Exercises

• $\beta = 0.25$: $\lambda_\pm = (1 \pm 0.5)^2 \in [0.25, 2.25]$
• $\beta = 0.5$: $\lambda_\pm = (1 \pm 0.707)^2 \in [0.086, 2.914]$
• $\beta = 1$: $\lambda_\pm = (1 \pm 1)^2 \in [0, 4]$
• $\beta = 2$: $\lambda_\pm = (1 \pm 1.414)^2 \in [0.172, 5.828]$, plus atom at 0

The support touches zero at $\beta = 1$. For $\beta > 1$, there is an atom at $0$ with mass $1 - 1/\beta$, and the continuous part is supported away from zero.

$-0.5 m^2 + (-1 + 0.5) m + 1 = 0$, i.e., $-0.5 m^2 - 0.5 m + 1 = 0$, or $m^2 + m - 2 = 0$.

$m = (-1 \pm 3)/2 \in \{1, -2\}$. The physical root (positive since $z < \lambda_-$) is $m(-1) = 1$.

$\text{SNR} = 100$, $\beta = 1$. Equation: $\gamma(1 + \gamma/(1+\gamma)) = 100$, i.e., $\gamma(1+\gamma) + \gamma^2 = 100(1+\gamma)$, which gives $2\gamma^2 + \gamma - 100 - 100\gamma = 0$, i.e., $2\gamma^2 - 99\gamma - 100 = 0$.

$\gamma^* = (99 + \sqrt{99^2 + 800})/4 = (99 + \sqrt{10601})/4 \approx (99 + 102.96)/4 \approx 50.5$. Per-stream rate $\approx \log_2(51.5) \approx 5.69$ bits/s/Hz. SISO rate $= \log_2(101) \approx 6.66$ bits/s/Hz. The loss is about 0.97 bits per stream, but total multiplexing gain is $n_t$ streams.

$\frac{1}{N}\operatorname{tr}(\mathbf{W} + \alpha\mathbf{I})^{-1} \to -m_\mu(-\alpha)$ almost surely.

$-\alpha\beta m^2 + (-\alpha - (\beta-1)) m + 1 = 0$. Quadratic formula gives $m = \frac{-(\alpha + \beta - 1) \pm \sqrt{(\alpha + \beta - 1)^2 + 4\alpha\beta}}{-2\alpha\beta}$. Choose the root such that $m(-\alpha) > 0$ for $\alpha > 0$.

As $\alpha \to \infty$, $m \to -1/\alpha \to 0$ (consistent). As $\alpha \to 0^+$, $m \to (1 - \beta^{-1})/z$ for $\beta > 1$ reflecting the zero eigenvalue atom.

$\gamma = \text{SNR}/(1 + \beta\gamma/(1+\gamma))$ $\Leftrightarrow \beta = \frac{(\text{SNR}/\gamma - 1)(1 + \gamma)}{\gamma}$.

With $\text{SNR} = 10^{1.5} \approx 31.62$, $\gamma = 3.162$: $\text{SNR}/\gamma - 1 = 10 - 1 = 9$. $(1+\gamma)/\gamma = 4.162/3.162 \approx 1.316$. $\beta^* \approx 9 \cdot 1.316 \approx 11.85$.

So up to $\beta^* \approx 11.85$ streams per receive antenna can be supported. The asymptotic formula is very permissive; in finite $n_r$, stay well below.

If $\delta = 0.3$, $M = 1500$. Required $\rho = s/M = 100/1500 \approx 0.067$. $\rho^*(0.3) = 0.343 \gg 0.067$. Comfortably in the recovery region.

To be tight against the cliff at a given $\delta$: $\rho^*(\delta) \cdot M = s$ gives $M \approx 100/\rho^*(\delta)$. For small $\delta$ the cliff drops; e.g., at $\delta = 0.05$, $\rho^* \approx 0.13$, so $M_{\min} \approx 770$.

The system is substantially overprovisioned. Could compress more aggressively (fewer measurements) by targeting smaller $\delta$.

For the semicircle on $[-2\sigma, 2\sigma]$, $m(z) = (-z + \sqrt{z^2 - 4\sigma^2})/(2\sigma^2)$.

Setting $m = w$ and solving $\sigma^2 w^2 + zw + 1 = 0$ for $z$: $z = -1/w - \sigma^2 w$. Thus $K(w) = -(1/w + \sigma^2 w)$, taking the branch with $K(0) = -\infty$. Drop sign convention: $K(z) = 1/z + \sigma^2 z$.

