A random point $(X,Y)$ has a normal distribution on a plane with circular scattering with $E[X]=E[Y]=0$ and var$[X]$=var$[Y]$=$\sigma^2$.
The distance of the point $(X,Y)$ from the centre of scattering is $R$. Find $E(R)$.
Centre = $(0,0)$ and $R=\sqrt{X^2+Y^2}$
I was wondering if we can use var$(R)=E(R^2)-\{E(R)\}^2$.
Given var$(X^2)=E(X^2)-\{E(X)\}^2=\sigma^2$ and var$(Y^2)=E(Y^2)-\{E(Y)\}^2=\sigma^2$
So, $E(R^2)=E(X^2+Y^2)=E(X^2)+E(Y^2)=\sigma^2 + \sigma^2=2\sigma^2$
var(R)=var$(\sqrt{X^2+Y^2})$
$E(R)=E(\sqrt{X^2+Y^2})$
What now ? Please help.
The probability distribution in the plane is the product of the $x$ and $y$ distributions: $$P(X,Y \in dx,dy) = {1 \over \sigma\sqrt{2\pi}} e^{-x^2/2\sigma^2} \cdot {1 \over \sigma\sqrt{2\pi}} e^{-y^2/2\sigma^2} dx dy = {1\over 2\pi\sigma^2}e^{-(x^2+y^2)/2\sigma^2}$$ which converts easily to the corresponding pdf for the function $r = \sqrt{x^2 + y^2}$. This lets us calculate $E[R]$:
$$ E[R] = \int_0^{2\pi} \int_0^{\infty} r \cdot P(R = dr)d\theta = {1 \over \sigma^2} \int_0^{\infty} r e^{-r^2/2\sigma^2} dr = [-e^{-r^2/2\sigma^2}]_0^{\infty} = 1.$$