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Comparison of signal detectability for single-phase and two-phase searches

The previous Section [*] described optimal filtering searches in two cases - looking for: With the choice of filter normalizations made previously, the expected signal produced by a source $\alpha T_0$ would be the same for both searches, but the expected (noise)${}^2$ was higher in the two-phase case. One might wonder if this reduced SNR means that a two-phase search reduces ones ability to identify signals. The answer turns out to be ``not significantly".

The reason for this is that the distribution of signal values produced by detector noise alone in the single- and two-phase cases are quite different. In order to answer the question: ``what is the smallest signal detectable" we need to fix a false alarm rate. For a given time-duration of data, this is equivalent to fixing a false alarm probability. Let us assume that this probability has been fixed to be a small value $\epsilon$, and compare the single- and two-phase searches.

In the single-phase case, in the absence of a source, the values of the signal $S$ ([*]) are Gaussian random variables with a mean-squared value of 1. Hence the threshold $S_0$ determined by the false alarm rate must be set so that there is probability $\epsilon$ of $S$ falling outside the range $[-S_0,S_0]$. This means that

\begin{displaymath}\epsilon = 2 {1 \over
\sqrt{2 \pi}} \int_{S_0}^\infty \exp(-x^2/2) \> dx = {\rm\ erfc} (S_0/\sqrt{2}) .
\end{displaymath} (6.15.88)

The solution to this equation is the threshold as a function of the false alarm probability:
\begin{displaymath}
S_0^{\small\textrm{single-phase}}(\epsilon)=\sqrt{2}
{\rm\ erfc}^{-1}(\epsilon).
\end{displaymath} (6.15.89)

Thus, for example, to obtain a false alarm probability of $\epsilon =
10^{-5}$ we need to set a threshold $S^{\small\textrm{single-phase}}_0
= \sqrt{2} \textrm{\ erfc}^{-1}(10^{-5}) = 4.417$. In this case, our minimum detectable signal has amplitude $\alpha = 4.417$.

In the two-phase case, the probability distribution of the signal in the absence of a source is different, because in this case the signal ([*]) is described by the probability distribution of a random variable $r$, where $r^2 = x^2 + y^2$ and $x$ and $y$ are independent random Gaussian variables with unit rms. Here, $x$ and $y$ are the real and imaginary parts of the signal ([*] in the absence of a source. Their probability distribution is:

$\displaystyle P(x) dx P(y) dy$ $\textstyle =$ $\displaystyle {1 \over 2 \pi} {\rm e}^{-x^2/2} {\rm e}^{-y^2/2} dx dy$  
  $\textstyle =$ $\displaystyle {1 \over 2 \pi} {\rm e}^{-r^2/2} r dr d\phi$ (6.15.90)
  $\textstyle \Rightarrow$    
$\displaystyle P(r)dr$ $\textstyle =$ $\displaystyle {\rm e}^{-r^2/2} r dr.$ (6.15.91)

In the final line, we have integrated over the irrelevant angular variable $\phi \in [0,2\pi)$. So in the two-phase case, as before, the threshold value of the signal is set by requiring that the false alarm probability be $\epsilon$:
\begin{displaymath}
\epsilon=\int_{S_0}^\infty \exp(-r^2/2) r dr = \exp({-S_0^2/2}).
\end{displaymath} (6.15.92)

The solution here is that the threshold is $S^{\small\textrm{two-phase}}_0 = \sqrt{-2 \ln
\epsilon}$. For example, to obtain a false alarm probability of $\epsilon =
10^{-5}$ we need to set a threshold $S^{\small\textrm{two-phase}}_0 = 4.799$. In this case, our minimum detectable (0-phase) signal has amplitude $\alpha = 4.799$, which is only slightly higher than in the single-phase case.

It is not a coincidence that for a given false alarm rate, the amplitude of the minimum detectable signals are almost the same. Although the expected value of the single-phase signal${}^2$ in the absence of a source is smaller than the expected value of the two-phase $\vert{\rm signal}\vert^2$ in the absence of a source, the tails of the two probability distributions are almost identical. For the same false alarm probability $\epsilon$ the thresholds in the two instances are related by

\begin{displaymath}
\epsilon=
\sqrt{2 \over \pi} \int_{S^{\small\textrm{single-p...
...\int_{S^{\small\textrm{two-phase}}_0}^\infty \exp(-r^2/2) r dr
\end{displaymath} (6.15.93)

But for thresholds of reasonable size (small $\epsilon$) both integrals are dominated by the region just to the right of $S_0$, and in this neighborhood the integrands differ by a small factor of approximately $\sqrt{\pi S_0 \over 2}$. Since $\epsilon$ varies exponentially with the threshold, there is a logarithmically small difference between the thresholds $S^{\small\textrm{single-phase}}_0$ and $S^{\small\textrm{two-phase}}_0$.

For a fixed false alarm probability, we can write the the two-phase threshold $S^{\small\textrm{two-phase}}_0$ as a function of the one-phase threshold $S^{\small\textrm{single-phase}}_0$:

\begin{displaymath}
S^{\small\textrm{two-phase}}_0 = \sqrt{ - 2 \ln \left(\textr...
...small\textrm{single-phase}}_0 \over \sqrt{2}}\right) \right)}.
\end{displaymath} (6.15.94)

The plot of this relationship in Figure [*] shows clearly that once the thresholds are reasonably large, they are very nearly equal.

Figure: The threshold for a two-phase search $S^{\small \textrm{two-phase}}_0$ is shown as a function of the threshold for the single-phase search $S^{\small \textrm{single-phase}}_0$ which gives the same false alarm rate. When the false alarm rates are small, they are very nearly equal.
\begin{figure}\begin{center}
\epsfig{file=Figures/thresholds.ps,angle=0,width=3.5in}\end{center}\end{figure}


next up previous contents
Next: Function: correlate() Up: GRASP Routines: Gravitational Radiation Previous: Wiener (optimal) filtering   Contents
Bruce Allen 2000-11-19