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I'm looking for an expression of the variance of a single component of a point chosen from within a uniformly distributed n-ball with radius r for any n.

There are a few proofs showing that components of a point chosen this way become increasingly normally distributed as n tends to infinity. I can't seem to find expressions for any n.

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2 Answers 2

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I didn't totally follow @kimchi lover's answer (particularly his assumption of normality), so here is a similar approach.

Suppose X is hyperspherically uniformly distributed with r=1, so each $X_i$ has $E[X_i] = 0$, and we want to compute $VAR[X_i]$. Since $E[X_i]=0$, by the variance-squares identity, we have $VAR[X_i] = E[X^{2}_i] - E[X_i]^2 = E[X^{2}_i]$, so we really just need to compute $E[X^{2}_i]$.

Notice that $r^2 = \Sigma_{i=1}^n X^{2}_i$, so we have $E[r^2] = nE[X^{2}_i]$. Since we know that r has pdf $p_n (r) = nr^{n-1}$, we can trivially compute $E[r^2] = \int_0^1 r^2 nr^{n-1}dr = \frac{n}{n+2}$, so then $E[X^{2}_i] = VAR[X_i] = \frac{E[r^2]}{n} = \frac{1}{n+2}$.

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    $\begingroup$ How do you get obtian a formula for $p_n(r)$ for a generic $n$? $\endgroup$ Aug 24, 2020 at 21:24
  • $\begingroup$ You can probably be more rigorous about it, but off-hand I know that volume of a hypersphere is proportional to $r^n$, so $V_n (r) \propto r^n$, so then the surface area of the shell at a given radial distance is $\frac{dV_n (r)}{dr} \propto nr^{n-1}$, i.e., $p_n (r) \propto nr^{n-1}$. To get the constant of proportionality, we just integrate $\int_0^R k*p_n (r) dr = 1$, which yields k=1, so $p_n (r)$ is actually equal to $nr^{n-1}$, not just proportional. $\endgroup$
    – user49404
    Aug 24, 2020 at 22:01
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    $\begingroup$ Ah I see. For the 3-D case, the way I would get $p_3(r)$ is to divide the volume of a sphere of radius $r$ (enclosed within the sphere of radius $R$), and we see that the CDF is $F_3(r) = \frac{r^3}{R^3}$, and then we differentiate this with respect to $r$ and get $3r^2/R^3$. $\endgroup$ Aug 24, 2020 at 23:04
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    $\begingroup$ I think this approach might actually generalize to higher $n$. I just don't know what the volume of an n-th order hypersphere is, but it seems the CDF $F_n(r)$ will always be $r^3 / R^3$. I think the constant of proportionality is always $1$ for any $n$ and for any radius. $\endgroup$ Aug 24, 2020 at 23:05
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    $\begingroup$ Yeah, that makes total sense, since the constants of proportionality in the numerator and denominator always just cancel to give 1. $\endgroup$
    – user49404
    Aug 25, 2020 at 16:10
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Assume first the radius of the ball is $r=1$.

Write your point $X$ as a scaled point of the surface of the unit sphere: $X=R\sigma$, where $R=\|X\|$ and $\sigma=X/R$. It should be easy to convince yourself (using spherical symmetry) that $R$ and $\sigma$ are independent, and that what you want is $E[R^2]E[\sigma_1^2]$. The density function for $R$ is $n\rho^{n-1}$, and $E[\sigma_1^2] = E[Z_1^2/(Z_1^2+\cdots+Z_n^2)]= 1/n$, where the $Z_i$ are iid $N(0,1)$ variates. Thus, $$\text{Var}(X_1) = E[X_1^2] -(E[X_1])^2 = E[X_1^2]-0^2 = \int_0^1 n\rho^{n-1} \rho^2\,d\rho \times \frac 1 n = \frac {n}{n+2} \times\frac 1 n = \frac 1 {n+2}.$$

Relaxing the assumption that $r=1$, a simple rescaling give the general expression that $\text{Var}(X_1)=\frac{r^2}{n+2}$. (Because if $X$ is uniformly distributed on the $1$-ball then $rX$ is uniformly distributed on the $r$-ball and for any random variable $Y$, we always know $\text{Var}(rY) = r^2\text{Var}(Y)$.)

Note we distinguish here between the non-random $r$, the radius of the problem-statement ball; $R$ the length of the random point in the ball; and $\rho$, the variable of integration when taking the expectation of $R^2$.

Going beyond the question asked here, this formulation can help in evaluating higher moments $E\|X_1\|^k$ for $k>2$, at the price of added complexity. One has $E\|X_1\|^k=r^k E[R^k]E[\sigma_1^k]$. The first expectation is just $\int_0^1n\rho^{n-1}\rho^k d\rho$ but the second involves the ``Beta'' distribution, and is given by a ratio of gamma functions.

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    $\begingroup$ You're a star @kimchi; thank you. Marked answered. If you get a moment, could you elaborate on how you got $\frac{r^2}{n+2}$? $\endgroup$
    – Russell
    Apr 24, 2019 at 21:10
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    $\begingroup$ He got it from $VAR[cX] = c^2 VAR[X]$. $\endgroup$
    – user49404
    May 15, 2019 at 18:43
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    $\begingroup$ @Zwiddletwix Thanks for the catch! $\endgroup$ Feb 22, 2020 at 22:19

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