Signal in the noise: a Bayesian look at GW150914

On 14 September 2015 the two LIGO detectors recorded a 0.2 s "chirp" from two black holes merging 1.3 billion light years away. The strain data are public. This notebook asks what a Bayesian practitioner with NumPy, SciPy and PyMC only can learn from them.

What this notebook is, and is not. It is a simplified, pedagogical analysis: analytic toy waveforms (a damped sinusoid, a leading-order chirp), a noise model estimated from 30 s of data, no detector calibration uncertainty, no sky location or polarisation model. It is not a reproduction of the LIGO-Virgo analysis, which uses numerical-relativity calibrated waveforms with 15 parameters and specialised samplers. Where our numbers agree with the published ones we will say "consistent with", not "confirms"; where they differ we will say so and look for the reason.

The plan:

1. See the signal - spectral estimation, whitening, band-passing, a time-frequency map.
2. Fit the ringdown - the one piece of the waveform with a closed-form model - jointly in both detectors, and turn its frequency and damping time into a mass and spin.
3. Is there really a signal? - off-source fits, a Bayes factor with a background distribution, calibration checks, and the classical matched filter.
4. The inspiral chirp (stretch) - a sampler failure, its fix, and a bias the posterior cannot warn you about.

#Part 1 · Seeing the signal

#1.1 · The data

The files are HDF5, straight from GWOSC, so we open them with h5py rather than data.load. Strain is dimensionless (relative arm-length change) and absurdly small, so everything below is in units of $10^{-21}$. Times are seconds relative to GPS 1126259462.4, the nominal event time, about 15.4 s into the segment.

Nothing to see. The raw strain wanders by a few hundred units (L1 also carries a constant offset of about -1000), while the signal we are after peaks at about one unit. The wandering is low-frequency noise (seismic motion, suspension resonances); the detector is only quiet in a band from a few tens to a few hundred hertz. To see anything we must weight each frequency by how noisy it is.

#1.2 · The noise spectrum

The power spectral density (PSD) $S(f)$ says how much noise variance there is per hertz. Welch's method estimates it by averaging periodograms of overlapping windowed chunks. We use only off-source data, more than 2 s away from the event, so the signal cannot leak into its own noise estimate.

The amplitude spectral density falls by a factor of several thousand - some seven orders of magnitude in power - between 10 Hz and the quiet "bucket" near 100-300 Hz, and there is a forest of narrow lines on top (mains harmonics at multiples of 60 Hz, suspension "violin" modes near 500 Hz and their harmonics near 1 kHz, calibration lines). This is what coloured noise looks like.

#1.3 · Whitening and band-passing

Whitening divides the Fourier transform of the data by $\sqrt{S(f)}$. Every frequency then carries the same noise power: the loud low frequencies are turned down, the quiet bucket is turned up. With the normalisation below, stationary Gaussian noise with PSD $S(f)$ becomes (approximately) unit-variance white Gaussian noise. We then band-pass to 35-350 Hz, where the signal lives, using a zero-phase Butterworth response applied in the frequency domain.

Whitened H1 noise has a standard deviation of 0.99, as designed. L1 comes out at 1.15; the excess sits entirely below 10 Hz (slow drifts that a PSD built from 4 s chunks cannot resolve) and disappears once we band-pass. After the band-pass both detectors are at 0.37: we kept roughly $(350-35)/2048 \approx 15\%$ of the bandwidth and hence of the variance ($\sqrt{0.15} = 0.39$). Hold on to that number - it matters in section 2.8.

Now the famous picture. The source was closer to Livingston, and the two detectors are oriented almost oppositely, so we cross-correlate to find the shift and sign that best overlay L1 on H1.

There it is: a handful of cycles growing in amplitude and frequency, a peak, and a rapid decay - the same shape in two instruments 3000 km apart, with Livingston about 7 ms ahead and with the opposite sign (a correlation of -0.72 between two independent detectors). The LIGO papers quote an arrival-time difference of $6.9^{+0.5}_{-0.4}$ ms; our crude cross-correlation estimate of 7.3 ms is in the same place. It is one number from one filter choice with no uncertainty attached; we return to the time shift in 2.5. The correlation has a single clean dip because the chirp is broadband: every frequency agrees on one time shift. Remember that when we get to the narrow-band ringdown.

