# Distribution of dot product of two unit random vectors

Consider $\mathbf{u}, \mathbf{v}\in \mathcal{C}^M$ to be two independent unit norm random vectors on the $M-1$ dimensional complex sphere $\mathcal{S}^{M-1}$. In addition, $\mathbf{u}$ follows an isotropic distribution (i.e., $\mathbf{u}$ is uniformly distributed on the complex sphere $\mathcal{S}^{M-1}$. What is the distribution of $Z=|\mathbf{u}\cdot\mathbf{v}|^2$?

I know that if $\mathbf{v}$ is also uniformly distributed on the complex sphere $\mathcal{S}^{M-1}$, then $Z$ follows Beta$(1, M-2)$ distribution. (I don't know how to prove this!) Does the same result hold if $\mathbf{v}$ follows an arbitrary distribution?

• First, consider the distribution of $|u\cdot v|^2$ when $v$ has one fixed value (and $u$ is uniformly distributed). That distribution is the same, regardless of the fixed value that you chose for $v$. So you also get the same distribution when you first pick $v$ any way you like, for example randomly with respect to any probability distribution you choose. – Andreas Blass Jun 10 '15 at 17:26
• @AndreasBlass Could you please tell me what is the distribution of $\mathbf{u} \cdot \mathbf{v}$ ? – tam Nov 19 '15 at 10:22
• Have you proved it? Why do I get that Z follows Beta(1,M−1) distribution – QiangLi Nov 20 '18 at 12:16
• You stated that the distribution of $|u\cdot v|^2$ is Beta. Do you have any book in mind that proves this, please? – Chrysanthi Paschou Nov 26 '19 at 14:55

## 2 Answers

If $u$ is uniformly distributed over the sphere, we can write it as $u=Uv$, where $U$ is a unitary transformation uniformly distributed over the unitary group. Then the quantity $|u\cdot v|^2$ is just the modulus square of a matrix element of $U$. So your question is: what is the distribution of the modulus square of a matrix element of a random unitary matrix?

Consider a column of the matrix as a random vector. The only constraint is must satisfy is that its norm must be unit, $\sum_{i=1}^N|z_i|^2=1$. In other words, the joint distribution of the entries is simply

$$\delta(1-\sum_{i=1}^N|z_i|^2)=\int ds e^{is(1-\sum_{i=1}^N|z_i|^2)}$$

The distribution of a single element, $z_1$ say, is obtained integrating over the remaining ones. Indeed, you get something proportional to $(1-|z_1|^2)^{N-2}$ in dimension $N$, which is your Beta distribution.

This is independent of $v$, so the distribution of $v$ is irrelevant.

A very similar argument applies to the real case and random orthogonal matrices.

This is discussed in several articles, such as

Kus, Mostowski and Haake, J. Phys. A 21, L1073 (1988)

Haake and Zyczkowski, Phys. Rev. A 42, 1013 (1990)

K. Zyczkowski, H.-J. Sommers, J. Phys. A 33, 2045 (2000)

$$\newcommand{\bu}{\boldsymbol{u}}$$ $$\newcommand{\bv}{\boldsymbol{v}}$$ $$\newcommand{\bP}{\mathbb{P}}$$ $$\newcommand{\bR}{\mathbb{R}}$$

Here are the details in Andrea Blass' comment. We define

$$F:S^{m-1}\times S^{m-1} \to \bR, \;\;F(\bu,\bv):=|\bu\cdot\bv|^2.$$ Note that the range of $$F$$ is $$[0,1]$$. Denote by $$p(d\bv)$$ the probability distribution of $$\bv$$. For every interval $$[a,b]\subset (0,1)$$ we have

$$\bP[a\leq f\leq b] =\int_{S^{m-1}} \bP[a\leq F\leq b| \bv=\bv_0] p(d\bv_0)$$

(use the independence of $$\bu$$ and $$\bv$$)

$$= \int_{S^{m-1}} \Bigl(\;{\rm Area}\;\{ \sqrt{a}\leq |\bu\cdot\bv_0|\leq \sqrt{b}\}\;\Bigr)\; p(d \bv_0)$$

$$=2 \int_{S^{m-1}} \underbrace{ \Bigl(\;{\rm Area}\;\{ \sqrt{a}\leq \bu\cdot\bv_0\leq \sqrt{b}\}\;\Bigr)}_{=:I(a,b,\bv_0)}\; p(d \bv_0)$$

Due to the rotational symmetry, the integrand $$I(a,b,\bv_0)$$ is independent of $$\bv_0$$ so I will denote it by $$I(a,b)$$. Hence

$$\bP[a\leq f\leq b]= 2 I(a,b).$$

To compute $$I(a,b)$$ use the coarea formula exactly as in Example 9.1.10 of these notes. We have $$\newcommand{\bsi}{\boldsymbol{\sigma}}$$

$$I(a,b) = \bsi_{m-2}\int_{\sqrt{a}}^{\sqrt{b}}(1-t^2)^{\frac{m-3}{2}} dt,$$

where $$\bsi_k$$ denotes the area of the unit $$k$$-dimensional sphere

$$\bsi_k=\frac{2\pi^{\frac{k+1}{2}}}{\Gamma\left(\frac{k+1}{2}\right)}.$$

If $$\rho_F(x)$$ denotes the probability density of $$F$$, then we deduce that

$$\rho_F(x)=2\frac{d}{dh}\Bigl|_{h=0} I(x,x+h)=\bsi_{m-2} x^{-1/2}(1-x^2)^{\frac{m-3}{2}}.$$

Thus the answer is independent of the distribution of $$\bv$$.