# Detecting a stochastic gravitational wave signal

This note discusses trying to detect a generic gravitational wave with an unknown waveform emitted from a particular sky position in data from two separate gravitational wave detectors. We define two slightly different approaches to this problem.

## The signal

First we will define the gravitational wave signal at one timestamp, $$i$$, observed in one detector, $$L$$. We envisage two possible methods for this analysis with slightly different model definitions. The first uses

$$\label{eq:signal1} \label{eq:signal1}h_{i}^{L}=h_{0}\left({A_{+}}_{i}{F_{+}}_{i}^{L}(\psi_{i})+{A_{\times}}_{i}{F_{\times}}_{i}^{L}(\psi_{i})\right),\\$$

where $${A_{+}}_{i}$$ and $${A_{\times}}_{i}$$ are the signal amplitudes scale factors (which could be positive or negative) in the plus and cross polarisations at timestamp $$i$$ (which would be the same for different detectors), $${F_{+}}_{i}^{L}(\psi_{i})$$ and $${F_{\times}}_{i}^{L}(\psi_{i})$$ are the detector’s antenna response to the plus and cross polarisations for a given polarisation angle $$\psi_{i}$$11Note that $$\psi$$ could change between data points, so this is also indexed for the current timestamp., and $$h_{0}$$ is an overall underlying gravitational wave amplitude. The second uses

$$\label{eq:signal2} \label{eq:signal2}h_{i}^{L}={A_{+}}_{i}{F_{+}}_{i}^{L}(\psi_{i})+{A_{\times}}_{i}{F_{\times}}_{i}^{L}(\psi_{i}),\\$$

where, in this case, the $${A_{+}}_{i}$$ and $${A_{\times}}_{i}$$ are the actual signal amplitudes in the plus and cross polarisations at timestamp $$i$$.

## First Method

\label{sec:method1}

Here we will examine the details of the first method, which uses the signal model defined in Equation \ref{eq:signal1}.

Now, if we had one data point for detector $$X$$, $$d_{i}^{X}$$, and assuming the noise in the detector is Gaussian with zero mean and standard deviation of $$\sigma_{i}^{X}$$, then the likelihood for the data given the model is

$$\label{eq:singlelikelihood} \label{eq:singlelikelihood}p(d_{i}^{X}|h_{0},{A_{+}}_{i},{A_{\times}}_{i},\psi,I)=\frac{1}{\sqrt{2\pi}\sigma_{i}^{X}}\exp{\left(-\frac{(d_{i}^{X}-h_{i}^{X})^{2}}{2{\sigma_{i}^{X}}^{2}}\right)}.\\$$

We now add another detector, $$Y$$, with data point $$d_{i}^{Y}$$, where the $$i$$ timestamp index in detector $$Y$$ is actually indexing a time that is shifted with respect to that in detector $$X$$ based on the time delay between detectors for the known signal position. So, e.g. $$t_{i}^{Y}=t_{i}^{X}+\Delta t_{i}(\alpha,\delta)$$. This gives a joint likelihood of the data for the two detectors of

$$\label{eq:jointlikelihood} \label{eq:jointlikelihood}p(\{d_{i}^{X},d_{i}^{Y}\}|h_{0},{A_{+}}_{i},{A_{\times}}_{i},\psi_{i},I)=\frac{1}{2\pi\sigma_{i}^{X}\sigma_{i}^{Y}}\exp{\left(-\sum_{L=X,Y}\frac{(d_{i}^{L}-h_{i}^{L})^{2}}{2{\sigma_{i}^{L}}^{2}}\right)}.\\$$

We would like to get a posterior probability distribution on $$h_{0}$$ alone (and in fact we also want the evidence marginalised over $$h_{0}$$ too). If we assume that the $$A$$ scale factors and $$\psi$$ change on the timescale of individual data points then we want to marginalise over them for each point, e.g.

