Why do we assert Hulse–Taylor binary system's orbital decay to gravitational waves and not radiation?

The Hulse–Taylor system's orbit has decayed since the binary system was initially discovered, in precise agreement with the loss of energy due to gravitational waves. The ratio of observed to predicted rate of orbital decay is calculated to be 0.997±0.002.

But also

The total power of the gravitational radiation (waves) emitted by this system presently, is calculated to be 7.35 × 1024 watts. For comparison, this is 1.9% of the power radiated in light by our own Sun.

So how do we know the Hulse–Taylor system's loss of energy is not (at least partially) due to electromagnetic radiation too? Especially if we don't even see the other pulsar due to its unfavorable inclination?

• AFAIK, the other star in the Hulse-Taylor system isn't a pulsar, it's a neutron star (though all pulsars are NS, not all NS are pulsars). – Kyle Kanos Feb 11 '16 at 13:32
• You are right, but my point holds: Pulses from the companion neutron star have not been detected, but this might only be the result of an unfavorable viewing angle. – daniel.sedlacek Feb 11 '16 at 13:34
• Probably useful reading: ned.ipac.caltech.edu/level5/ESSAYS/Boughn/boughn.html – Kyle Kanos Feb 11 '16 at 13:37
• Kind of amusing that the Wikipedia authors characterize a 1.5 sigma deviation from theory as "in precise agreement." – Rococo Feb 16 '16 at 19:59

The emission of gravitational waves causes the separation $r$ between the two binary components to decrease. As they do so, the power emitted in gravitational waves increases as $r^{-5}$. Thus the rate of change of the orbital period is very non-linear, with $dr/dt \propto r^{-3}$.