#1.4 · A time-frequency map

A short-time Fourier transform has one fixed time resolution. For a chirp it is better to use windows that get shorter as frequency rises (constant $Q = f/\Delta f$), which is the idea of the "Q-transform" used in LIGO's own plots. A bare-bones version is a Gaussian band-pass around each frequency, applied in the Fourier domain:

The textbook chirp: the frequency sweeps from about 35 Hz to about 250 Hz in 0.2 s, and the sweep accelerates. In both detectors. That upward sweep is two masses spiralling together (Part 4); the abrupt end is the merger; what follows the peak is the newborn black hole settling down - the ringdown (Part 2).

#1.5 · Why whitening gives a likelihood, and where the shortcut breaks

If the noise is stationary and Gaussian, its Fourier coefficients at different frequencies are independent with variance proportional to $S(f)$. That is the Whittle likelihood:

$$\log p(d \mid \theta) = -\tfrac{1}{2}\sum_k \frac{|\tilde d_k - \tilde h_k(\theta)|^2}{\sigma_k^2} + \text{const}, \qquad \sigma_k^2 \propto S(f_k).$$

Dividing by $\sqrt{S}$ makes every $\sigma_k$ equal, and by Parseval's theorem the sum over frequencies becomes a sum over time samples: whitened data minus whitened template is unit-variance white noise, i.e. pm.Normal(mu=whitened_template, sigma=1). That is the attraction. The shortcut has three traps, though:

1. The template must be whitened too. The filter changes the shape of any signal. You cannot fit an un-whitened damped sinusoid to whitened data.
2. Band-passed samples are not independent. After the band-pass the noise has std 0.37 and neighbouring samples are strongly correlated. If you "helpfully" set or fit sigma = 0.37 with independent samples you count the same information several times.
3. The filters are acausal and long. Whitening and zero-phase band-passing smear every feature over tens of milliseconds in both directions. For the ringdown we want to use data after a start time $t_0$ and nothing before it - but the filtered data after $t_0$ contain smeared-in power from the much louder merger just before it.

Trap 3 is the reason ringdown analyses in the literature (Isi & Farr 2021, and the ringdown package) work in the time domain with the full noise covariance matrix. That is what we will do. It is exact for stationary Gaussian noise, it handles the truncation at $t_0$ with no leakage, and it whitens data and template with the same operator by construction. In section 2.8 we come back and measure what the shortcut would have cost.

#Part 2 · The ringdown: weighing a black hole with a damped sinusoid

A perturbed black hole rings like a struck bell, in damped sinusoids called quasi-normal modes. Keeping only the dominant one,

$$h(t) = A\,e^{-(t-t_0)/\tau}\cos\!\big(2\pi f\,(t-t_0) - \varphi\big), \qquad t \ge t_0 .$$

General relativity says $f$ and $\tau$ depend only on the mass and spin of the final black hole, so measuring them weighs the remnant.

#2.1 · A noise model in the time domain

Stationary Gaussian noise is fully described by its autocovariance function $\rho(k) = \mathrm{cov}(n_i, n_{i+k})$, the inverse Fourier transform of the PSD. For a window of $N$ samples the noise is $\mathcal{N}(0, C)$ with $C_{ij} = \rho(|i-j|)$. With the Cholesky factor $C = LL^\top$,

$$ L^{-1} d \;\sim\; \mathcal{N}\big(L^{-1} h(\theta),\; I\big). $$

$L^{-1}$ is a whitening filter too - the time-domain, finite-window counterpart of dividing by $\sqrt{S(f)}$ - but it is lower-triangular, hence causal: nothing before the window leaks in, and the template is whitened by the same matrix as the data.

Conditioning, following common practice: high-pass at 20 Hz (below that there is only noise, and it would wreck the conditioning of $C$), down-sample to 2048 Hz (the ringdown is at ~250 Hz; the top octave only adds cost and an anti-aliasing roll-off that $C$ would have to model), then estimate the PSD from off-source data as before.

Test the noise model on noise before trusting it with a signal. Whitening non-overlapping off-source windows should give independent standard-normal numbers. (The same data informed the PSD, so the overall variance is nearly 1 by construction; the absence of correlation and the Gaussian shape are the non-trivial checks.)

