= Every second Poisson arrival has underdispersed counts
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{title2=$M(t)=\lfloor N(t)/2\rfloor$}
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 $(4x-8xe^{-2x}-e^{-4x}+1)/16$. The variance is strictly smaller than the mean for $x>0$: sixteen times their difference has derivative $4[-1+4xe^{-2x}+e^{-4x}]<0$ and is zero at zero, because $\sinh(2x)>2x$. Therefore this process is not a <Cox process>.
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