to maximize ${\mathsf{I}(X;Y)}$, we find the optimal parameter $\theta$ by setting $\xi(\theta) = 1$. This can be seen by setting the derivative of ${\mathsf{I}(X;Y)}$ to zero, with $$\begin{gathered}
\frac{\partial}{\partial\theta} {\mathsf{I}(X;Y)} = \\
 \log_2{\mathopen{}\mathclose\bgroup\originalleft}(\sum_k{\operatorname{e}^{-\theta T(\lambda_k)}}{\aftergroup\egroup\originalright}){\mathopen{}\mathclose\bgroup\originalleft}(\frac{\sum_k{\operatorname{e}^{-\theta T(\lambda_k)}}}{\sum_kT(\lambda_k){\operatorname{e}^{-\theta T(\lambda_k)}}}{\aftergroup\egroup\originalright})^2\mathop{\mathrm{var}}(T(\lambda))\, ,\end{gathered}$$ where $\mathop{\mathrm{var}}(T(\lambda))$ denotes the variance of $T(\lambda_k)$ for the given $\theta$. By assumption, as all $T(\lambda_k)$ are different, the middle part of this expression is strictly positive and $\mathop{\mathrm{var}}(T(\lambda)) > 0$. Hence, it is easy to see that this derivative can only be zero if $\sum_k{\operatorname{e}^{-\theta T(\lambda_k)}} = 1$. The optimal $\theta$ is hence found by setting $\xi(\hat{\theta}) = 1$. Consider the polynomial $$f(x) = \sum_{k=1}^Mx^{-T(\lambda_k)} - 1.$$ As this polynomial is monotonically decreasing for positive $x$, with $\lim_{x\to 0^+}f(x) = +\infty$ and $\lim_{x\to+\infty}f(x)=-1$, $f(x)$ has exactly one positive real root. Let $r$ be the unique positive real root of $f(x)$. Then $\hat{\theta} = \ln(r)$. Inserting $\hat{\theta}$ into ${\mathsf{I}(X;Y)}$ and  proves the lemma.

![Time-scaled of the optimal distribution for linearly spaced constellations with $M$ points (colored with markers solid), and of a system as in [@shevchenko2018tcom] (with markers dotted). As a reference, the capacity ${\mathsf{C}}$ is plotted as well (black solid without markers). For the cutoff parameter, $\delta=0.005$ was used.[]{data-label="fig:shaping:mi_comparison"}](mi_comparison)

We clearly see 