# Information-Theoretic Analysis of a Family of Improper Discrete Constellations
## 4 Information-Theoretic Analysis
### Shaping Loss  
Next, we compute the shaping loss defined as the ratio of the powers required to achieve the same transmission rate when using symbols that are either uniformly distributed in \(\mathcal{A}(\phi,\kappa,M)\) (by assuming \(M\) to be an unbounded large number) or Gaussian distributed. The shaping loss in the improper case can also be interpreted as the power penalty due to using codewords uniformly distributed inside a multidimensional parallelogram rather than inside a multidimensional ellipsoid.  
**Theorem 5**.: _The shaping loss \(\gamma_{s}\) of the proposed family of improper constellations with respect to improper Gaussian signaling is:_  
\[\gamma_{s}=\frac{\pi\,e}{6}\ \left(1.53\,dB\right)\]  
_which is independent of \(\kappa\) and \(\phi\) and coincides with the shaping loss of proper M-QAM constellations [25]._  
Proof.: The shaping gain can be obtained as follows. Let us consider two memoryless complex sources. The first source generates discrete values \(V=V_{I}+jV_{Q}\) uniformly distributed over the constellation \(\mathcal{A}(\phi,\kappa,M)\) where \(M\) is assumed to be a very large number. Under this assumption, its uniform distribution can be approximated by the probability density function:  
\[f_{V}(v_{I},v_{Q})=\left\{\begin{array}{ll}\frac{1}{A(\mathcal{P})}&\text{if }(v_{I},v_{Q})\in\mathcal{P};\\ 0&\text{otherwise},\end{array}\right.\]  
where \(\mathcal{P}\) denotes the parallelogram determined by the boundaries of the constellation \(\mathcal{A}(\phi,\kappa,M)\) and \(A(\mathcal{P})\) denotes its area. Based on the Jacobian of the transformation \(\mathbf{H}(\phi,\kappa)\), this area can be computed as:  
\[A(\mathcal{P})=\det(\mathbf{H}(\phi,\kappa))A(\mathcal{R})\]  
where \(A(\mathcal{R})=1\) denotes the area of the square \(\mathcal{R}=[-0.5,0.5]^{2}\). Note that the second-order moment of \(\mathcal{R}\) is given by \(P_{\mathcal{R}}=1/6\) and so will be the second-order moment, \(P_{V}\), of the parallelogram \(\mathcal{P}\), since the transformation \(\mathbf{H}(\phi,\kappa)\) keeps its input-output power invariant. Therefore, \(A(\mathcal{P})\) can also be written as:  
\[A(\mathcal{P})=\det(\mathbf{H}(\phi,\kappa))A(\mathcal{R})=\sqrt{1-\kappa^{2} }\,6\,P_{V}.\]  
The second source, \(G=G_{I}+jG_{Q}\), generates continuous values in \(\mathbb{C}\), with the improper complex Gaussian distribution:  
\[f_{G}(g_{I},g_{Q})=\mathcal{N}\left(0,\mathbf{K}_{G}\right),\]  
where \(\mathbf{K}_{G}=\text{diag}\left(\frac{P_{G}}{2}(1+\kappa),\frac{P_{G}}{2}(1- \kappa)\right)\) and \(P_{G}\) denotes the second-order moment of this Gaussian source.  
Note that in order to generate the same number of bits per source symbol (i.e., the same information rate), the entropy of both sources should be the same. Since:  
\[h_{V}(V_{I},V_{Q})=\int\int_{A(\mathcal{P})}\log A(\mathcal{P})\frac{1}{A( \mathcal{P})}dv_{I}dv_{Q}=\log A(\mathcal{P})=\log\left(\sqrt{1-\kappa^{2}}\, 6\,P_{V}\right)\]  
and:  
\[h_{G}(G_{I},G_{Q})=\log\left((2\pi e)\sqrt{\det(\mathbf{K}_{G})}\right)=\log \left((2\pi e)\frac{P_{G}}{2}\sqrt{1-\kappa^{2}}\right),\]  
we find that \(\gamma_{s}\) is:  
\[\gamma_{s}=\frac{P_{V}}{P_{G}}=\frac{\pi e}{6},\]  
as we wanted to prove.