4
$\begingroup$

How can I get a good estimate of parameter $\nu$ of the Rice distribution based on a set of $(x,y)$-coordinates? Edit: Given whuber's excellent comment, I'm not looking for an unbiased estimate. Instead, I'd like to know whether there is an estimate that takes advantage of having available the 2D points themselves (not just their Euclidean norm) as well as an estimate of $\sigma$.

Background:

Given an uncorrelated bivariate normal variable $X$ with mean $\mu \neq 0$ with covariance matrix $\Sigma = \sigma^{2}I$, the Euclidean norm $R = \sqrt{X'X}$ (the radius, or distance to the origin) follows a Rice distribution with parameters $\nu = \sqrt{\mu'\mu}$ and $\sigma$.

When given a set of $N$ 2D-points $x = \mu + e$ from a Rice distribution, getting an unbiased estimate of $\sigma$ is straightforward and works as it does for the Rayleigh distribution. The obvious idea to estimate $\nu$ is $\sqrt{\bar{x}'\bar{x}}$, but if I'm not mistaken, the square has expectation $E(\bar{x}'\bar{x}) = \mu'\mu + E(e'e) = \mu'\mu + tr(\frac{1}{N}\Sigma) = \mu'\mu + \frac{2}{N} \sigma^{2}$. And this is before taking the square-root which introduces another bias.

The Wikipedia page mentions methods for parameter estimation, but they seem to be for the situation when we don't have the original 2D points, but just their Euclidean norm. It's possible to apply these methods after calculating the distance of the 2D points, they are complicated and based on fewer information than what I have.

$\endgroup$
  • 2
    $\begingroup$ I do not see why an unbiased non-negative estimator should exist or be useful. If one did exist, then its expectation when $\mu=0$ must equal $0$, but since all finite configurations of points have finite density even when $\mu=0$, that would imply that the estimator is constantly zero. Thus, any unbiased estimator of $\mu$ must have some chance of giving negative values. That seems like a high price to pay for unbiasedness. $\endgroup$ – whuber Aug 6 '14 at 15:40
  • 1
    $\begingroup$ @whuber Excellent point! I changed my question to accomodate your argument. $\endgroup$ – caracal Aug 6 '14 at 17:54

Your Answer

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

Browse other questions tagged or ask your own question.