I have a moving platform which moves in orthogonal axis to move a light detector to take account of positional uncertainty. I have a systematic error in space that is well described by a circular normal distribution (actually its not quite circular but the covariance matrix is almost exacty diagonal). I can think of two ways I could determine the amount by which my platform needs to move. I can either look at the probability as a function of radius so that my platform needs to move by 2.15 times the standard deviation to achieve 90%, this was my first though. The motors on the platform however move in orthogonal axes so I also though that I could look at the marginal distributions in which case for each axis the platform needs to be able to move by 1.64 times the standard deviation to achieve 90% coverage. As it happens this isn't really a concern as the range is much greater than required but it got me thinking.

If I had a positional uncertainty distribution that was a circular normal distribution which method should I use to detemine the amount of movement I need to cover 90% of events. If I think just about the motion in a single axis then, thinking radially, I need to move by 2.15 times the standard deviation however thinking 'marginally' then I only need 1.64 times the standard deviation. Which method is correct?


Here's my understanding of your question: you have a circular covariance matrix, $$C = \left( \begin{array}{cc} \sigma^{2} & 0 \\ 0 & \sigma^{2} \end{array} \right)$$ together with a position vector $ \overrightarrow{X} = [x,y]$ and a 2D Gaussian pdf defined by those two items, i.e., $$p(x,y) = \frac{1}{2\pi\sigma^{2}} e^{-\frac{X^{T} C^{-1} X}{2}} dx dy = \frac{1}{2\pi\sigma^{2}} e^{-\frac{(x^{2} + y^{2})}{2 \sigma^{2}}} dx dy$$ and you want to know what encircling radius $r$, in polar coordinates, will enclose 90% of the volume inside the pdf.

If that is indeed your question, you may answer it by first transforming to polar coordinates, and then performing a pair of integrals. Using the fact that this transformation requires a Jacobian (see example 3 in the link) and the fact that $r^{2} = x^{2} + y^{2}$, then in polar coordinates, the equivalent expression becomes: $$p(r,\theta) = \frac{1}{2 \pi \sigma^{2}} e^{-r^{2}/2 \sigma^{2}} r dr d \theta $$ You can marginalize this over $ \theta $ to get $$ p(r) = \int_{\theta=0}^{\theta=2\pi} \frac{1}{2 \pi \sigma^{2}} e^{-r^{2}/2 \sigma^{2}} r dr d \theta = \frac{r}{\sigma^{2}} e^{-r^{2} / 2 \sigma^{2}} dr $$ which you may recognize as a Rayleigh distribution. The Rayleigh distribution has a cdf: $$ p_{c}(r) = \int_{0}^{r} \frac{z}{\sigma^{2}} e^{-z^{2} / 2 \sigma^{2}} dz = 1 - e^{-r^{2} / 2 \sigma^{2}} $$

You can solve for the 90% encircling radius by setting: $$ 0.9 = 1 - e^{-r^{2} / 2 \sigma^{2}}$$ or $$ r = \sqrt{-2 \ln (0.1)} \sigma \approx 2.15 \sigma $$ so your first instinct appears to have been the correct one.

Phrased a little differently, the marginal approach is wrong because a square measuring $2 \times 1.64 \sigma$ on each side will cover a smaller area than a circle of radius $2.15 \sigma $ and will enclose a volume consisting of less than 90% of the probability. Thus, the marginal approach is wrong.

Alternatively, you can think of it this way: if you assign a coverage area of $\pm 1.64 \sigma $ in each dimension, then the probability for each coordinate of the $(x, y)$ arrival vector to lie separately within the coverage is indeed 0.9, but the joint probability for both components to fall simultaneously within the coverage area is smaller than that: it's $0.9^{2} = 0.81$.

  • $\begingroup$ Thank you for answering a very badly worded question - you are quite right I was neglecting the joint probability $\endgroup$ – Bowler Dec 19 '13 at 17:20

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