Distribution of scalar products of two random unit vectors in $D$ dimensions

Question

If $\mathbf{x}$ and $\mathbf{y}$ are two independent random unit vectors in $\mathbb{R}^D$ (uniformly distributed on a unit sphere), what is the distribution of their scalar product (dot product) $\mathbf x \cdot \mathbf y$?

I guess as $D$ grows the distribution quickly (?) becomes normal with zero mean and variance decreasing in higher dimensions $$\lim_{D\to\infty}\sigma^2(D) \to 0,$$ but is there an explicit formula for $\sigma^2(D)$?

Update

I ran some quick simulations. First, generating 10000 pairs of random unit vectors for $D=1000$ it is easy to see that the distribution of their dot products is perfectly Gaussian (in fact it is quite Gaussian already for $D=100$), see the subplot on the left. Second, for each $D$ ranging from 1 to 10000 (with increasing steps) I generated 1000 pairs and computed the variance. Log-log plot is shown on the right, and it is clear that the formula is very well approximated by $1/D$. Note that for $D=1$ and $D=2$ this formula even gives exact results (but I am not sure what happens later).

Answer 1

Because (as is well-known) a uniform distribution on the unit sphere $S^{D-1}$ is obtained by normalizing a $D$-variate normal distribution and the dot product $t$ of normalized vectors is their correlation coefficient, the answers to the three questions are:

1. $u= (t+1)/2$ has a Beta$((D-1)/2,(D-1)/2)$ distribution.

2. The variance of $t$ equals $1/D$ (as speculated in the question).

3. The standardized distribution of $t$ approaches normality at a rate of $O\left(\frac{1}{D}\right).$

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Method

The exact distribution of the dot product of unit vectors is easily obtained geometrically, because this is the component of the second vector in the direction of the first. Since the second vector is independent of the first and is uniformly distributed on the unit sphere, its component in the first direction is distributed the same as any coordinate of the sphere. (Notice that the distribution of the first vector does not matter.)

Finding the Density

Letting that coordinate be the last, the density at $t \in [-1,1]$ is therefore proportional to the surface area lying at a height between $t$ and $t+dt$ on the unit sphere. That proportion occurs within a belt of height $dt$ and radius $\sqrt{1-t^2},$ which is essentially a conical frustum constructed out of an $S^{D-2}$ of radius $\sqrt{1-t^2},$ of height $dt$, and slope $1/\sqrt{1-t^2}$. Whence the probability is proportional to

$$\frac{\left(\sqrt{1 - t^2}\right)^{D-2}}{\sqrt{1 - t^2}}\,dt = (1 - t^2)^{(D-3)/2} dt.$$

Letting $u=(t+1)/2 \in [0,1]$ entails $t = 2u-1$. Substituting that into the preceding gives the probability element up to a normalizing constant:

$$f_D(u)du \; \propto \; (1 - (2u-1)^2)^{(D-3)/2} d(2u-1) = 2^{D-2}(u-u^2)^{(D-3)/2}du.$$

It is immediate that $u=(t+1)/2$ has a Beta$((D-1)/2, (D-1)/2)$ distribution, because (by definition) its density also is proportional to

$$u^{(D-1)/2-1}\left(1-u\right)^{(D-1)/2-1} = (u-u^2)^{(D-3)/2} \; \propto \; f_D(u).$$

Determining the Limiting Behavior

Information about the limiting behavior follows easily from this using elementary techniques: $f_D$ can be integrated to obtain the constant of proportionality $\frac{\Gamma \left(\frac{D}{2}\right)}{\sqrt{\pi } \Gamma \left(\frac{D-1}{2}\right)}$; $t^k f_D(t)$ can be integrated (using properties of Beta functions, for instance) to obtain moments, showing that the variance is $1/D$ and shrinks to $0$ (whence, by Chebyshev's Theorem, the probability is becoming concentrated near $t=0$); and the limiting distribution is then found by considering values of the density of the standardized distribution, proportional to $f_D(t/\sqrt{D}),$ for small values of $t$:

$$\eqalign{ \log(f_D(t/\sqrt{D})) &= C(D) + \frac{D-3}{2}\log\left(1 - \frac{t^2}{D}\right) \\ &=C(D) -\left(1/2 + \frac{3}{2D}\right)t^2 + O\left(\frac{t^4}{D}\right) \\ &\to C -\frac{1}{2}t^2 }$$

where the $C$'s represent (log) constants of integration. Evidently the rate at which this approaches normality (for which the log density equals $-\frac{1}{2}t^2$) is $O\left(\frac{1}{D}\right).$

