I have the following problem I'm having trouble starting with:
Consider a probability function $p(\textbf{x})$ in $d$ dimensions which is a function only of radius $r=\|\textbf{x}\|$ and which has a Gaussian form
$$p(\textbf{x}) = \frac{1}{(2\pi\sigma^2)^{d/2}}\exp\left(-\frac{\|\textbf{x}\|^2}{2\sigma^2}\right).$$
By changing variables from Cartesian to polar coordinates, show that the probability mass inside a thin shell of radius $r$ and thickness $\epsilon$ is given by $\rho(r)\epsilon$ where
$$\rho(r) = \frac{S_dr^{d-1}}{(2\pi\sigma^2)^{d/2}}\exp\left(-\frac{r^2}{2\sigma^2}\right)$$
where $S_d$ is the surface area of a unit sphere in $d$ dimensions. That is:
$$S_d=\frac{2\pi^{d/2}}{\Gamma(d/2)},$$
where $$\Gamma(x)\equiv\int_0^\infty u^{x-1}e^{-u}\;du$$
So my problem here is that I'm not quite getting the picture where I'm supposed to start. What I understand from the problem statement is that I need to first, do the change the variables in function $p(\textbf{x})$, and then integrate this function over an interval of thickness $\epsilon$ to get the probability mass?
How should I start? I could try doing this problem myself after I get the picture what I'm supposed to do. The problem statement is not 100% unambiguous for me.
P.S. This problem is from the book: Neural Networks for Pattern Recognition, Christopher M. Bishop, page 29, problem 1.4.
UPDATE:
in the problem statement in Bishop's book the normalizing constant is NOT $1/(2\pi\sigma^2)^{d/2}$, but $1/(2\pi\sigma^2)^{1/2}$. I think this is a typo mistake, because otherwise the integral wouldn't sum into $1$. What do you think? I corrected $1/(2\pi\sigma^2)^{1/2}\; \rightarrow \; 1/(2\pi\sigma^2)^{d/2}$ in my original question
Thank you!
I think I found the answer:
My idea is that, if I can show that (n-sphere wiki):
$$\int_{\theta_{d-1}=0}^{2\pi}\int_{\theta_{d-2}=0}^\pi\cdots\int_{\theta_1=0}^\pi\int_{r=0}^\infty p(r)\,r^{d-1}\sin^{d-2}(\theta_{1}) \sin^{d-3}(\theta_{2}) \cdots \sin(\theta_{d-2})\,dr\,d\theta_1\,d\theta_2\,\cdots d\theta_{d-1}$$
$$=\int_\Omega d\Omega_{d-1} \int_0^\infty p(r)\,r^{d-1}\,dr=1,$$
where $\int_\Omega d\Omega_{d-1}$ is the angular components of the integral and is equal to the surface area of $d$-dimensional unit sphere $S_d$, that is:
$$S_d=\int_\Omega d\Omega_{d-1}=\frac{2\pi^{d/2}}{\Gamma(d/2)},$$
where $\Gamma(d/2)$ is the gamma function. If I can show that the above integral is equal to $1$, then
$$\int_\Omega d\Omega_{d-1} \int_0^\infty p(r)\,r^{d-1}\,dr=S_d\int_0^\infty p(r)\,r^{d-1}\,dr = \int_0^\infty \rho(r)\,dr=1,$$
which would mean that $\rho(r)\epsilon$ is the probability mass (probability mass = probability density $\times$ volume ) of shell of radius $r$ and thickness $\epsilon=dr$, which is what I'm supposed to show. So I do the integral:
$$\int_\Omega d\Omega_{d-1} \int_0^\infty p(r)\,r^{d-1}\,dr=S_d \int_0^\infty p(r)\,r^{d-1}\,dr=S_d\int_0^\infty\frac{1}{(2\pi\sigma^2)^{d/2}}\exp\left({-\frac{r^2}{2\sigma^2}}\right)\,r^{d-1}\,dr$$
$$=\frac{S_d}{(2\pi\sigma^2)^{d/2}}\int_0^\infty\exp\left({-\frac{r^2}{2\sigma^2}}\right)\,r^{d-1}\,dr=\frac{S_d}{(2\pi\sigma^2)^{d/2}}\cdot\frac{(2\pi\sigma^2)^{d/2}}{S_d}=1.$$
The last integral is solved here. Because of the final result, I conclude that the problem is done. $\blacksquare $