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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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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 that $X$ 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 at 21:10
  • $\begingroup$ He got it from $VAR[cX] = c^2 VAR[X]$. $\endgroup$ – user49404 May 15 at 18:43
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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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