Generalised Mutual Information

Fix $s > 0$ and $R$. Draw $2^{NR}$ codewords $\{\mathbf{x}_m\}$ i.i.d.\ uniform on $\mathcal{X}^N$. The decoder outputs $$\hat m \;=\; \arg\max_m \sum_{n=1}^N \log q_{\rm BICM}(y_n, x_{m,n})^s.$$ Conditional on transmitting $\mathbf{x}_1$, the pairwise error probability of $\mathbf{x}_1 \to \mathbf{x}_{m}$ ($m \ne 1$) is $$P_{\rm pair} = \Pr\!\left[\sum_n \log q(Y_n, \bar X_n)^s \ge \sum_n \log q(Y_n, X_n)^s \,\middle|\, \mathbf{X}_1\right].$$

For any $t > 0$, by Markov: $$P_{\rm pair} \le \mathbb{E}\!\left[ \exp\!\left(t \sum_n [\log q(Y_n, \bar X_n)^s - \log q(Y_n, X_n)^s]\right) \right].$$ Choosing $t = 1$ and using the i.i.d.\ codeword letters plus memoryless channel gives $$P_{\rm pair} \le \prod_n \mathbb{E}_{P_{X,Y}}\!\left[\frac{\mathbb{E}_{\bar X}[q(Y, \bar X)^s]}{q(Y, X)^s}\right]^{?},$$ and after standard manipulation (see Ganti-Lapidoth-Telatar 1999 or Ch. 3 of the monograph) the per-letter factor is exactly $e^{- I^{\mathrm{GMI}}(s)}$.

Union-bounding over the $2^{NR} - 1$ incorrect codewords, the ensemble-average error probability is bounded by $$\bar P_e(R) \le 2^{NR} \cdot e^{-N I^{\mathrm{GMI}}(s)} = 2^{N(R - I^{\mathrm{GMI}}(s) / \ln 2)}.$$ This tends to zero iff $R < I^{\mathrm{GMI}}(s) / \ln 2 = I^{\mathrm{GMI}}(s)$ (with a $\log_2$ convention on the GMI). Optimising over $s > 0$ gives the largest achievable rate $R^\star_{\rm BICM} = \sup_s I^{\mathrm{GMI}}(s)$.

At $s = 1$, the inner expectation $\mathbb{E}_{\bar X}[q(y, \bar X)]$ is, for the BICM product metric and uniform inputs, $$\mathbb{E}_{\bar X}[q_{\rm BICM}(y, \bar X)] = \prod_\ell \mathbb{E}_{\bar B_\ell}[p_{W_\ell}(y\mid \bar B_\ell)] = \prod_\ell \tfrac{1}{2}(p_{W_\ell}(y\mid 0) + p_{W_\ell}(y\mid 1)).$$ Therefore $$I^{\mathrm{GMI}}_{\rm BICM}(1) = \mathbb{E}_{P_{X,Y}}\!\left[ \sum_\ell \log \frac{p_{W_\ell}(Y\mid B_\ell)} {\tfrac{1}{2}(p_{W_\ell}(Y\mid 0) + p_{W_\ell}(Y\mid 1))}\right] = \sum_\ell I(Y; B_\ell).$$ This is the Caire-Taricco-Biglieri formula. The $s = 1$ GMI equals the BICM capacity of Ch. 5 exactly — proving that the CTB formula is a rigorous mismatched-decoding achievability result.

For Gray labelling on square $M$-QAM over AWGN at high SNR, the per-position bit channels $W_\ell$ become asymptotically symmetric and Gaussian, so the GMI of the product metric approaches the GMI of the genuine channel, and $\partial I^{\mathrm{GMI}}/\partial s \big|_{s = 1} = 0$ — i.e., $s^\star \to 1$. For non-Gray labellings the asymmetry between the per-position channels forces $s^\star \ne 1$. $\blacksquare$

$I^{\mathrm{GMI}}(1) = \mathbb{E}_{P_{X,Y}}\!\left[ \log \frac{q_{\rm BICM}(Y, X)}{\mathbb{E}_{\bar X}[q_{\rm BICM}(Y, \bar X)]}\right].$$

$q_{\rm BICM}(y, x) = \prod_{\ell=0}^{L-1} p_{W_\ell}(y\mid b_\ell)$ by definition; hence $\log q_{\rm BICM}(y, x) = \sum_\ell \log p_{W_\ell}(y\mid b_\ell)$.

With uniform input $P_X$ on $\mathcal{X}$, the marginal over the $\ell$-th bit is $\Pr(B_\ell = 0) = \Pr(B_\ell = 1) = \tfrac{1}{2}$, and the bits are marginally independent. Therefore $$\mathbb{E}_{\bar X}[q_{\rm BICM}(y, \bar X)] = \prod_\ell \mathbb{E}_{\bar B_\ell}[p_{W_\ell}(y\mid \bar B_\ell)] = \prod_\ell \tfrac{1}{2}\!\left[p_{W_\ell}(y\mid 0) + p_{W_\ell}(y\mid 1)\right].$$

$I^{\mathrm{GMI}}(1) = \mathbb{E}_{P_{X,Y}}\!\left[ \sum_\ell \log \frac{p_{W_\ell}(Y\mid B_\ell)} {\tfrac{1}{2}(p_{W_\ell}(Y\mid 0) + p_{W_\ell}(Y\mid 1))}\right] = \sum_\ell I(Y; B_\ell).$$The inner expectation per$\ell$is precisely the mutual information$I(Y; B_\ell)$of the$\ell$-th BICM bit channel. Summing gives the CTB formula.$\blacksquare$