\symProbability(\symRegValVariate' = \symRegVal\vert\symPoissonRate)
=
\symNumReg\,e^{-\frac{\symPoissonRate}{\symNumReg 2^{\symRegVal}}}\left(1-e^{-\frac{\symPoissonRate}{\symNumReg 2^{\symRegVal}}}\right).$$ According to we have $\symExpectation(\symCountVariate_0/\symNumReg)=
e^{-\frac{\symPoissonRate}{\symNumReg}}$ and $\symExpectation(1-\symCountVariate_{\symRegRange+1}/\symNumReg)=
e^{-\frac{\symPoissonRate}{\symNumReg 2^\symRegRange}}$ which gives $$\begin{aligned}
\symExpectation(\symCountVariate'_\symRegVal)
&=
\symNumReg\,
(\symExpectation(\symCountVariate_0/\symNumReg))^{2^{-\symRegVal}}
\left(1-\left(\symExpectation(\symCountVariate_0/\symNumReg)\right)^{2^{-\symRegVal}}\right),
\\
&=
\symNumReg\,
(\symExpectation(1-\symCountVariate_{\symRegRange+1}/\symNumReg))^{2^{\symRegRange-\symRegVal}}
\left(1-(\symExpectation(1-\symCountVariate_{\symRegRange+1}/\symNumReg))^{2^{\symRegRange-\symRegVal}}\right).\end{aligned}$$ These two expressions for the expectation suggest to use $$\begin{aligned}
\hat{\symCount}'_\symRegVal
&=
\symNumReg\,
(\symCountVariate_0/\symNumReg)^{2^{-\symRegVal}}
\left(1-(\symCountVariate_0/\symNumReg)^{2^{-\symRegVal}}\right)\quad\text{and}
\\
\hat{\symCount}'_\symRegVal
&=
\symNumReg\,
(1-\symCountVariate_{\symRegRange+1}/\symNumReg)^{2^{\symRegRange-\symRegVal}}
\left(1-(1-\symCountVariate_{\symRegRange+1}/\symNumReg)^{2^{\symRegRange-\symRegVal}}\right)\end{aligned}$$ as estimators for $\symRegVal \leq 0$ and $\symRegVal \geq \symRegRange+1$, respectively. Both estimators conserve the mass of zero-valued and saturated registers, because is satisfied, if $\symCountVariate'_\symRegVal$ is replaced by $\hat{\symCount}'_\symRegVal$. Plugging all these estimators into as replacements for $\symCountVariate'_\symRegVal$ finally gives $$\label{equ:correctedestimator}
\symPoissonRateEstimate 
= 
\frac{\symAlpha_\infty\symNumReg^2}
{
\symNumReg\,\symSmallCorrectionFunc(\symCountVariate_0/\symNumReg) + \sum_{\symRegVal=1}^\symRegRange \symCountVariate_\symRegVal 2^{-\symRegVal} + \symNumReg\, \symLargeCorrectionFunc(1-\symCountVariate_{\symRegRange+1}/\symNumReg) 2^{-\symRegRange}
}$$ which we call the improved estimator. Here $\symNumReg\,\symSmallCorrectionFunc(\symCountVariate_0/\symNumReg)$ and $2\symNumReg\,\symLargeCorrectionFunc(1-\symCountVariate_{\symRegRange+1}/\symNumReg)$ are replacements for $\symCountVariate_0$ and $\symCountVariate_{\symRegRange+1}$ in the raw estimator , respectively. The functions $\symSmallCorrectionFunc$ and $\symLargeCorrectionFunc$ are defined as $$\begin{aligned}
\label{equ:sigma}
\symSmallCorrectionFunc(\symX) &:= 
\symX
+
\sum_{\symRegVal=1}^\infty
\symX^{2^\symRegVal} 2^{\symRegVal-1},
\\
\label{equ:tau}
\symLargeCorrectionFunc(\symX)
&:=
\frac{1}{3}
\left(
1-\symX
-
\sum_{\symRegVal=1}^\infty
\left(
1-
\symX^{2^{-\symRegVal}}
\right)^{\!2}
2^{-\symRegVal}
\right).\end{aligned}$$ We can cross-check the new estimator for the linear counting case with $\symRegRange=0$. Using the identity $$\label{equ:sigma_tau_relationship}
\symSmallCorrectionFunc(\symX) + \symLargeCorrectionFunc(\symX) = \symAlpha_\infty\symPowerSeriesFunc(\log_2(\log(1/\symX)))/\log(1/\symX),$$ we get $$\symPoissonRateEstimate 
= 
\frac{\symAlpha_\infty\symNumReg}{
\symSmallCorrectionFunc(\symCountVariate_0/\symNumReg)
+
\symLargeCorrectionFunc(\symCountVariate_0/\symNumReg)
}
=
\frac{\symNumReg\log\!\left(\symNumReg/\symCountVariate_0\right)}{\symPowerSeriesFunc\!\left(\log_2\!\left(\log\!\left(\symNumReg/\symCountVariate_0\right)\right)\right)}$$ which is as expected almost identical 