The conditioned noise is as far from white as it gets. Once the low frequencies are removed, about 96% of the remaining variance is in narrow lines - near 1 kHz in H1, just below the new Nyquist frequency, which is why its neighbouring samples are anti-correlated at -0.93; near 515 Hz in L1 - and lines ring for a long time. After $L^{-1}$ all of that is gone: the lag-1 correlation is zero to three decimals, the variance is 1, and the tails follow the Gaussian over four decades. For these 30 seconds, "stationary and Gaussian" is a good description.

#2.2 · The model

Phase as quadratures. Writing $A\cos(\omega t - \varphi) = a_c\cos\omega t + a_s\sin\omega t$ with $a_c = A\cos\varphi,\ a_s = A\sin\varphi$ replaces a wrapped angle by two unbounded numbers. NUTS moves through a Euclidean space: it cannot step from $\varphi = 3.1$ to $-3.1$ even though they are neighbours on the circle, so a posterior that straddles the wrap looks bimodal to it, and at $A \to 0$ the phase is undefined (a funnel). In quadratures the posterior is a single blob - and, given $(f, \tau)$, the template is linear in $(a_c, a_s)$ with Gaussian priors, which we will exploit in Part 3. Independent $\mathcal{N}(0, s)$ priors on the quadratures imply a uniform phase and a Rayleigh amplitude.

Two detectors. Each detector gets its own quadratures. Why not "one amplitude and phase, plus a time shift and a relative amplitude"? Because for a damped sinusoid a time shift $\delta$ is a phase shift $2\pi f\delta$ and a rescaling $e^{\delta/\tau}$: the two parameterisations describe the same family of curves, but the time-shift version has a ridge that repeats every $1/f \approx 4$ ms. So we align the L1 window using the cross-correlation shift from Part 1, keep the well-behaved parameters, and derive the relative amplitude, the sign and the implied time shift afterwards (2.5).

Start time. The model only holds once the remnant has settled into linear ringing. Published analyses put the peak strain at Hanford at about GPS 1126259462.423 (T_PEAK, t = 23 ms here; our whitened envelope peaks 1-2 ms later, which is within the smearing of the filters). Guided by the LIGO-Virgo testing-GR paper, where a single damped sinusoid starts to agree with the full analysis about 3 ms after the merger, we start 3 ms after the peak, and treat that as a choice to be stress-tested (2.6), not a fact. Start times are rounded up to the next sample (at most 0.5 ms).

Priors. $f \sim \text{Uniform}(150, 400)$ Hz: the ringdown should sit at or above the ~150-250 Hz reached at the end of the chirp in the time-frequency map. $\tau \sim \text{LogNormal}(\log 4\text{ ms}, 0.75)$, i.e. roughly 1-17 ms: a bell that rings for a handful of cycles. Quadratures $\sim \mathcal{N}(0, 2)$ in units of $10^{-21}$, because the published peak strain is about 1 in those units.

#2.3 · Prior predictive check

The derived snr is $\lVert L^{-1}h\rVert$ summed over detectors: how many noise standard deviations the template stands out by. It is the natural scale on which to ask what the priors claim.

Damped wiggles of a few cycles with amplitudes of a few units: the right kind of object. The implied SNR runs from about 5 (marginal) to about 40 (unmissable), with a median of 15. So this prior leans towards signals that are louder than a ringdown 3 ms after the peak is likely to be - acceptable as a weakly-informative choice, but worth knowing; in Part 3 we will see how much the Bayes factor cares.

#2.4 · Fit, diagnose, check

Clean: no divergences, r_hat at most 1.01, ESS above a thousand. Six parameters, 204 data points and a 102 x 102 matrix product per gradient - the whole fit takes a few seconds.

For the posterior predictive check we look at the data the way the likelihood sees them: whitened by $L^{-1}$. If the model is adequate, whitened data minus whitened template should be standard normal noise.

The whitened template is not a pretty damped sinusoid - $L^{-1}$ reshapes it, which is exactly why the template must go through the same operator as the data. The first few samples carry the signal (several standard deviations), and after subtracting the template the residual mean square is compatible with 1 (for 102 samples it scatters by about 0.14; in H1 it drops from 1.40 to 1.04). A ringdown starting 3 ms after the peak is a weak signal: a network SNR of about 8, spread over a handful of samples.

#2.5 · Do the two detectors tell the same story?

