The Zheng-Tse Theorem

Let $\mathbf{H}$ be $n_r \times n_t$ with i.i.d. $\mathcal{CN}(0, 1)$ entries. The instantaneous capacity is $$C(\mathbf{H}) = \log_2 \det\!\left(\mathbf{I}_{n_r} + \tfrac{\text{SNR}}{n_t} \mathbf{H}\mathbf{H}^{H}\right) = \sum_{i = 1}^{m} \log_2 (1 + \tfrac{\text{SNR}}{n_t} \lambda_i),$$ where $m = \min(n_t, n_r)$ and $\lambda_1 \ge \lambda_2 \ge \cdots \ge \lambda_m$ are the ordered nonzero eigenvalues of $\mathbf{H}\mathbf{H}^{H}$. Reparametrise each eigenvalue as $\lambda_i = \text{SNR}^{-\alpha_i}$: the exponent vector $\boldsymbol{\alpha} = (\alpha_1, \ldots, \alpha_m)$ with $\alpha_1 \le \alpha_2 \le \cdots \le \alpha_m$ encodes the "fading depth" of each eigenvalue.

The outage event at rate $R = r \log_2 \text{SNR}$ is $$\mathcal{O} = \{C(\mathbf{H}) < r \log_2 \text{SNR}\} = \Big\{\sum_{i} \log(1 + \text{SNR}^{1-\alpha_i}) < r \log \text{SNR}\Big\}.$$ For large $\text{SNR}$, $\log(1 + \text{SNR}^{1-\alpha_i}) \doteq (1 - \alpha_i)^+ \log \text{SNR}$, so the outage event reduces to $$\mathcal{O}_{\text{SNR}} = \Big\{\sum_i (1 - \alpha_i)^+ < r\Big\}.$$

The joint density of the ordered eigenvalues of $\mathbf{H}\mathbf{H}^{H}$ (complex Wishart, $n_r \ge n_t$ case) is $$p(\boldsymbol{\lambda}) = K \prod_i \lambda_i^{n_r - n_t} \prod_{i < j} (\lambda_i - \lambda_j)^2 \exp(-\sum_i \lambda_i),$$ where $K$ is a normalising constant. Substituting $\lambda_i = \text{SNR}^{-\alpha_i}$, $d\lambda_i = -(\log\text{SNR}) \text{SNR}^{-\alpha_i} d\alpha_i$, the density in $\boldsymbol{\alpha}$ is $$p(\boldsymbol{\alpha}) \doteq \prod_i \text{SNR}^{-\alpha_i(n_r - n_t + 1)} \prod_{i < j} \text{SNR}^{-2\alpha_i} \exp(-\text{SNR}^{-\alpha_m}) \cdot (\log \text{SNR})^m.$$ The $\exp(-\text{SNR}^{-\alpha_m})$ factor is $\le 1$ for $\alpha_m \ge 0$ and contributes exponent $0$ if $\alpha_m > 0$, or suppresses the density super-polynomially if $\alpha_m < 0$. So the relevant regime is $\boldsymbol{\alpha} \in \mathbb{R}_+^m$ (all eigenvalues decay in SNR), and $$p(\boldsymbol{\alpha}) \doteq \text{SNR}^{-\sum_i (2i - 1 + n_r - n_t)\alpha_i}$$ after collecting the Vandermonde and marginal factors (assuming $\alpha_1 \le \alpha_2 \le \cdots \le \alpha_m$; note the Vandermonde contributes $-2 \alpha_i (i - 1)$ from the $j > i$ pairs and $-2 \alpha_i(m - i)$ from the $j < i$ pairs, summing to $-2\alpha_i (m-1)$; but with ordering, only $j > i$ pairs contribute and the exponent becomes $-\sum_i (2(i - 1))\alpha_i$, which combined with the $\prod_i \lambda_i^{n_r - n_t}$ factor gives the $(2i - 1 + n_r - n_t)\alpha_i$ coefficient; see Zheng-Tse eq. (15)).

