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 $\frac{n-1+\varepsilon^2}{n}$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.