$R(z) = K(z) - 1/z = \sigma^2 z$. By additivity, $R_{\mu_1 \boxplus \mu_2}(z) = (\sigma_1^2 + \sigma_2^2) z$, so the sum is Wigner with variance $\sigma_1^2 + \sigma_2^2$. Free convolution of semicircles gives a semicircle — analogous to Gaussian convolution in classical probability.

Given $\beta z m^2 + (z - \beta + 1) m + 1 = 0$, solve for $z$: $z(\beta m^2 + m) = \beta m - m - 1$, so $z = (\beta m - m - 1)/(\beta m^2 + m) = ((β-1)m - 1)/(m(βm+1))$. Hence $K(m) = z$ evaluated at $w = m$.

Replace $m$ by $z$ (by slight abuse): $K(z) = \frac{1}{z} + \frac{\beta}{1-z}$ (after algebraic simplification). So $R_{\text{MP}}(z) = \beta/(1-z)$.

By additivity, $R_{\mu_{\text{MP}} \boxplus \mu_W}(z) = \beta/(1-z) + \sigma^2 z$. The resulting measure is obtained by functional inversion of $K(z) = 1/z + \beta/(1-z) + \sigma^2 z$. No closed form in general.

$\delta = 0.2$, $\rho = 30/200 = 0.15$. $\rho^*(0.2) \approx 0.277$.

$\rho = 0.15 < \rho^*(\delta) = 0.277$: operating below the cliff. Recovery succeeds almost surely in the asymptotic limit.

$20/M = 0.38 M / 1000$, so $M^2 = 20 \cdot 1000/0.38 \approx 52{,}632$, hence $M \approx 229$.

$\delta = 0.229$, $\rho = 20/229 \approx 0.087$, $0.38\delta \approx 0.087$. Consistent. Add safety margin: use $M \approx 300$.

$S(z) = \frac{1+z}{(\alpha+z)(1+\beta z)}$.

$S(z) = (1+z)/((1+z)(1+\beta z)) = 1/(1+\beta z) = S_{\text{MP}}(z)$. ✓ When the projector is the identity, the spectrum is unchanged.

Write $F(\gamma, s) = \gamma(1 + \beta\gamma/(1+\gamma)) - s$ where $s = \text{SNR}$. Then $\partial\gamma^*/\partial s = 1 / \partial_\gamma F$.

$\partial_\gamma F = 1 + \beta\gamma/(1+\gamma) + \gamma \cdot \beta/((1+\gamma)^2)$. At $\gamma^* = 6.96$, $\beta = 0.5$: $1 + 0.5 \cdot 6.96/7.96 + 6.96 \cdot 0.5/63.36 \approx 1 + 0.437 + 0.055 = 1.492$. Derivative $\approx 1/1.492 \approx 0.67$ per dB (naturally, in linear SNR units).

A unit increase in SNR translates to about 0.67 units in SINR. The sub-unity slope reflects the cost of multi-stream interference even as the channel becomes cleaner.

$\text{SNR} = 10^{1.5} \approx 31.62$, $\beta = 0.05$. $\gamma = 31.62/(1 + 0.05\gamma/(1+\gamma))$. Iterating from $\gamma_0 = 31.62$: $\gamma_1 = 31.62/(1 + 0.05 \cdot 0.969) \approx 30.15$. One more iteration gives $\gamma^* \approx 30.17$.

$R = \log_2(1 + \gamma^*) \approx \log_2(31.17) \approx 4.96$ bits/s/Hz per user. Total: $K \cdot R \approx 49.6$ bits/s/Hz.

With $\beta = 0.05$, the multi-user interference penalty is small (loss of ~0.03 bits per user versus the single-user rate). This is the massive-MIMO favorable-propagation regime.

The map $T(\gamma) = \text{SNR}/(1 + \beta\gamma/(1+\gamma))$ is strictly decreasing in $\gamma$. Hence $T \circ T$ is increasing, and $\gamma_{2k}$ and $\gamma_{2k+1}$ are monotone.

$T(\gamma) \in [\text{SNR}/(1+\beta), \text{SNR}]$ for all $\gamma > 0$. So the iterates are bounded; combined with monotonicity, both subsequences converge.

Any limit point satisfies $\gamma = T(\gamma)$; we showed this equation has a unique positive root. Hence both subsequences converge to the same limit $\gamma^*$.

Fix $N$ (say 1000). Vary $\delta \in \{0.05, 0.1, \ldots, 0.5\}$ and $\rho \in \{0.05, 0.1, \ldots, 0.5\}$. For each $(\delta, \rho)$, generate $s = \rho M$-sparse signals, random partial Fourier matrices, run $\ell_1$-recovery, record success (small $\ell_2$-error). Repeat with $N = 2000, 4000$ to check sharpening.

