What's the expected radius of a hypersphere

Question

I would like to compute the expected radius of a hypersphere (dimension $N$) given these conditions:

• radius $R\in[R_{min}, R_{max}]$,
• radius is acquired by uniformly chosing point from $S_N(R_{max})\backslash{S_N(R_{min})}$, and taking its radius,
• $A+B$ number of points $x = (x_1,x_2,...,x_N)$ have been chosen uniformly from the same subspace $S_N(R_{max})\backslash{S_N(R_{min})}$. $A$ points are from the space $S_N(R)\backslash{S_N(R_{min}})$, and $B$ points are from $S_N(R_{max})\backslash{S_N(R)}$,
• $E(R)=f(R_{min},R_{max},A,B,N)$.

Done so far: I have writted distribution function of $R$ as follows: $$f_R(r)=\binom{A+B}{B}\frac{(r^N-R_{min}^N)^A(R_{max}^N-r^N)^B}{(R_{max}^N-R_{min}^N)^{A+B}}\text{ .}$$ I've got the function by comparing volumes between hyperspheres for given radius. I've had the intention of trying to compute the expected value from this distribution function, but it gets a bit messy if I try to crack it and integrate polynomials that I would get.

Questions:

• (of very very minor importance) is general approach alright. I suppose it is.
• (of major importance) is there a nice way to compute the $E(R)$?

Probable way: If I would write $$p(r,N)=\frac{r^N-R_{min}^N}{R_{max}^N-R_{min}^N}\text{ ,}$$ then the density function would be like this $$f_R(r)=\binom{A+B}{B}p^A (1-p)^B\text{ .}$$ It looks rather similar to binomial distribution to me...

Answer 1

Here is my interpretation of your question:

$n=a+b+1$ points are uniformly chosen in $S_d(R_\max)\backslash S_d(R_\min)$, and are sorted by their radius. You are interested in the expectation of $a+1$-th smallest radius.

The radius $R$ of a point, uniformly chosen in $d$-dimensional ball $S_d(R_\max)$, has a known probability density: $$ \mathrm{d} F_R(r) = \frac{d \cdot r^{d-1}}{R_\max^d} \mathbf{1}_{0 < r < R_\max} \mathrm{d} r $$ Truncating the distribution to $(R_\min, R_\max)$, the density becomes $$ \mathrm{d} \bar{F}_R(r) = \frac{d \cdot r^{d-1}}{R_\max^d - R_\min^d} \mathbf{1}_{R_\min < r < R_\max} \mathrm{d} r $$ We are now interested in finding the mean of $a+1$-th out of $a+b+1$ order statistics, whose probability density is well-known: $$ \begin{eqnarray} \mathrm{d} F_{a+b+1}^{(a+1)}(r) &=& (a+b+1) \binom{a+b}{a} f_R(r) F_R(r)^a (1-F_R(r))^b \mathrm{d} r \\ &=& (a+b+1) \binom{a+b}{a} \cdot d \cdot r^{d-1} \cdot \frac{\left(r^d - R_\min^d\right)^a \cdot\left(R_\max^d - r^d \right)^b }{\left(R_\max^d-R_\min^d\right)^{a+b+1}} \mathrm{d} r \\ &=& (a+b+1) \binom{a+b}{a} \cdot \frac{\left(r^d - R_\min^d\right)^a \cdot\left(R_\max^d - r^d \right)^b }{\left(R_\max^d-R_\min^d\right)^{a+b+1}} \mathrm{d} r^d \\ &=& (a+b+1) \binom{a+b}{a} \cdot u^a \cdot\left(1 -u \right)^b \mathrm{d} u \\ &=& \frac{1}{B(a,b)} u^a \cdot\left(1 -u \right)^b \mathrm{d} u \end{eqnarray} $$ where $u = \frac{r^d - R_\min^d}{R_\max^d-R_\min^d} = \bar{F}_R(r)$, meaning that $U = \bar{F}_R(R)$ follows $\mathrm{Beta}(a+1,b+1)$ distribution. Thus $$ \begin{eqnarray} \mathbb{E}(R) &=& \mathbb{E}\left( \left( R_\min^d + U (R_\max^d-R_\min^d) \right)^{1/d} \right) \\ &=& \int_0^1 \frac{u^a (1-u)^b}{\mathrm{B}(a+1,b+1)} \left( R_\min^d + u (R_\max^d-R_\min^d) \right)^{1/d} \mathrm{d} u \\ &\stackrel{u\to 1-u}{=}& R_\max \int_0^1 \frac{u^b (1-u)^a}{\mathrm{B}(b+1,a+1)} \left( 1 - u \left( 1 -\frac{R_\min^d}{R_\max^d}\right) \right)^{1/d} \mathrm{d} u \\ &=& R_\max \cdot {}_2 F_1\left(a+1, -\frac{1}{d} ; a+b+2; 1-\frac{R_\min^d}{R_\max^d}\right) \end{eqnarray} $$ where the last equality is direct application of Euler's integral.