mean and variance of norm of normal random variables If $x$ and $y$ are independent and normally distributed:$$x\sim N(\mu_x,\sigma_x)$$
 $$y\sim N(\mu_y,\sigma_y)$$
and $r$ is a random variable with the following relationship to $x$ and $y$
$$r = \sqrt{x^2 + y^2}$$
is it possible to derive expressions for the mean and variance of $r$ in terms of the parameters of $x$ and $y$?
$$\mu_r = E[r] = ?$$
$$\sigma_r = E[r^2] -  E[r]^2 =?$$
 A: The distribution you are looking for relates to the convolution of two non-central chi-squared distributions each with one degree-of-freedom (which is a nasty one).  Squaring the norm gives you:
$$\begin{align}
R^2 
&= X^2 + Y^2 \\[6pt]
&= \sigma_x^2 \Big( \frac{X}{\sigma_x} \Big)^2 + \sigma_y^2 \Big( \frac{Y}{\sigma_y} \Big)^2 \\[6pt]
&\sim \sigma_x^2 \cdot \text{ChiSq}\Big(\frac{\mu_x}{\sigma_x}, 1\Big)^2 + \sigma_y^2 \cdot \text{ChiSq}\Big(\frac{\mu_y}{\sigma_y}, 1\Big)^2. \\[6pt]
\end{align}$$
There is no closed form expression for this distribution.  However, it has the characteristic function:
$$\phi_{R^2}(t) = \frac{1}{1-2it} \cdot \exp \Big( \frac{it}{1-2it} \Big(\frac{\mu_x}{\sigma_x} + \frac{\mu_y}{\sigma_y} \Big) \Big).$$
A: If $x$ and $y$ are independent, $\mu_x=\mu_y=0$ and $\sigma_x=\sigma_y=\sigma$, $r$ follows a Rayleigh distribution, thus:
$$ \begin{align}
& E[r] = \sigma \sqrt{\frac{\pi}{2}}
\\[6pt]
& V[r] = \frac{4-\pi}{2}\sigma^2
\end{align} $$
A: Use the polar coordinates transformation if you're good at integrating. 
Define
$$
\left[\begin{array}{c}
R \\
\theta
\end{array} \right]
=
\left[\begin{array}{c}
\sqrt{X^2 + Y^2} \\
\text{arctan}(\theta)
\end{array} \right],
$$
which means
$$
\left[\begin{array}{c}
X \\
Y
\end{array} \right]
=
\left[\begin{array}{c}
R \cos(\theta) \\
R\sin(\theta)
\end{array} \right].
$$
Use the transformation theorem to get the joint density $g_{R,\theta}(r,t)$, then integrate out $t$. Using that, you can get whatever moment you want. At the moment I'm having a hard time integrating out $t$, though.
I get a joint densty of 
$$
g_{R,\theta}(r,t) = \frac{r}{2\pi \sigma_x \sigma_y}\exp\left[-\frac{h(r,t)}{2} \right]
$$
with
$$
h(r,t) = \frac{r^2\cos^2(t) + \mu_x^2 - 2\mu_xr\cos(t)}{\sigma^2_x} + \frac{r^2\sin(t) + \mu_y^2 -2\mu_yr\sin(t)}{\sigma^2_y},
$$
and $0 < r$ and $0 < t < 2\pi$. You can see how that will simplify if the variances are the same, and if the means are zero. This was mentioned in the other answer. Those two fractions inside the exponential will add nicely and a lot of terms will cancel, and you can use a trigonometric property--the whole function will be free of $t$.
But you're interested in the general case. Not sure, maybe someone good at mathematica can step in here.
NB: a few differences in notation: I'm using capital letters, and $\sigma^2$s for the variance.
