= Shot noise counting model
{title2=$\operatorname{Var}(\overline i_T)=qI/T$}
Let $q>0$ be the magnitude of each transferred <electric charge>, and let arrivals form a <Homogeneous Poisson process> of rate $\lambda$. Extend the stream over past and future times when considering stationary filtering. For <Electrons>, $q=e$, the <elementary charge>. Choose the positive <electric current> direction to follow these transfers, so $I=q\lambda>0$ is its mean magnitude. If $N_T$ counts arrivals in a gate of duration $T>0$, then
$$
N_T\sim\operatorname{Poisson}(\lambda T),\qquad
\mathbb E[N_T]=\operatorname{Var}(N_T)=\lambda T.
$$
The gate's transferred <electric charge> is $Q_T=qN_T$, and its average <electric current> is $\overline i_T=Q_T/T$. Scaling the <expected value> and <variance> gives
$$
\mathbb E[\overline i_T]=I,\qquad
\operatorname{Var}(Q_T)=qIT,\qquad
\operatorname{Var}(\overline i_T)=\frac{qI}{T},\qquad
\frac{\sqrt{\operatorname{Var}(\overline i_T)}}{I}=\frac1{\sqrt{\lambda T}}.
$$
More arrivals increase the absolute fluctuation but decrease its size relative to the mean. For an original count example with $\lambda T=4$, the mean count and its <variance> are both $4$, the <standard deviation> is $2$, and the relative <standard deviation> is $1/2$. The <probability> of no transfer is $e^{-4}\approx0.01832$; the <probability> of four is $4^4e^{-4}/4!\approx0.19537$. A constant arrival rate does not prescribe four arrivals in every gate.
These are fluctuations of a time average, not the <variance> of an ideal instantaneous <electric current>. A model of point transfers has arbitrarily sharp impulses; it acquires a finite measured <variance> only after specifying a gate or <linear filter>. Actual collection pulses have finite duration and <electrical circuits> have finite response times.
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