quantum Fourier kernel can be written in the form

$K_N(A,B) = 1 + \sum\limits_{d=1}^\infty \frac{i^d}{d!}\sum\limits_{\alpha,\beta \in \mathsf{P}_N^d} \frac{p_\alpha(a_1,\dots,a_N)}{N^{\ell(\alpha)}}\frac{p_\beta(b_1,\dots,b_N)}{N^{\ell(\beta)}}K_N(\alpha,\beta),$

where the coefficients satisfy $K_N(\alpha,\beta)=K_N(\beta,\alpha).$ This complicated-looking expression is actually quite meaningful. For any vector $A \in \mathbb{R}^N$ , the normalized power sum polynomial of degree $d$ in the coordinates of $A$ is

$\frac{p_k(a_1,\dots,a_N)}{N} = \int\limits_{\mathbb{R}} x^k \eta_A(\mathrm{d}x),$

where

$\eta_A = \frac{1}{N} \sum\limits_{i=1}^N \delta_{a_i}$

is the discrete probability measure on $\mathbb{R}$ which places equal mass at each coordinate of $A.$ This is called the empirical distribution of $A \in \mathbb{R}^N.$ Thus, the expansion of the quantum Fourier kernel $K_N(A,B)$ is a generating function for the moments of the empirical distributions of its arguments. Consequently, for $B_N$ a random vector, the quantum characteristic function $\psi_{B_N}(A)$ is a generating function for expectations of polynomials in the random variables