expected valuenormal distributionprobabilityrandom variabletransform
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 $y$?
$$\mu_r = E[r] = ?$$
$$\sigma_r = E[r^2] – E[r]^2 =?$$
The question readily reduces to the case $\mu_X = \mu_Y = 0$ by looking at $X-\mu_X$ and $Y-\mu_Y$.
Clearly the conditional distributions are Normal. Thus, the mean, median, and mode of each are coincident. The modes will occur at the coordinates of a local maximum of the bivariate PDF of $X$ and $Y$ constrained to the curve $g(x,y) = x+y = c$. This implies the contour of the bivariate PDF at this location and the constraint curve have parallel tangents. (This is the theory of Lagrange multipliers.) Because the equation of any contour is of the form $f(x,y) = x^2/(2\sigma_X^2) + y^2/(2\sigma_Y^2) = \rho$ for some constant $\rho$ (that is, all contours are ellipses), their gradients must be parallel, whence there exists $\lambda$ such that
$$\left(\frac{x}{\sigma_X^2}, \frac{y}{\sigma_Y^2}\right) = \nabla f(x,y) = \lambda \nabla g(x,y) = \lambda(1,1).$$
It follows immediately that the modes of the conditional distributions (and therefore also the means) are determined by the ratio of the variances, not of the SDs.
This analysis works for correlated $X$ and $Y$ as well and it applies to any linear constraints, not just the sum.
In that case, you have to write (with hopefully clear notations) $$ \left(\begin{matrix}X\\Y \end{matrix}\right) \sim \mathcal{N}\left[ \left(\begin{matrix}\mu_X\\\mu_Y\end{matrix}\right), \Sigma_{X,Y} \right] $$ (edited: assuming joint normality of $(X,Y)$) Then $$ AX+BY=\left(\begin{matrix}A& B \end{matrix}\right) \left(\begin{matrix}X\\Y \end{matrix}\right) $$ and $$ AX+BY+C \sim \mathcal{N}\left[ \left(\begin{matrix}A& B \end{matrix}\right) \left(\begin{matrix}\mu_X\\\mu_Y\end{matrix}\right) + C, \left(\begin{matrix}A & B \end{matrix}\right)\Sigma_{X,Y} \left(\begin{matrix}A^T \\ B^T \end{matrix}\right)\right] $$ i.e. $$ AX+BY+C \sim \mathcal{N}\left[A\mu_X + B\mu_Y +C, A\Sigma_{XX}A^T+B\Sigma_{XY}^TA^T+A\Sigma_{XY}B^T+B\Sigma_{YY}B^T \right] $$
Best Answer
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:
$$\phi_{R^2}(t) = \frac{1}{1-2it} \cdot \exp \Big( \frac{it}{1-2it} \Big(\frac{\mu_x}{\sigma_x} + \frac{\mu_y}{\sigma_y} \Big) \Big).$$