Telling a real circle match from a lucky one

You should know the circle geometry from the previous chapter and what a mean, a variance and a Gaussian distribution are.

The circle comparison statistic

Sample the temperature at $N$ equally spaced points around each circle, $a_i$ on the first and $b_i$ on the second, with $i = 0, \ldots, N-1$. Temperatures are measured from the sky average. Because we do not know where the matching point starts, compare the second circle rotated by $s$ samples:

Indices wrap around the circle. This is the statistic of Cornish, Spergel and Starkman. Three properties follow from $2ab \le a^2 + b^2$, which is $(a-b)^2 \ge 0$ rearranged:

1. $S = 1$ only when $b_{i+s} = a_i$ for every $i$, a perfect match.
2. $S = -1$ only when $b_{i+s} = -a_i$, a perfect anti-match.
3. Every other pair of circles gives a value in between.

A search computes $S$ for every shift, every pair of circle centres and every radius it tries, and keeps the maximum. That maximum is what has to be judged.

How far unrelated circles scatter

Suppose the $2N$ temperatures are independent Gaussian numbers with the same variance $\sigma^2$, the simplest model of two circles that have nothing to do with each other. Fix $s = 0$. Introduce

This is a rotation by 45 degrees in the plane of $a_i$ and $b_i$, so $u_i$ and $v_i$ are again independent Gaussians with variance $\sigma^2$. In the new variables $a_i^2 + b_i^2 = u_i^2 + v_i^2$ and $2a_ib_i = u_i^2 - v_i^2$. With $U = \sum_i u_i^2$ and $V = \sum_i v_i^2$,

$U/\sigma^2$ and $V/\sigma^2$ are independent chi-squared variables with $N$ degrees of freedom each. The ratio $X$ then follows a beta distribution with both parameters equal to $N/2$, whose variance is $1/[4(N+1)]$. Therefore

Changing variables from $X$ to $S$ turns the beta density $X^{N/2-1}(1-X)^{N/2-1}$ into a density proportional to $(1-S^2)^{(N-2)/2}$, which is equation 6 of the 1998 paper.

Worked example. With $N = 64$, the standard deviation is $1/\sqrt{65} \approx 0.124$. A value of $0.37$ is three standard deviations out, and a single unrelated pair rarely reaches it.

Why the real null is wider

The derivation treats every sample as independent. On a real sky, neighbouring points share the same hot or cold spot, so the samples are correlated. Cornish, Spergel and Starkman noted that correlations move the centre of the distribution towards larger $S$ and widen it. A rough way to see it is an effective number of samples: the formula still holds with $N$ replaced by the number of independent spots around the circle.

The 1998 paper also estimates how many independent points a circle offers: roughly $N \simeq 2\pi\sin\alpha/\Delta\theta$ for a map with angular resolution $\Delta\theta$. Small circles have fewer of them, so their null distribution is wider, and below some radius a true match can no longer be told from luck. That radius is the $\alpha_{\min}$ of the Planck analysis.

In the saved tutorial run for a box of side $1.2$, 300 unrelated back-to-back pairs sampled at 64 points, with noise at half the signal level, scatter with a standard deviation of $0.197$ at a fixed shift. That is much wider than the independent-sample value $0.124$. In the effective-number picture it corresponds to about $1/0.197^2 - 1 \approx 25$ independent samples.

How noise lowers a true match

Now let the circles be a true pair. Each carries the same signal $t_i$ plus independent detector noise: $a_i = t_i + n_i$ and $b_i = t_i + m_i$, with signal variance $\sigma_s^2$ and noise variance $\sigma_n^2$. For large $N$, numerator and denominator approach their averages:

The cross terms vanish because signal and noise are independent. The ratio is

equation 5 of the 1998 paper. Equal signal and noise give $S \approx 0.5$. Noise at one fifth of the signal gives $0.96$.

Worked example. The tutorial run with $\sigma_n/\sigma_s = 0.5$ predicts $S \approx 0.8$. Its nine true pairs average $0.795$ at the correct shift. With $\sigma_n/\sigma_s = 2$ the prediction is $0.2$, and the nine pairs average $0.200$.

A threshold from false pairs

A blind search does not know the right shift, centre or radius, so it reports the best score over all of them. Even unrelated circles then look good. In the tutorial run with noise at half the signal, the best shift for each of 300 unrelated pairs averages $0.36$, and one percent of them exceed $0.674$. The Lab's grey band shows the same effect: at each radius it is the best score among unrelated axes, 86 of them on the default sky.

The only fair threshold comes from running the full search on simulated skies that share the real map's resolution, noise and mask but have no topology. Planck did exactly that. It processed 300 Monte Carlo simulations of its maps in the same way as the data and set the false detection level so that fewer than 1 percent of simulations would yield a false event.

Compare the nine true pairs of the same run with that threshold. Their best scores lie between $0.69$ and $0.88$, all above $0.674$. With noise at twice the signal, the true pairs fall to between $0.27$ and $0.51$ while the one percent false level is $0.43$, and only one of nine pairs clears it.

The microwave sky adds its own noise for this test. Part of the temperature pattern does not come from the gas density at last scattering but from the gas velocity and from changing gravitational potentials along the way. These parts differ between the two paths to the same point. The COMPACT preprint counts them as noise for the circle signal. Planck also searched polarization maps, which come mainly from the moment of last scattering.

Lab tasks

1. Predict the box-axis score at a noise ratio of 1 from the formula above.
2. Set the noise ratio to 1 and read the box-axis match. Repeat at 0.5 and 1.5.
3. Find the noise ratio at which the box-axis score sinks into the grey chance band.
4. Explain why the unrelated pair in the circle profiles still scores above zero.

Exercises

1. Show that $|S| \le 1$ for any two circles.
2. For $N = 256$, what is the standard deviation of $S$ for independent unrelated circles?
3. A detector has noise equal to $0.3$ of the signal. What score should a true pair reach?
4. A search tries 1000 independent circle pairs, each with a null standard deviation of $0.124$. Roughly how large is the largest score expected from chance alone, if the null is Gaussian?

For the first task: $\xi = 1$ gives $0.5$. One pair scatters around that value, because the signal variance along a single circle differs from the sky average. On the default sky the Lab reads about $0.41$.

Sources: Cornish, Spergel and Starkman 1998, equations 4 to 6 and the effective sample count, Planck 2015 results XVIII, section 4.2 on the threshold from 300 simulations, Cornish et al. 2004, equation 1, weighting each Fourier mode by m, COMPACT collaboration preprint on Doppler and ISW noise.

Raise the noise until the true pair sinks into the chance band.