Each detector has its own amplitude and phase. If both see the same wave, with Livingston inverted and ahead by $\Delta t$, then after our alignment the phases should differ by $\pi$, and any leftover phase is a leftover time shift: $\Delta t = \Delta t_\text{used} - (\Delta\varphi - \pi)/(2\pi f)$.

The phase difference is compatible with $\pi$ - the sign flip; you can also see it in the trace plot above, where a_cos is about -1.1 in H1 and +0.9 in L1. Read as a time shift, the ringdown alone gives 7.1 ms with a 90% interval of 6.8-7.5 ms, overlapping both the cross-correlation value and the published $6.9^{+0.5}_{-0.4}$ ms. Two caveats: this is the solution nearest to the alignment we imposed (add any multiple of $1/f \approx 4$ ms and the fit is identical - the interval is conditional on having picked the right cycle), and "same wave, inverted" is only approximately true because the detectors are not exactly anti-aligned. The amplitude ratio is compatible with 1 and poorly measured.

#2.6 · The start time is a modelling choice - so vary it

Start too early and the data still contain the non-linear merger, which a single damped sinusoid cannot describe (bias). Start too late and the signal has decayed into the noise (variance). There is no objectively correct $t_0$, so we show the dependence.

Read the three panels together:

• SNR falls steadily as we start later - the signal decays with a ~4 ms time constant.
• At 0-1 ms the posterior is narrow but sits at a lower frequency (233-234 Hz) and a longer damping time (about 5.5 ms) than the Kerr mode of the published remnant. Precise and biased: a single mode is being asked to describe the merger. (Published work argues that adding overtones repairs this; see "Try it yourself".)
• At 3 ms the posterior is centred near 247 Hz and 4.3 ms, and comfortably contains the expected mode. The LIGO-Virgo testing-GR paper reports the same qualitative pattern: its damped-sinusoid fit becomes consistent with the full-waveform prediction from about 3 ms after the merger.
• By 5-7 ms the posterior for $f$ spreads over much of its prior and $\tau$ drifts towards short values: there is little signal left, and what the fit finds is increasingly noise. Notice the snr panel cannot tell you that by itself - a flexible template always finds something (Part 3).

The dashed reference line was computed with the fitting formulas we meet next, so let us do that properly.

#2.7 · The payoff: mass and spin of the remnant

For a Kerr black hole of mass $M$ and dimensionless spin $\chi$, Berti, Cardoso & Will (2006) give accurate fits for the dominant ($\ell = m = 2$, $n = 0$) mode:

$$2\pi f\,\frac{GM}{c^3} = f_1 + f_2(1-\chi)^{f_3}, \qquad Q \equiv \pi f\tau = q_1 + q_2(1-\chi)^{q_3}.$$

The quality factor $Q$ depends on spin only; given the spin, the frequency sets the mass. The check above (68 $M_\odot$, $\chi = 0.69$ gives 252 Hz and 4.1 ms) is our sanity test of the constants and units. Inverting is one line per draw - but it is only possible when $Q > q_1 + q_2 = 2.12$ (the value at $\chi = 0$). Draws with a lower $Q$ ring down faster than any Kerr black hole would, and have no mass or spin.

A few percent of the draws are "not a Kerr black hole". That is a symptom of a prior that was flat in $(f, \tau)$ and knew nothing about black holes. If we believe the remnant is a Kerr black hole, the honest thing is to put the prior where the belief is: sample $M$ and $\chi$ directly and compute $(f, \tau)$ from them inside the model. With a likelihood this weak, the two priors will not give the same answer - that is the point of doing both.

No divergences and r_hat at 1.00-1.01, but the ESS for mass and spin is a few hundred rather than a thousand-plus: the curved mass-spin ridge below is harder going for NUTS than the blob in $(f, \tau)$. Adequate for the 90% intervals we quote; run longer before quoting anything finer.

