Remember that PAC is computing the Fourier coefficients and reporting on those individually. Consider applying a 1Vp sine wave to the input. If the output is a clocked 1Vp sine wave, people often assume that the gain of the circuit is 1 because that is the way they think of it. But SpectreRF is computing the Fourier coefficients and coming to a different conclusion. The clocked sine wave often as a duty cycle of 50% or less, and the Fourier coefficients will be scaled by the duty cycle. So while you might look at the clocked output and see a 1Vp sine wave, the fundamental Fourier coefficient would be less than 0.5V. Thus, the gain is actually less that 0.5.
The basic problem stems from the fact that designers are thinking of the switched-capacitor circuit as a discrete-time circuit, and so discount the behavior in between the sample points. By default SpectreRF operates on continuous time signals and so considers the whole signal.
To get the results you expect, you will need to perform a sampling operation on the output. SpectreRF provides this sampling as one of its options.
This situation is described more fully in
http://www.designers-guide.org/Analysis/sc-filters.pdf.
-Ken