The Polarization Identity and Wick's Formula

Posted on April 3, 2016
Tags: math

Let \(V\) be a real vector space. Recall that a symmetric bilinear form on \(V^2\) is a function \(T: V^2 \to \mathbf{R}\) such that \begin{equation} \label{eq:1} T(\alpha x + \beta y, z) = \alpha T(x,z) + \beta T(y,z) \quad \text{and} \quad T(x,y) = T(y,x) \end{equation} for every \(\alpha, \beta \in \mathbf{R}\) and \(x,y,z \in V\). Define \(Q(x) = T(x,x)\). Then, \begin{equation} \label{eq:2} \frac{1}{8}[Q(x+y) - Q(x-y) - Q(-x+y) + Q(x+y)] = T(x,y) \end{equation}

for every \(x,y \in V\). In particular, \(T\) is determined completely by the values \(\{Q(x) = T(x,x) : x \in V\}\). This is the general case of the following identity.

Theorem 1 (Polarization Identity). Let \(V\) be a real vector space, and let \(T: V^n \to \mathbf{R}\) be a symmetric \(n\)-linear form. Then, \begin{equation} \label{eq:3} T(x_1, \dots, x_n) = \frac{1}{2^n n!} \sum_{(\epsilon_1, \dots, \epsilon_n) \in \{-1,1\}^n} \epsilon_1 \dots \epsilon_n Q(\epsilon_1 x_1 + \dots + \epsilon_n x_n) \end{equation} for every \(x_1, \dots, x_n \in V\), where \begin{equation} \label{eq:4} Q(x) = T(x, \dots, x). \end{equation}

In particular, \(T\) is completely determined by \(Q\).

The Polarization Identity yields a quick proof of the following formula due to Wick.

Proposition (Wick’s Formula). Let \((X_1, \dots, X_n)\) be a centered Gaussian vector. Then \begin{equation} \label{eq:5} \mathbf{E}[X_1 \dots X_n] = \sum \prod \mathbf{E}[X_i X_j] \end{equation}

where the RHS is taken over all partitions of \(\{1, \dots, n\}\) into pairs \(\{(i_1, j_1), \dots, (i_{n/2}, j_{n/2})\}\).

Proof. It is clear that both the LHS and RHS are symmetric and \(n\)-linear on the vector space \(V\) of centered Gaussian variables. By Theorem 1, it suffices to establish (\ref{eq:5}) when \(X_1 = \ldots = X_n =: X \sim N(0,\sigma^2)\). This follows from the familiar formula \begin{equation} \label{eq:6} \mathbf{E}[X^n] = \begin{cases} 0 & \text{if $n$ is odd}\\ (n-1)!! \sigma^2 & \text{if $n$ is even} \end{cases} \end{equation} where \begin{equation} \label{eq:7} (n-1)!! = 1 \times 3 \times 5 \times \dots \times (n-1) = \frac{n!}{(n/2)!} \frac{1}{2^{n/2}} \end{equation}

is the number of partitions of \(\{1, \dots, n\}\) into pairs (e.g. by induction on \(m = n/2\)).