From 50 milliseconds of data and a damped sinusoid: a black hole of very roughly 70 solar masses, spinning at some large fraction of the maximum. The published values from the full analysis are a detector-frame mass of about 68 $M_\odot$ (62-63 $M_\odot$ in the source frame, after removing the cosmological redshift $z \approx 0.09$ - a ringdown alone cannot do that, it only ever measures the redshifted mass) and a spin of 0.67-0.69. Those sit well inside our posterior. But be clear about how little that says:

• Our 90% intervals are broad - roughly 52-88 $M_\odot$, and a spin anywhere from 0.15 to 0.9 depending on the prior. Almost all of the published precision comes from the inspiral and merger, which we threw away. Broad is the correct answer for a ringdown-only analysis at this SNR.
• Mass and spin are strongly correlated (the banana in the right panel): a heavier, faster-spinning hole rings at the same frequency as a lighter, slower one.
• The answer depends on the prior: the flat-in-$(M,\chi)$ fit prefers a lower spin (about 0.6) than the mapped flat-in-$(f,\tau)$ fit (about 0.7), because equal areas in $(f,\tau)$ are not equal areas in $(M,\chi)$ - high spins occupy a large slice of $\tau$. Neither is "the" answer; reporting both is more honest than reporting one.
• And it depends on $t_0$ (2.6).

#2.8 · What would the whitening shortcut have cost?

Back to section 1.5. Suppose we had fitted the Part 1 data - FFT-whitened, band-passed, at 4096 Hz - with independent Normal errors, from the same $t_0$. To be fair to the shortcut we whiten the template correctly: the combined whitening + band-pass filter is a convolution, so on a window it is a matrix built from its impulse response, and the very same PyMC model applies with a different W. We try two noise levels: sigma = 1 (the level before band-passing, which is what Parseval's theorem justifies) and sigma = 0.37 (the measured std of the band-passed noise, which is what most of us would reach for).

Two different failures, neatly separated:

• sigma = 0.37 shrinks the posterior standard deviations by a factor of two to four relative to sigma = 1 (just under 3 for the amplitudes, which scale with sigma). Nothing about the data improved; we told the model that 205 strongly correlated samples were independent measurements at the band-passed noise level. The rule of thumb: band-limited noise of bandwidth $B$ has only $2B$ independent samples per second (here 630, not 4096), so the information is over-counted by $4096/630 \approx 6.5$ and widths shrink by about $\sqrt{6.5} \approx 2.5$ - the same ballpark. Overconfident by a factor of three is the kind of error that produces false "tensions" with general relativity.
• sigma = 1 gives sensible widths, as the Parseval argument promised, but the answer has moved: a damping time under 3 ms instead of 4.5 ms, and amplitudes more than half as large again as in the exact fit. The likely culprit is trap 3: the acausal filters smear the loud merger into the first milliseconds after $t_0$, and the model explains that extra power with a bigger, faster-decaying ringdown. We have not proven that here (an injection study would), but it is the reason the ringdown literature gives for working in the time domain.

So the exact time-domain likelihood is not pedantry. It is also no harder to write.

#Part 3 · Is there really a signal?

A damped sinusoid with free frequency, decay, amplitude and phase will fit something in any stretch of noise. Before believing Part 2 we should ask how the same analysis behaves when there is nothing there - and we have 30 seconds of "nothing there" in the same file.

#3.1 · The same model on off-source data

On noise, the frequency posterior simply returns its prior (an sd of 67-81 Hz, against 72 Hz for Uniform(150, 400)) and the fitted SNR has a median of 2-4 - not zero, because a norm is positive and a flexible template can always soak up a little noise. On-source the SNR posterior sits clearly apart and the draws form one tight cluster in the right panel. Note what does not discriminate: the amplitude. Off-source medians are 0.5-1, not far below the on-source 1.2, because a large amplitude with a very short damping time costs almost nothing in SNR. The question "is the amplitude zero?" is the wrong one; "does the template explain variance?" is the right one. Several off-source fits show divergences (up to 11 here) and an r_hat up to 1.03: with no signal, the posterior is the awkward union of "small template, any frequency" and a few weak noise bumps. That is a property of the question, not a bug to tune away.

#3.2 · A Bayes factor that actually computes

"Signal versus noise" is a comparison between our model and the same model with all four amplitudes set to zero, which has no free parameters at all. Bayes factors are notoriously hard to compute, but this problem has structure: given $(f, \tau)$, the whitened template is linear in the four quadratures, with Gaussian priors and Gaussian noise. The quadratures can be integrated out in closed form (it is Bayesian linear regression with design matrix $X = L^{-1}[e^{-t/\tau}\cos\omega t,\ e^{-t/\tau}\sin\omega t]$):

$$\log\frac{p(d \mid f,\tau)}{p(d \mid \text{noise})} = \tfrac12\, b^\top\!\big(G + s^{-2}I\big)^{-1} b - \tfrac12\log\det\!\big(I + s^2 G\big), \qquad G = X^\top X,\; b = X^\top L^{-1}d .$$

What is left is a two-dimensional integral over $(f, \tau)$, which a grid does to high accuracy in a few seconds (most of it spent whitening the basis once). No sampler, no tuning, no stochastic error.

