Radioactive Decay Process
The average disintegration rate is given by $\bar n=\sum_{n=0}^{N_0} nP(n)$ where $P$ is the binomial distribution. If $M$ is the average number of events in time $t$ then $\displaystyle M=N_0(1-e^{-\lambda t})$. For times smaller than the lifetime $1/\lambda$  the $\displaystyle e^{-\lambda t} \to 1-\lambda t$ then $M=N_0\lambda t$ and as the disintegration rate is $M/t$ this leads to $\displaystyle \frac{dN}{dt}=-\lambda N$. Thus provided the measuring time-bins are small compared to the lifetime this rate law should hold. 

As the decay proceeds the number of counts becomes small, perhaps $10^6$ at short times to just 1 or 2 at long times then of course the signal to noise becomes huge, zero counts in some time bins for example and 1 or 2 counts in others. (This becomes an issue in analysing data, see Anal. Chem., 2003, 75 (16), pp 4182–4187).

The expected variance is, as you quote, $V=\sigma^2=N_0e^{-\lambda t}(1-e^{-\lambda t})$ and can be rewritten as $V=Me^{-\lambda t}$ where $M$ is the mean number of events recorded in time $t$. As the time bins are normally short so that $\lambda t &lt;&lt;1$ then $\sigma=\sqrt{M}$ so that $\displaystyle \frac{\sigma}{M}=\frac{1}{\sqrt{M}}$