I have an uncorrelated bivariate normal process: $X \sim N(0, \sigma_1), Y \sim N(0, \sigma_2), \rho = 0$, and I'm interested in the distribution of $Z = \sqrt{X^2 + Y^2}$.

I know that if I'm willing to normalize X and Y by their variance then Z has the chi distribution with 2 degrees of freedom.

But I am looking for closed-form expressions of the variance and CDF of Z that incorporate the difference in variance in the two dimensions. I.e., without factoring out the sigmas, and knowing $\sigma_1 \neq \sigma_2$,

  • What is the variance of Z?

  • What is the CDF -- i.e., Pr($Z \leq \alpha$)? (I know this is also an elliptic distribution, but the general form for those is far too complicated.)

  • 1
    $\begingroup$ I doubt there is a closed form because the radial distribution is proportional to the Laplace transform of $\sqrt{r}I_0(r\tau)$ where $I_0$ is the modified Bessel function of the first kind and $\tau$ is a constant (depending on $\sigma_1$ and $\sigma_2$): this simplifies only when $\sigma_1=\sigma_2$. $\endgroup$
    – whuber
    Commented Dec 20, 2013 at 15:40
  • $\begingroup$ Shoot, I hate it when these simple problems have complex solutions ;) Can you point me towards numerical methods for computing the probabilities? $\endgroup$
    – feetwet
    Commented Dec 20, 2013 at 20:56
  • 1
    $\begingroup$ I use Mathematica for general-purpose numerical integration. $\endgroup$
    – whuber
    Commented Dec 20, 2013 at 21:36

1 Answer 1


If $X \sim N(0, \sigma_1^2)$ and $Y \sim N(0, \sigma_2^2)$ are independent random variables, then the joint pdf of $(X,Y)$ is say $f(x,y)$:

enter image description here

Given $Z = \sqrt{X^2 + Y^2}$, you seek $\text{Var}(Z)$:

enter image description here

where Var is the Variance function from the mathStatica add-on to Mathematica (to compute the pleasantries), and EllipticE is the complete elliptic integral: http://reference.wolfram.com/mathematica/ref/EllipticE.html

Here is a plot of the solution $\text{Var}(Z)$, as a function of $\sigma_1$ and $\sigma_2$, as you desired:

enter image description here

For the cdf calculation, I would suggest that a transformation to polar coordinates should do the trick.

  • 1
    $\begingroup$ +1. The transformation to polar coordinates leads to the results I cited in a comment to the question. $\endgroup$
    – whuber
    Commented Dec 20, 2013 at 15:53
  • 2
    $\begingroup$ @whuber The distribution of $Z$ is the Hoyt distribution which can be written in closed form. The Mathematica reference also gives the moments in terms of its parametrization $q$ and $\omega$. $\endgroup$
    – caracal
    Commented Mar 4, 2014 at 9:06

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