This plot shows the densities of the dot product for $D=4, 6, 10$, as standardized to unit variance, and their limiting density. The values at $0$ increase with $D$ (from blue through red, gold, and then green for the standard normal density). The density for $D=1000$ would be indistinguishable from the normal density at this resolution.

Answer 2

Let's find the distribution and then the variance follows by standard results. Consider the vector product and write it on it's cosine form, i.e. note that we have $$P(x'y\leq t)=P(|x||y|\cos\theta\leq t)=P(\cos\theta\leq t)=\mathbb{E}P(\cos\theta\leq t\mid y),$$ where $\theta$ is the angle between $x$ and $y$. In the last step I have used that for any events $A$ and $B$ $$\mathbb EP(A\mid B):=\mathbb{E}[\mathbb{E}[\chi_A\mid B]]=\mathbb{E}\chi_A=P(A).$$ Now consider the term $P(\cos\theta\leq t\mid y)$ . It is clear that since $x$ is choosen uniformly with respect to the sphere surface, it does not matter what $y$ actually is, only the angle between $x$ and $y$ matters. Thus, the term inside the expectation is actually constant as a function of $y$ and we can w.l.o.g. assume that $y=[1,0,0,\dots ]'.$ Then we get that $$P(x'y\leq t)=P\left( x_1\leq t\right).$$ but since $x_1$ is the first coordinate of a normalized Gaussian vector in $\mathbb{R}^n,$ we have that $x'y$ is Gaussian with variance $1/n$ by invoking the asymptotic result of this paper.

For an explicit result of the variance, use the fact that the dot product is mean zero by independence and, as shown above, distributed like the first coordinate of $x$. By these results, finding $\text{Var}(x'y)$ amounts to finding $\mathbb{E}x_1^2$. Now, note that per construction $x'x=1$ and so we can write $$1=\mathbb{E}x'x=\mathbb{E}\sum_{i=1}^nx_i^2=\sum_{i=1}^n\mathbb{E}x_i^2=n\mathbb{E}x_1^2,$$ where the last equality follows from that the coordinates of $x$ are identically distributed. Putting things together, we have found that $\text{Var}(x'y)=\mathbb{E}x_1^2=1/n$

Answer 3

To answer the first part of your question, denote $Z = \langle X,Y \rangle = \sum X_i Y_i$. Define $$ f_{Z_i}(z_i) = \int_{-\infty}^\infty f_{Z_1,\ldots,Z_D}(z_1,\ldots,z_D) \: d z_i $$ The product of the $i^{th}$ elements of $X$ and $Y$ denoted here as $Z_i$ will be distributed according to the joint distribution of $X_i$ and $Y_i$. $$ f_{Z_i}(z_i) = \int_{-\infty}^\infty f_{X_i,Y_i}(x,\frac{z_i}{x})\frac{1}{|x|}dx $$ then since $Z = \sum Z_i$, $$ f_Z(z) = \int_{-\infty}^\infty \ldots \int_{-\infty}^\infty f_{Z_1,\ldots,Z_D} (z_1,\ldots,z_d) \: \delta(z - \sum z_i)\: dz_1\ldots d z_d $$

For the second part, I think that if you want to say anything interesting about the asymptotic behaviour of $\sigma$ you need to at least assume independence of $X$ and $Y$, and then apply a CLT.

For instance if you were willing to assume that the $\{Z_1,\ldots,Z_D\}$ are i.i.d with $\mathbb{E}[Z_i] = \mu$ and $\mathbb{V}[Z_i] = \sigma^2$ you could say that $\sigma^2(D) = \frac{\sigma^2}{D}$ and $\lim_{D\to\infty} \sigma^2(D) = 0$.