5
$\begingroup$

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.)

$\endgroup$
  • 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 Dec 20 '13 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 Dec 20 '13 at 20:56
  • 1
    $\begingroup$ I use Mathematica for general-purpose numerical integration. $\endgroup$ – whuber Dec 20 '13 at 21:36
5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ +1. The transformation to polar coordinates leads to the results I cited in a comment to the question. $\endgroup$ – whuber Dec 20 '13 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 Mar 4 '14 at 9:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.