\begin{align} \label{eq:h0likelihood} p(\{d_{i}^{X},d_{i}^{Y}\}|h_{0},I)= & \int_{{A_{+}}_{i}}\int_{{A_{\times}}_{i}}\int_{0}^{\pi}p(\{d_{i}^{X},d_{i}^{Y}\}|h_{0},{A_{+}}_{i},{A_{\times}}_{i},\psi_{i},I)\times\notag \\ & \label{eq:h0likelihood} p({A_{+}}_{i}|I)p({A_{\times}}_{i}|I)p(\psi_{i}|I){\rm d}{A_{+}}_{i}{\rm d}{A_{\times}}_{i}{\rm d}\psi_{i},\\ \end{align}

where $$p({A_{+}}_{i}|I)$$ and $$p({A_{\times}}_{i}|I)$$ are the priors on the scale factors and $$p(\psi_{i}|I)$$ is the prior on $$\psi_{i}$$.

To get the joint likelihood over all the data we can just use the product of Equation \ref{eq:h0likelihood}, such that we have

$$\label{eq:fulljoint} \label{eq:fulljoint}p(\{\vec{d}^{X},\vec{d}^{Y}\}|h_{0},I)=\prod_{i=1}^{N}p(\{d_{i}^{X},d_{i}^{Y}\}|h_{0},I),\\$$

where $$N$$ is the total number of data points. We can then get the posterior on $$h_{0}$$ as

$$\label{eq:posterior1} \label{eq:posterior1}p(h_{0}|\{\vec{d}^{X},\vec{d}^{Y}\},I)=\frac{p(\{\vec{d}^{X},\vec{d}^{Y}\}|h_{0},I)p(h_{0}|I)}{p(\{\vec{d}^{X},\vec{d}^{Y}\}|I)},\\$$

where $$p(h_{0}|I)$$ is the prior on $$h_{0}$$ and $$p(\{\vec{d}^{X},\vec{d}^{Y}\}|I)$$ is the evidence for the data. The evidence, $$Z$$, is given by

$$\label{eq:evidence1} \label{eq:evidence1}Z=\int p(\{\vec{d}^{X},\vec{d}^{Y}\}|h_{0},I)p(h_{0}|I){\rm d}h_{0}.\\$$

### Practical evaluation

Depending on the choice of prior some of the marginalisations in Equation \ref{eq:h0likelihood} are analytical, but others will need to be evaluated numerically. To start with we can set a Gaussian priors on $${A_{+}}_{i}$$ and $${A_{\times}}_{i}$$, both with zero mean and standard deviations of $$\sigma_{A_{+,\times}}$$

$$\label{eq:priorAp} \label{eq:priorAp}p({A_{+,\times}}_{i}|I)=\frac{1}{\sqrt{2\pi}\sigma_{A_{+,\times}}}\exp{\left(-\frac{{A_{+,\times}}_{i}^{2}}{2\sigma_{A_{+,\times}}^{2}}\right)}.\\$$

We can get a marginalised likelihood on $$h_{0}$$ and $$\psi_{i}$$ by multiplying Equation \ref{eq:h0likelihood} by these priors and integrating over $${A_{+}}_{i}$$ and $${A_{\times}}_{i}$$ between $$-\infty$$ and $$\infty$$

\begin{align} \label{eq:margApAc} \label{eq:margApAc}p(\{d_{i}^{X},d_{i}^{Y}\}|h_{0},\psi_{i},I) & =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}p(\{d_{i}^{X},d_{i}^{Y}\}|h_{0},{A_{+}}_{i},{A_{\times}}_{i},\psi_{i},I)p({A_{+}}_{i}|I)p({A_{\times}}_{i}|I){\rm d}{A_{+}}_{i}{\rm d}{A_{\times}}_{i}, \\ & =\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{\exp{\left(-\sum_{L=X,Y}\frac{(d_{i}^{L}-h_{i}^{L})^{2}}{2{\sigma_{i}^{L}}^{2}}\right)}\exp{\left(-\frac{1}{2}\left[\frac{{A_{+}}_{i}^{2}}{\sigma_{A_{+}}^{2}}+\frac{{A_{\times}}_{i}^{2}}{\sigma_{A_{\times}}^{2}}\right]\right)}}{4\pi^{2}\sigma_{i}^{X}\sigma_{i}^{Y}\sigma_{A_{+}}\sigma_{A_{\times}}}{\rm d}{A_{+}}_{i}{\rm d}{A_{\times}}_{i}\\ \end{align}

