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This is the exercise of 1.18 on page 62 of pattern recognition and machine learning.

We can use the result ($\int_{-\infty}^\infty \exp(-\frac{x^2}{2\sigma^2} \, \mathrm d x) = (2\pi\sigma^2)^{1/2}$) to derive an expression for the surface area $S_D$, and the volume $V_D$, of a sphere of unit radius in $D$ dimensions. To do this, consider the following result, which is obtained by transforming from Cartesian to polar coordinates:

(1) $\prod_{i=1}^D \int_{-\infty}^\infty e^{-x_i^2} \, dx_i = S_D\int_0^\infty e^{-r^2} r^{D-1} \, dr$

Using the definition ($\Gamma(x)=\int_0^\infty u^{x-1}e^{-u} \, du$), together with ($\int_{-\infty}^\infty \exp(-\frac{x^2}{2\sigma^2} \, \mathrm d x)=(2\pi\sigma^2)^{1/2}$), evaluate both sides of the equation, and hence show that:

(2) $S_D=\frac{2\pi^{D/2}}{\Gamma(D/2)}$

Next, by integrating with respect to radius from $0$ to $1,$ show that the volume of the unit sphere in $D$ dimensions is given by:

(3) $V_D=\frac{S_D}{D}$

I have completed all the proofs but am confused about the meaning of these proofs:

  1. What is the meaning of equation (1), and how to convert it from left to right?
  2. By requiring to integrating equation (1) on both sides to obtain volume, I guess the left side of (1) is volume, but why volume is not $\pi^{D/2}$ then?
  3. Can someone give me the hint of the idea here to solve the problem of volumes of $n$-ball please?

Thank you very much in advance.

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Tengerye
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1 Answers1

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Equation $1$ is the transformation from Cartesian coordinates to $D$ dimensional spherical coordinates. The left side is the multiplication of $D$ dimensions of the normal distribution. The right side is transforming the integral to doing it by spherical shells. A more concrete example is to do it in $D=2$, which I do in this answer. For $D=2$ the $r^{D-1}$ factor allows a $u-$substitution to solve the integral on the right.

Both sides of $1$ are integrals over all space of this function. That comes from the limits on the left being $\pm \infty$ in each axis and on the right by integrating from $r=0$ to $r=+\infty$. The left side is $\pi^{D/2}$ but it is not the volume of the $D-$ball.

$2$ gives you an explicit equation for the surface of the unit $D-$ball. $3$ tells you how to convert that to the volume of the unit $D-$ball. A $D-$ball of radius $R$ then has volume $$V_D(R)=\frac{2\pi^{D/2}}{D\Gamma(D/2)}R^D$$

Ross Millikan
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