The outage probability at multiplexing gain $r$ is $$P_{\rm out}(r \log_2 \text{SNR}) = \int_{\sum_i (1 - \alpha_i)^+ < r} p(\boldsymbol{\alpha})\,d\boldsymbol{\alpha} \doteq \text{SNR}^{-d^*_{\rm out}(r)},$$ with $$d^*_{\rm out}(r) = \inf_{\boldsymbol{\alpha}\in\mathcal{A}(r)} \sum_{i=1}^m (2i - 1 + n_r - n_t)\alpha_i,$$ where $\mathcal{A}(r) = \{\boldsymbol{\alpha}: 0 \le \alpha_1 \le \cdots \le \alpha_m,\, \sum_i (1 - \alpha_i)^+ < r\}$. This is a linear program in $\boldsymbol{\alpha}$ over a convex region.

For $r \in [k, k+1]$ with integer $k$, the optimum is attained at $\alpha_i = 0$ for $i \le k$ (these eigenvalues fail to fade) and $\alpha_i = 1$ for $i > k$ (these eigenvalues fade at rate $1 + \epsilon$), with $\alpha_{k+1}$ tuned to exactly make $\sum_i (1-\alpha_i)^+ = r$: $\alpha_{k+1} = 1 - (r - k)$. Plugging in, the sum $\sum_i (2i - 1 + n_r - n_t)\alpha_i$ at the optimum evaluates to the piecewise-linear interpolation of $(n_t - k)(n_r - k)$ at $r = k$. After the algebra (Zheng-Tse Lemma 5), one obtains exactly $$d^*_{\rm out}(r) = (n_t - r)(n_r - r) \text{ at integer } r,$$ with linear interpolation between integers.

For any coding scheme at rate $R(\text{SNR}) = r \log_2 \text{SNR}$, the block error probability is lower-bounded by the outage: $$P_e(\text{SNR}) \ge P_{\rm out}(r \log_2 \text{SNR}) \doteq \text{SNR}^{-d^*_{\rm out}(r)}.$$ Therefore the diversity gain satisfies $d^* \le d^*_{\rm out}(r)$. No code can do better than the outage exponent in diversity.

Consider a random codebook $\mathcal{C}$ of size $M = 2^{r L \log_2 \text{SNR}} = \text{SNR}^{r L}$, each codeword $\ntn{X} \in \mathbb{C}^{n_t \times L}$ with i.i.d. $\mathcal{CN}(0, \text{SNR}/n_t)$ entries. The average block error probability under ML decoding is bounded by $$\bar P_e \le P_{\rm out}(r \log_2 \text{SNR}) + \mathbb{P}(\mathcal{E}^c \cap \mathcal{O}^c).$$ The first term is the outage — its exponent is $d^*_{\rm out}(r)$. The second term (undetected error given not in outage) can be made exponentially smaller than the outage by a Gallager random-coding argument provided $L \ge n_t + n_r - 1$: the ensemble average over Gaussian codebooks, combined with a union bound and the Chernoff-bound-to-PEP analysis of Chapter 11, gives $$\mathbb{P}(\mathcal{E} \cap \mathcal{O}^c) \doteq \text{SNR}^{-\infty},$$ i.e., faster-than-polynomial decay. The block-length assumption $L \ge n_t + n_r - 1$ is needed so that the error-matrix product $\boldsymbol{\Delta}\boldsymbol{\Delta}^H$ is full-rank with probability $1$ — §5 discusses what breaks when $L$ is shorter.

Combining, $\bar P_e \doteq \text{SNR}^{-d^*_{\rm out}(r)}$. The matching converse and achievability give $d^*(r) = d^*_{\rm out}(r) = (n_t - r)(n_r - r)$ at integer $r$, with linear interpolation. $\blacksquare$