# The average length of vectors on spherical caps

Let $U$ denote the uniform distribution on the $n$-ball of radius $1$. What is the expected square-length of a vector under this distribution:

$$\mathbf{E}_U[\|x\|^2]$$

By standard concentration-of-measure results almost all the probability measure is on vectors of length at least $(1-\nu)$ for any constant $\nu>0$, hence one can bound the above by $[1-\nu,1]$ for any constant $\nu$. However, is there a more precise statement?

More generally, let $U_{\epsilon}$ denote the uniform distribution on the set of vectors $x = (x_1,...,x_n)$ with $\|x\|\leq 1$ for which $x_1 \geq \epsilon$ ($\epsilon$ can depend on $n$). What is the expected square length of such vectors $x$ (i.e. conditioned on being at least $\epsilon$ away from the equator)

$$f(\epsilon) := \mathbf{E}_{U_{\epsilon}}[\|x\|^2]$$

Note that here, concentration of measure does not apply in general because for fixed $\epsilon$ almost all vectors are located around the "equator", i.e. $|x_1| \leq \epsilon$, as $n \to \infty$.

For the first question: $$\mathbb{E}_U(\|x\|^2)=\int_0^1 {\rm prob}\,(\|x\|>t)dt=\int_0^1 (1-t^n)dt=\frac{n}{n+1}.$$
For the second question: integrating at first by $x_1=x$ and using previous formula we get that expectation equals $$\frac{\int_\varepsilon^1 (x^2+\frac{n-1}n(1-x^2))(1-x^2)^{(n-1)/2}dx}{\int_\varepsilon^1 (1-x^2)^{(n-1)/2}dx}.$$ This ratio may be bounded from below by $\frac{n-1+\varepsilon^2}{n}$.