optimization is performed over all $r \ge 0$, it can be easily seen that the above maximization problem can be recast as follows: $$\begin{gathered}
\label{eq:R_Etheta0equiv}
{{\sf{R}}}_E({{\text{\footnotesize{SNR}}}},0) = \max_{x\ge 0} \,\,x \, P\left\{z >
\frac{2^x-1}{{{\text{\footnotesize{SNR}}}}}\right\}.\end{gathered}$$ &gt;From (\[eq:R\_Etheta0equiv\]), we note that ${{\sf{R}}}_E({{\text{\footnotesize{SNR}}}},0)$ depends on $B$ only through ${{\text{\footnotesize{SNR}}}}= \frac{{\bar{P}}}{N_0 B}$. Therefore, increasing $B$ has the same effect as decreasing ${\bar{P}}$. Hence, low-power and wideband regimes are equivalent when $\theta = 0$. Consequently, the result of Theorem \[theo:minenergywideband\], which shows that the minimum bit energy is achieved as $B \to
\infty$, implies that the minimum bit energy is also achieved as ${\bar{P}}\to 0$.

Note that ${{\sf{R}}}_E({{\text{\footnotesize{SNR}}}},\theta) \le {{\sf{R}}}_E({{\text{\footnotesize{SNR}}}},0)$ for $\theta > 0$. Therefore, the bit energy required when $\theta > 0$ is larger than that required when $\theta = 0$. On the other hand, as we have proven in Theorem \[theo:stacond\], ${{\alpha^{\ast}_{\text{opt}}}}$ is unique and the bit energy required as ${\bar{P}}\to 0$ is the same for all $\theta \ge 0$ when $z$ has a Gamma density in the form given in (\[eq:gammadensity0\]). Since the minimum bit energy in the case of $\theta = 0$ is achieved as ${\bar{P}}\to 0$, and the same bit energy is attained for all $\theta > 0$, we immediately 