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$X$ and $Y$ arise from observations contaminated by i.i.d. additive Gaussian noise $\sigma$.

I seek the approximate variance of the angle from the origin to $(X,Y)$.

What I've tried:

The answer (variance of angle) is invariant to

  • scaling of the plane about the origin.
  • rotation of the plane about the origin, since $X-\mu_X$ and $Y-\mu_Y$ are i.i.d. Gaussian rv's.

So we can transform $(X,Y)$ to $(X',Y')$ by

  • rotating onto +x axis so that $\mu_{Y'}=0$ and $\mu_{X'}>0$.
  • scaling such that $\sigma'=1$.

At this point, the problem is to approximate the variance of $\arctan_2(Y',X')$ where

$$X'\sim N(r,1)$$ $$Y'\sim N(0,1)$$ where $r=\frac{\sqrt{\mu_X^2+\mu_Y^2}}{\sigma}$ is given from the original observations.

Aside:

This answer shows how to arrive at a p.d.f. for $\arctan(Y'/X')$ with key term

$$ \exp(-\frac{2r^2\tan^2\theta}{2+\tan^2\theta}) $$

but this p.d.f. repeats over a period of $\pi$ instead of $2\pi$ as it counts the opposite direction as well. My intention is to consider the variance of the angle's distribution from $-\pi$ to $\pi$ (in the rotated scenario where $(X',Y')$ is on +x axis).

I only seek the (approximate, say for $r>2$) variance, not the distribution.

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Museful
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    It is troubling that the angle is not well-defined. Choosing the cut at $\pm \pi$, although conventional, is not adapted to the location $(\mu_X,\mu_Y)$ of the distribution. This lends an element of arbitrariness to the question (and it will surely emerge in any good answer). Could you explain the purpose of computing this variance? I wonder about this issue because I suspect that your circumstances might be such where you could model an angle with a von Mises distribution and apply methods of [tag:circular-statistics] instead of the variance. – whuber Oct 20 '17 at 18:25
  • @whuber The cut is at $\pm\pi$ only for the simplified $(X',Y')$ scenario. So in general the cut would be at $\arctan_2(\mu_Y,\mu_X)\pm\pi$. The variance is intended for quality-guided phase unwrapping in a 2D phase map. I am unfamiliar with circular statistics but thank you for the pointer. – Museful Oct 20 '17 at 20:57

1 Answers1

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The distribution has been studied in the digital communications literature in the context of phase-locked loops and Rician fading channels. Suppose that the phase angle $\Theta$ of the point $(X,Y)$ is measured not with respect to the $x$ axis but rather as the deviation from the preferred direction (the straight line passing through the origin and the mean point $(\mu_X, \mu_Y)$). This is equivalent to the OP's rotation of the axes till the mean point lies on the $x$ axis and then measuring angles with respect to the $x$ axis. From this viewpoint, $\Theta$ is a random variable taking on values in $(-\pi, \pi)$ with density function $$f_{\Theta}(\theta) = \frac{\exp(\alpha \cos \theta)}{2\pi I_0(\alpha)} \,\mathbf 1_{\theta \in (-\pi, \pi)}$$ where $\alpha = \frac{\mu_X^2+\mu_Y^2}{\sigma^2}$. The details can be found in Viterbi's Principles of Coherent Communication, McGraw-Hill 1966.

Dilip Sarwate
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  • Thank you! What are $1_{\theta\in(-\pi,\pi)}$ and $I_0()$? – Museful Oct 21 '17 at 13:53
  • Museful, these are the indicator function and modified Bessel function, respectively. @Dilip Sarwate, I'm reasonably certain that the angle of two i.i.d. gaussians has the projected normal distribution, and not, as you write, the von Mises distribution. These distributions are in fact distinct. The first is obtained by marginalizing over the length of the vector $(X, Y)$, the second is obtained by conditioning on a specific length of that vector. See Mardia & Jupp, 2000, p.41-43. – Kees Mulder Oct 24 '17 at 08:11