Following, e.g. Appendix B of Haasteren et al. (2012) (and expanding out the model terms) we can put this equation into vector and matrix form to give

$$\label{eq:vh1} \label{eq:vh1}p(\mathbf{d}|h_{0},\psi_{i},I)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{\exp{\left(-\frac{1}{2}\left(\mathbf{d}^{T}-\boldsymbol{\xi}\mathbf{M}\right)^{T}\mathbf{C}^{-1}\left(\mathbf{d}^{T}-\boldsymbol{\xi}\mathbf{M}\right)\right)}\exp{\left(-\frac{1}{2}\boldsymbol{\xi}^{T}\boldsymbol{\Sigma}_{0}^{-1}\boldsymbol{\xi}\right)}}{4\pi^{2}\sqrt{\left|\boldsymbol{\Sigma}_{0}\right|\left|\mathbf{C}\right|}}{\rm d}{A_{+}}_{i}{\rm d}{A_{\times}}_{i},\\$$

where we a data vector $$\mathbf{d}=\{d_{i}^{X},d_{i}^{Y}\}$$, a design matrix

$$\mathbf{M}=h_{0}\left[\begin{array}{cc}{F_{+}}^{X}_{i}(\psi_{i})&{F_{+}}^{Y}_{i}(\psi_{i})\\ {F_{\times}}^{X}_{i}(\psi_{i})&{F_{\times}}_{i}^{Y}(\psi_{i})\end{array}\right]\\$$

and parameter vector $$\boldsymbol{\xi}=\{{A_{+}}_{i},{A_{\times}}_{i}\}$$, and a data covariance matrix (and its inverse) of

$$\mathbf{C}=\left[\begin{array}{cc}\left(\sigma_{i}^{X}\right)^{2}&0\\ 0&\left(\sigma_{i}^{Y}\right)^{2}\end{array}\right],\mathrm{}\mathbf{C}^{-1}=\left[\begin{array}{cc}\left(\sigma_{i}^{X}\right)^{-2}&0\\ 0&\left(\sigma_{i}^{Y}\right)^{-2}\end{array}\right].\\$$

For the Gaussian priors on the amplitudes we use a covariance matrix (and its inverse) of

$$\boldsymbol{\Sigma}_{0}=\left[\begin{array}{cc}\sigma_{A_{+}}^{2}&0\\ 0&\sigma_{A_{\times}}^{2}\end{array}\right],\mathrm{}\boldsymbol{\Sigma}_{0}^{-1}=\left[\begin{array}{cc}\sigma_{A_{+}}^{-2}&0\\ 0&\sigma_{A_{\times}}^{-2}\end{array}\right].\\$$

This can be rearranged (to isolated the $$\boldsymbol{\xi}$$ term) into the form given in equation (B2) of Haasteren et al. (2012) for which the solution is

$$\label{eq:margApAcmat} \label{eq:margApAcmat}p(\mathbf{d}|h_{0},\psi_{i},I)=\frac{\sqrt{\left|\boldsymbol{\Sigma}\right|}}{4\pi^{2}\sqrt{\left|\boldsymbol{\Sigma}_{0}\right|\left|\mathbf{C}\right|}}\exp{\left(-\frac{1}{2}\left[\mathbf{d}^{T}\mathbf{C}^{-1}\mathbf{d}+\boldsymbol{\chi}^{T}\boldsymbol{\Sigma}^{-1}\boldsymbol{\chi}\right]\right)},\\$$

where

$$\label{eq:chi} \label{eq:chi}\boldsymbol{\chi}=\left(\mathbf{M}^{T}\mathbf{C}^{-1}\mathbf{M}+\boldsymbol{\Sigma}_{0}^{-1}\right)^{-1}\left(\mathbf{M}^{T}\mathbf{C}^{-1}\mathbf{d}\right)\\$$

and

\boldsymbol{\Sigma}^{-1}=\mathbf{M}^{T}\mathbf{C}^{-1}\mathbf{M}+\boldsymbol{\S