If the 50% success contour matches the Gaussian DT curve as $N$ grows, the claim is supported (universality). If not, the partial Fourier ensemble has a distinct phase transition — this is what happens in practice, with partial Fourier being slightly more conservative.

$\mathbf{W} = \sum_{k=1}^M \mathbf{h}_k\mathbf{h}_k^H$ where $\mathbf{h}_k$ are the columns of $\mathbf{H}$ (i.i.d. with entries of variance $1/N$).

For the resolvent $\mathbf{Q}(z) = (\mathbf{W} - z\mathbf{I})^{-1}$, use $\mathbf{Q} = -z^{-1}\mathbf{I} + z^{-1}\mathbf{W}\mathbf{Q}$. Take normalized trace: $m_N(z) = -z^{-1} + z^{-1} \cdot \frac{1}{N}\operatorname{tr}(\mathbf{W}\mathbf{Q})$.

For each column $\mathbf{h}_k$, $\mathbf{h}_k^H\mathbf{Q}\mathbf{h}_k = \frac{\mathbf{h}_k^H\mathbf{Q}_{(k)}\mathbf{h}_k}{1 + \mathbf{h}_k^H\mathbf{Q}_{(k)}\mathbf{h}_k}$, where $\mathbf{Q}_{(k)}$ leaves out the rank-one piece $\mathbf{h}_k\mathbf{h}_k^H$.

$\mathbf{h}_k^H\mathbf{Q}_{(k)}\mathbf{h}_k - \frac{1}{N}\operatorname{tr}(\mathbf{Q}_{(k)}) \xrightarrow{a.s.} 0$, and by rank-one stability $\frac{1}{N}\operatorname{tr}(\mathbf{Q}_{(k)}) \approx m(z)$. Summing over $k$ (there are $M$ terms) and equating $\frac{1}{N}\operatorname{tr}(\mathbf{W}\mathbf{Q}) \to \beta \cdot \frac{m(z)}{1 + m(z)}$, substitute back: $m(z) = -z^{-1} + z^{-1} \cdot \frac{\beta m(z)}{1 + m(z)}$. Clearing fractions: $\beta z m(z)^2 + (z - (\beta-1)) m(z) + 1 = 0$. ✓

$R_{1 \boxplus 2}(z) = 2z + 1/(1-z)$.

$R_1 = 2z$ is a scaled Wigner semicircle; $R_2 = 1/(1-z)$ is Marchenko-Pastur with $\beta = 1$. The sum has support determined by inverting $K(z) = 1/z + 2z + 1/(1-z)$.

$\lambda_\pm = (1 \pm \sqrt{0.5})^2 = (1 \pm 0.707)^2 \in \{0.086, 2.914\}$.

Measured $[0.09, 2.91]$ matches theoretical $[0.086, 2.914]$ to within 0.5%. Very consistent with MP predictions. (Finite-$N$ fluctuations are typically in the third decimal at $N = 64$.)

The event "recovery fails" can be rewritten as the count of "bad faces" of a random polytope exceeding a threshold. The count is approximately a sum of $N$ weakly correlated indicators, whose variance is $O(N)$. The standard deviation is $O(\sqrt{N})$, while the mean jumps by $O(N)$ across the phase transition; hence the relative width is $O(1/\sqrt{N}) \to 0$.

At $N = 100$, the 10%-90% transition width in $\rho$ is roughly $1/\sqrt{100} = 0.1$ — quite visible. At $N = 10000$, the width shrinks to $\approx 0.01$. Finite-$N$ curves therefore should not be fit as sharp thresholds; fit as sigmoids with a width parameter.

Let $\{\lambda_i\}$ be the eigenvalues of $\boldsymbol{\Theta}$ (Tx correlation). Then $\frac{1}{n_r}\operatorname{tr}(\mathbf{H}\mathbf{H}^{H} + \sigma^2\mathbf{I})^{-1} \to m(-\sigma^2)$ where $m$ satisfies a system of equations.

$m(z) = -\frac{1}{z}\left(1 - \beta \cdot \frac{1}{n_t}\sum_i \frac{\lambda_i}{1 + \lambda_i \tilde m(z)}\right)$ and $\tilde m(z)$ satisfies a dual equation involving $m(z)$. In the $n_t \to \infty$ limit, the sum becomes an integral against the LSD of $\boldsymbol{\Theta}$.

The output SINR for a stream with channel energy $\lambda_i$ becomes $\gamma_i = \lambda_i \tilde m(-\sigma^2)$. System design now couples the antenna correlation structure to the achievable rate.