Proof. We present for \(\beta = 1\). By linear algebra, we can choose an orthogonal matrix \(U_1\), and transform the matrix \(M\) to \(U_1MU_1^\top \), whose the first row is \[ (x_{11}, \sqrt {\sum _{k=2}^n x_{ik}^2}, 0,...,0). \] Note that the eigenvalues do not change after the transformation.

We can inductively do the same operation on the rest sub-matrix. □

Proof. By SLLN, we have \begin{align*} &\lim _{N\to \infty } \frac {\chi _{\beta (N-1)}}{\sqrt {2N}} = \sqrt {\frac {\beta }{2}},\\ &\lim _{N\to \infty } \frac {\chi _{\beta (N-\alpha N)}}{\sqrt {2N}} = \sqrt {\frac {\beta (1-\alpha )}{2}}. \end{align*}

So by the definition of the trace, we have \begin{align*} {\operatorname{tr}} \ \left (\frac {T_\beta }{\sqrt {N}}\right )^k &= \sum _{m=1}^N \sum _{\substack {\text {k-step paths} \\ \text {from } m \text { to } m}} \left (\ \textcolor {gray}{ \text {some} \ \frac {\chi _{\beta (N-i)}}{\sqrt {2N}} } \right ) \left (\ \textcolor {gray}{ \text {some} \ \frac {\mathcal {N}(0,1)}{\sqrt {N}} }\right )\\ &=\sum _{m=1}^N \left ( \# {\text {k-step paths from } m \text { to } m} \right ) \sqrt {\frac {\beta }{2}\left ( \frac {N-m}{N} \right )^k} + o(N). \end{align*}

The second equation follows from that if there exists "some \(\frac {\mathcal {N}(0,1)}{\sqrt {N}}\)", the term vanishes. When \(2 \nmid k\), \(\# {\text {k-step paths from } m \text { to } m}\) is zero. When \(2\mid k\), \(\# {\text {k-step paths from } m \text { to } m} = \binom {k}{\frac {k}{2}}\). Approximate the summation by integration and we have \[ \frac {1}{N} {\operatorname{tr}} \ \left (\frac {T_\beta }{\sqrt {N}}\right )^k \to \binom {k}{\frac {k}{2}} \left (\frac {\beta }{2}\right )^{\frac {k}{2}} \int _0^1 x^{\frac {k}{2}}{\mathop{}\!\mathrm{d}} x = {\operatorname{Cat}} _{\frac {k}{2}} \left (\frac {\beta }{2}\right )^{\frac {k}{2}} \] □

Proof. (Method I) We just calculate \[ m_k = \int \frac {1}{\sqrt {2\pi }} \sqrt {4-x^2}x^k {\mathop{}\!\mathrm{d}} x. \] Let \(x = 2\cos \theta \), and by inductive calculation, we derive the results. □

Proof. (Method II) Introduce generating function: \[ G(z) = \sum _{k=0}^\infty m_k z^{-k-1}, \] \(m_0 =1\). By classical results, we have \[ \sum _{n=0}^\infty {\operatorname{Cat}} _n x^n = \frac {1-\sqrt {1-4x}}{2x} =: C(x). \] Now we can derive the expression of \(G(z)\) using \(C(x)\). \[ G(z) = \sum _{k=0}^\infty m_k z^{-k-1} = \sum _{l=0}^\infty {\operatorname{Cat}} _l z^{-2l-1} = \frac {z-\sqrt {z^2-4}}{2}. \] Next we introduce two propositions.

Return to the Theorem 8.10, by proposition 8.12 \[ \mu (x_0) = -\frac {1}{\pi } \lim _{y\to 0^+} \Im \left ( \frac {1}{2}\left [(x+iy) - \sqrt {(x+iy)^2 - 4)}\right ] \right ) = \begin {cases} 0 &, \quad |x|>2;\\ \frac {\sqrt {4-x^2}}{2\pi } &,\quad |x|\le 2. \end {cases} \] □

Proof. We prove the result in four steps.

Step 1 The results hold for \(f(x) = x^k\).

Step 2 The results hold for any polynomials.

Step 3 Take \(L>2\), the results hold for \(f(x) = \mathbb {1}_{|x|>L} x^k\). We have \begin{align*} \left |\int f(x)\mu _N({\mathop{}\!\mathrm{d}} x)\right | &= \left |\int x^k\mu _N({\mathop{}\!\mathrm{d}} x)\right |\\ &\le L^{-2m} \int _{|x|\ge L} x^{2k+2m}\mu _N({\mathop{}\!\mathrm{d}} x)\\ &\to L^{-2m} \int x^{2k+2m} \mu ({\mathop{}\!\mathrm{d}} x)\\ &\le L^{-2m} 2^{2k+2m} = 2^{2k} \left (\frac {2}{L}\right )^m \to 0. \end{align*}

Step 4 By step 3, we can restrict the support set of \(f\) on a compact set, i.e. \([-4, 4]\). Apply the Weierstrass theorem and we get the proof. □

Proof. We calculate the moments for \(\rho ^N\) and \(\mu _{\mathrm {MP}}\) separately

Step 1 We claim that \[ \lim _{N,S\to \infty } \frac {1}{NS^k}{\operatorname{tr}} [(XX^*)^k] = \beta ^k \sum _{r=0}^{k-1} \frac {\gamma ^{2r}}{r+1} \binom {k}{r} \binom {k-1}{r}. \]

