if $g \sim N(0,I_n)$ in polar form as $g=r \theta$ where $r = \|g\|_2$ is the length and $\theta = \frac{g}{\|g\|_2} $ is the direction
$I_n$ is the Identity matrix of dimension n, so each element in vector g follows the standard gaussian distribution (i.e. $g_i \sim N(0,1))$
I want to show that
$E[r]$ ~ $\sqrt{n}$ for n goes to $\infty$.
I computed the pdf of r to be
$f(r) = r e^{\frac{-r^{2}}{2}}$
My attempt:
$E[r] = \int_{0}^{\infty} r^{n-1} r e^{\frac{-r^{2}}{2}} dr$
$ = \int_{0}^{\infty} r^{n} e^{\frac{-r^{2}}{2}} dr$
$ = \int_{0}^{\infty} (\sqrt{2u})^{n-1} e^{-u} du$
$ = 2^{\frac{n-1}{2}} \Gamma{\frac{n+1}{2}}$
and then I used Stirling's approximation to expand Gamma function and simplified it to:
$\Gamma{\frac{n+1}{2}} \leq \frac{\sqrt{\pi}}{2} (n+1)^{\frac{n}{2}} e^{-\frac{n+1}{2}} e^{\frac{1}{6(n+1)}}$
I'm unable to simplify it more to be a factor of $\sqrt{n}$.
I think it is false. The $r$ variable is the square root of a $\chi^2(n)$. But it is know that a such variable has an expected value of
$$u_n :=\sqrt{2}\Gamma((n+1)/2))/\Gamma(n/2). $$
Using How to show $\frac{\Gamma((n-1)/2)}{\Gamma(n/2)} \approx \frac{\sqrt{2}}{\sqrt{n-2}}$
You have that :
$$\sqrt{2}\Gamma((n+1)/2))/\Gamma(n/2) \approx \frac{\sqrt{n-1}}{\sqrt{2}}. $$
Finally, you have the desired results, i.e :
$$u_n \approx \sqrt{n} $$