Tldr;
So basically the question is given a vector $t=(x,y,z)$ and you are given $x,y,z$ are completely random and that they follow boltzman distribution, and that
$ \bar x = 0 $
$\bar |x| = \frac{a}{\sqrt \pi}$
$\bar {x^2} = \frac{a^2}{2}$
can you find $< |t| >$
Actual motivation:
So I was studying some physics, the Kinetic theory and boltzman distribution stuff...
So for gasses, assuming the Kinetic theory, (ie. basically completely random motion)
I derived using the boltzman distribution that:
$<v_x>=0$
$<|v_x|> \ =\frac{a}{\sqrt \pi}$
$<(v_x)^2>\ =a^2/2$
where $<x>$ is ofcourse the mean of $x$. (and $a^2=\frac{2k_BT}{m}$ but that's not important)
So then when it came time for the means of speed, ie $|v|$, (in the following line i am representing speed by $v$
I clearly saw $<v^2> \ = \ <v_x^2+v_y^2+v_z^2> \ = \ 3<v_x^2> \ = \ \frac{3a^2}{2}$
But then similarly I thought I could find $< v > \ =\ \sqrt3 \cdot <|v_x|>$
Because $v = \sqrt{v_x^2+v_y^2+v_z^2}$ but it turns out the actualy answer is
$< v > \ = \ 2 \cdot <|v_x|> \ = \ \frac{2a}{\sqrt \pi}$
(then I saw the correct derivation for this was doing the spherical $4 \pi v^2dv$ thing but it got me thinking if we could derive it without going down that route, but since I haven't really studied much statistics yet, I can't answer that question, so i was hoping someone more experienced could...
So basically the question is given a vector $t=(x,y,z)$ and you are given $x,y,z$ are completely random and that they follow boltzman distribution, and that
$ \bar x = 0 $
$\bar |x| = \frac{a}{\sqrt \pi}$
$\bar {x^2} = \frac{a^2}{2}$
can you find $< |t| >$