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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.

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  • $\begingroup$ What does $N$ mean? $\endgroup$ – user158565 Dec 17 '18 at 17:23
  • $\begingroup$ @user158565: My bad, typo. $\endgroup$ – user14717 Dec 17 '18 at 17:25
  • $\begingroup$ Try delta method. But you need the covariance between that 2 items. $\endgroup$ – user158565 Dec 17 '18 at 23:22
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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 $E(a)\le \sqrt{E(\sum_{j=1}^{n-1}x_j^2)}=\sqrt{\frac{(n-1)}{3}}b.$

Similarly, one can also find an upper bound on the given quantity using Cauchy-Scwartz, for example.

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