What is the distribution of the random variable

$$X = a \cdot b$$

where $a, b$ are unit $m$-vectors independently drawn from the uniform distribution on the unit $m$-sphere? Is there a special name for this distribution?

EDIT: Thanks to Joriki for his answer. The distribution of the dot product of two vectors in $S^{p-1}$ is also the null distribution for Pearson's rho given $p+1$ observations from two independent normally distributed populations. Alternatively, $X^2 \sim Beta(\frac{1}{2},\frac{p-1}{2})$.


1 Answer 1


The surface area of the hyphersphere between $X$ and $X+\mathrm dX$ is $Y^{m-2}/Y\mathrm dX=Y^{m-3}\mathrm dX=(1-X^2)^{(m-3)/2}\mathrm dX$, where $Y=\sqrt{1-X^2}$ is the sine of the angle between $a$ and $b$. Up to normalization, this gives you the distribution of $X$. The normalization factor is the ratio of the surface area of a unit $(m-2)$-sphere and a unit $(m-1)$-sphere.

In particular, for $m=3$, the distribution is uniform. I don't know whether there's a name for it for $m\neq3$.

[Edit in response to the comment:] To get $Y^{m-2}/Y\mathrm dX$, it's easiest (at least for me) to think of the case of the $2$-sphere. Without loss of generality, let $a$ point along the positive $x$-axis, so the dot product is the $x$-coordinate of $b$. Then the dot product is constant on circles (in general: hyperspheres) orthogonal to the $x$ direction, and the sphere's surface is made up of small strips bounded by such circles. The surface area of the strip between $x$ and $\mathrm dx$ is the perimeter (in general: hypervolume) of the circle at $x$, which is proportional to $\sqrt{1-x^2}$ (in general: $\sqrt{1-x^2}^{m-2}$), times the width of the strip. The width of the strip is $\mathrm dx$ divided by the sine of the angle that the strip forms with the $x$ axis, which is $\sqrt{1-x^2}$, so the surface element is $\sqrt{1-x^2}^{m-3}\mathrm dx$.

  • $\begingroup$ Can you give more details? How does the $Y^{m−2}/YdX$ come out? $\endgroup$
    – Ziyuan
    Jun 3, 2011 at 15:29

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