Every second Poisson arrival has underdispersed counts ($M(t)=\lfloor N(t)/2\rfloor$)

For a rate-$\lambda$ Poisson process, keeping every second arrival gives a renewal process with independent interarrival times equal to sums of two exponential waiting times. Put $x=\lambda t$. Its count mean and variance are $(2x+e^{-2x}-1)/4$ and $(4x-8x{e}^{-2x}-e^{-4x}+1)/16$. The variance is strictly smaller than the mean for $x>0$: sixteen times their difference has derivative $4[-1+4x{e}^{-2x}+e^{-4x}]<0$ and is zero at zero, because $\sinh (2x)>2x$. Therefore this process is not a Cox process.