Cross-check 1: the grid reproduces NUTS. The same grid gives the exact marginal posterior of $(f, \tau)$, which should match the PyMC fit from 2.4.

Cross-check 2: Sequential Monte Carlo. pm.sample_smc tempers from the prior to the posterior and estimates the marginal likelihood as a by-product. It runs fine in PyMC 6; the estimate lives in sample_stats["log_marginal_likelihood"] as an object array with one entry per tempering stage, of which only the last is the final estimate. To turn it into a Bayes factor, subtract the log-likelihood of the noise model, which is just the standard normal log-density of the whitened data.

The four SMC chains agree with one another and with the grid to within about 0.2. Two independent methods, one number: $\ln \text{BF} \approx 19.7$, odds of a few hundred million to one. For a generic model you would have only SMC (or bridge sampling, or nested sampling); the lesson of the grid is to look for conditionally linear-Gaussian structure first, because integrating parameters out analytically beats any sampler.

But what is a Bayes factor of $10^8$ worth? It compares our two models: "Gaussian noise plus a damped sinusoid" versus "Gaussian noise". Real detectors glitch, and a glitch is neither. The empirical answer is to compute the same statistic where we know there is no astrophysical signal, and see what the noise can do: a background distribution. Because the grid is fast, we can afford every non-overlapping off-source window.

Every one of the off-source windows has a negative log Bayes factor: when there is nothing there, the evidence (mildly) prefers the simpler model - Occam's razor working as advertised. The on-source value is far outside anything the noise produced. With a couple of hundred windows that supports a false-alarm probability below about 1 in 200 for this statistic and no more; LIGO's "less than one in 200,000 years" comes from time-sliding weeks of data from the two sites against each other, which 32 seconds cannot imitate.

The right panel is the standard health warning. The Bayes factor depends on the prior width: make the amplitude prior absurdly wide and the signal model is penalised for all the loud signals it predicted and did not see (roughly $-4\ln s$ for four amplitudes). Here the conclusion survives any remotely sensible choice, but the number is only meaningful together with the prior that produced it.

#3.3 · Is the likelihood calibrated on real noise? (SBC, for free)

The exact grid posterior makes simulation-based calibration cheap: draw parameters from the prior, add the corresponding ringdown to a real off-source noise window, compute the posterior, and record where the truth falls in it. If the whole pipeline - noise model, whitening, priors - is right, those posterior CDF values are uniform. Unlike a test on simulated Gaussian noise, this one can fail if the real noise is not what our likelihood assumes.

Histograms compatible with flat, and 90% intervals that cover the truth 92% and 89% of the time, with 237 simulations (so a binomial standard error of 2 points) on real detector noise. Do not read more into it than that: for the loudest prior draws the posterior is narrower than a grid cell, which limits this check to a few points of coverage (repeating it with other seeds gives 88-94%). Within that resolution the machinery - noise covariance, whitening, marginalisation, priors - is calibrated. It says nothing about whether a single damped sinusoid is the right model for what GW150914 did 3 ms after its peak - that was section 2.6.

#3.4 · How detection is actually done: the matched filter

LIGO's search pipelines do not run MCMC on every second of data. They slide templates across the data and compute, for every arrival time at once (one FFT),

$$\rho(t) = \frac{|\langle d, h_t\rangle|}{\sqrt{\langle h, h\rangle}}, \qquad \langle a, b\rangle = 4\,\mathrm{Re}\!\int_0^\infty \frac{\tilde a(f)\,\tilde b^*(f)}{S(f)}\,df .$$

That inner product is the Whittle likelihood again: for fixed template shape, $\rho^2/2$ is the log-likelihood ratio maximised over amplitude, and taking the modulus of the complex filter output maximises over phase. The matched filter is the maximum-likelihood cousin of the marginalisation we just did - the same two quadratures, maximised instead of integrated. In Gaussian noise $\rho$ of a few is typical and 5 is already rare.

We use two templates: our ringdown (posterior medians from 2.4), and the inspiral chirp of Part 4 at leading order, for a small bank of chirp masses.