# Expectation of norm ratio

Suppose $$x = (x_0, x_1, \ldots, x_{n-1}) \in \mathbb{R}^{n}$$ with $$x_{i} \sim \mathrm{Unif}(0, b)$$.

How can I calculate or estimate $$E\left[\left\|x \right\|_{1}/\left\|x \right\|_{2}\right]$$?

I have found that $$E\left[\left\|x \right\|_{1}\right] = nb/2$$ and $$E\left[\left\|x \right\|_{2}^2\right] = nb^2/3$$, and I have run simulations which show that $$E\left[\left\|x \right\|_{1}/\left\|x \right\|_{2}\right]$$ is indeed very close to $$\sqrt{3}/2$$ = $$E\left[\left\|x \right\|_{1}\right]/\sqrt{E\left[\left\|x \right\|_{2}^2 \right]}$$, but I can't understand why this should be the case.

• What does $N$ mean? – user158565 Dec 17 '18 at 17:23
• @user158565: My bad, typo. – user14717 Dec 17 '18 at 17:25
• Try delta method. But you need the covariance between that 2 items. – user158565 Dec 17 '18 at 23:22

Note that, from symmetry, $$E(\|x\|_1/\|x\|_2)=nE(x_0/\|x\|_2).$$ Conditioning, on $$x_1,\cdots,x_{n-1}$$, one gets that $$E(x_0/\|x\|_2\mid x_1,\cdots,x_{n-1})=\frac{1}{b}\int_0^b\frac{x}{\sqrt{x^2+a^2}}dx=\frac{\sqrt{b^2+a^2}-a}{b}=\frac{b}{a+\sqrt{a^2+b^2}},$$ where $$a^2=\sum_{j=1}^{n-1}x_j^2.$$ I guess now at best upper and lower bounds on this quantity can be found. For example, since the function $$\frac{b}{x+\sqrt{x^2+b^2}}$$ is convex for any $$b\ge 0$$, using Jensen's inequality one can derive that $$E(x_0/\|x\|_2)\ge \frac{b}{\sqrt{(E(a))^2+b^2}+E(a)}\ge\frac{\sqrt{3}}{\sqrt{n-1}+\sqrt{n+2}},$$ where I have again used Jensen's inequality, using the fact that $$\sqrt{x}$$ is concave for $$x\ge 0$$, to obtain $$E(a)\le \sqrt{E(\sum_{j=1}^{n-1}x_j^2)}=\sqrt{\frac{(n-1)}{3}}b.$$