Lower bound on $E\|Z\|$ for normal $Z$

Question

Put $Z \sim N(0,I_n)$. Show that \begin{align*} E \|Z\| \geq \frac{n}{\sqrt{n+1}}. \end{align*} The CLT says that $n^{-1}\|Z\|^2 = 1 + O_p(n^{-1/2})$, so $n^{-1/2}\|Z\| = \sqrt{1+O_p(n^{-1/2})} \approx 1 + O_p(n^{-1/2})$. Multiplying by $\sqrt{n}$ and taking expectation gives \begin{align*} E\|Z\| \geq \sqrt{n} - O(1), \end{align*} which agrees with the above bound, but I don't see how to prove it.

Answer 1

It would help to define $Z$ and $I_n$. If $Z$ is one random variable. what has that to do with CLT?

As stated $E|Z|=\frac{2}{\sqrt{2\pi}I_n}\int_0^\infty xe^{-\frac{x^2}{2I_n^2}}dx=\sqrt{\frac{2}{\pi}}I_n$

Answer 2

$$||Z||=\sqrt{Z_1^2+Z_2^2+\cdots Z_n^2}$$

$Z_i$ are independent and $Z_i^2\sim \chi^2_{1}$

$$Y=Z_1^2+Z_2^2+\cdots Z_n^2\sim \chi^2_{n}$$

Chi-square_distribution#Noncentral_moments

$$E(Y^k)=\frac{\Gamma(\frac{n}{2}+k)}{\Gamma(\frac{n}{2})}2^{k}$$

$$E||Z||=E(Y^\frac{1}{2})$$