Notice that as \(S\to \infty \) \[ \frac {\chi _{aS}}{\sqrt {S}} \to \sqrt {a}, \] we have \begin{align*} \operatorname {LHS} &= \lim _{S,N\to \infty }\frac {1}{NS^k} \sum _{m=1}^N\sum _{\substack {\text {k-step paths} \\ \text {from } m \text { to } m}} (\text {diagonal entries})(\text {sub-diagonal entries})\\ &= \lim _{S,N\to \infty } \frac {1}{NS^k} \sum _{m=1}^N \sum _{i=0}^{[k/2]} \left ( \chi _{\beta (N-m)}\chi _{\beta (S-m+1)} \right )^{2i}\chi _{S+N-2m-2}^{2(k-2i)} \binom {k}{2i}\binom {2i}{i} \end{align*}

By replacing \(\chi \) by its limit, we have \[ \frac {\chi _{\beta (N-m)}\chi _{\beta (S-m+1)}}{S} \to \beta \sqrt {(\gamma ^2-\frac {m}{S})(1-\frac {m}{S})} \] \[ \frac {\chi ^2_{\beta (S+N-2m+2)}}{S} \to \beta (1+\gamma ^2 - 2\frac {m}{S}) \] Then we approximate the sum by integral \begin{align*} \operatorname {LHS} &= \frac {\beta ^k}{\gamma ^2} \sum _{i=0}^{[k/2]} \binom {k}{2i}\binom {2i}{i} \int _0^{\gamma ^2} (\gamma ^2-t)^i(1-t)^i (1+\gamma ^2-2t)^{k-2i}{\mathop{}\!\mathrm{d}} t\\ &=\frac {\beta ^k}{\gamma ^2} \sum _{i=0}^{[k/2]} \binom {k}{2i}\binom {2i}{i} \int _0^{\gamma ^2} u^i(1-\gamma ^2+u)^i (1-\gamma ^2+2u)^{k-2i}{\mathop{}\!\mathrm{d}} u \end{align*}

We claim that \[ \frac {1}{\gamma ^2}\sum _{i=0}^{[k/2]} \binom {k}{2i}\binom {2i}{i} \int _0^{\gamma ^2} u^i(1-\gamma ^2+u)^i (1-\gamma ^2+2u)^{k-2i}{\mathop{}\!\mathrm{d}} u = \sum _{r=0}^{k-1} \frac {\gamma ^{2r}}{r+1} \binom {k}{r} \binom {k-1}{r}. \] and we obtain the proof.

The proof of the claim: Denote the left hand side as A. First, we observe that \begin{equation} \operatorname {A} = \text {The sum of coefficients of all the }x^iy^i \text { in} \int _{0}^{\gamma ^2} [xu+y(1-\gamma ^2 +u)+(1-\gamma ^2+2u)]^k{\mathop{}\!\mathrm{d}} u \end{equation} Denote B as the integral of (25) \begin{align*} B &= \frac {1}{\gamma ^2}\frac {[(x+1)\gamma ^2+(y+1)]^{k+1} - [(y+1)(1-\gamma ^2)]^{k+1}}{(x+y+2)(k+1)}\\ &=\frac {1}{k+1}\sum _{l=0}^k [(y+1)(1-\gamma ^2)]^l [(x+1)\gamma ^2+y+1]^{k-l}. \end{align*}

To calculate the coefficients, we can let \(x = \frac {1}{y}\) and calculate the constant coefficient. \begin{align*} \operatorname {B} &= \frac {1}{k+1} \sum _{l=0}^k [(y+1) (1-\gamma ^2)]^l \left (\left (\frac {1}{y}+1\right )\gamma ^2 + y+1\right )^{k-l}\\ &=\frac {(y+1)^k}{k+1} \sum _{l=0}^k(1-\gamma ^2)^l(\frac {1}{y}\gamma ^2+1)^{k-l}\\ &=\frac {(y+1)^k}{k+1} \frac {(\frac {1}{y}\gamma ^2+1)^{k+1} - (1-\gamma ^2)^{k+1}} {\frac {1}{y}\gamma ^2 + \gamma ^2}\\ &=\frac {y(y+1)^{k-1}}{(k+1)\gamma ^2} \left (\left (\frac {1}{y}\gamma ^2+1\right )^{k+1} - (1-\gamma ^2)^{k+1}\right ) \end{align*}

Then the constant coefficient is \[ \frac {1}{(k+1)\gamma ^2} \sum _{r=0}^{k-1} \binom {k-1}{r} \binom {k+1}{r+1} \gamma ^{2r+2} = \sum _{r=0}^{k-1} \frac {\gamma ^{2r}}{r+1} \binom {k}{r} \binom {k-1}{r}. \]

Step 2 We claim that \(\lambda _{+} = \beta (1+\gamma )^2\) and \(\lambda _{-} = \beta (1-\gamma )^2\).

Define the density of the limit distribution \[ \mu (x) = \frac {1}{2\pi \gamma ^2}\frac {\sqrt {(\lambda _{+}-x)(x-\lambda _{-})}}{x} \mathbb {1}_{[\lambda _{-}, \lambda _{+}]} \] and \(\mu _N = \rho ^N\).

The calculation below is inspired by Zhidong Bai’s book. First, we calculate the moment of \(\mu (x)\), for \(m\ge 1\), \begin{align*} {\mathbb{E}} _{\mu } x^m &= \int x^m \mu (x){\mathop{}\!\mathrm{d}} x\\ &= \frac {1}{2\pi \gamma ^2}\int x^{m-1} \sqrt {(\lambda _{+}-x)(x-\lambda _{-})} {\mathop{}\!\mathrm{d}} x\\ &= \frac {1}{2\pi \gamma ^2}(2\gamma )^2\int _{-1}^1 (1+\gamma ^2+2\gamma u)^{m-1}\sqrt {1-u^2}{\mathop{}\!\mathrm{d}} u \\ &(\text {by setting }x =\beta (1+\gamma ^2+2\gamma u))\\ &= \beta ^m\frac {1}{\pi }\sum _{l=0}^{[(m-1)/2]} \binom {m-1}{2l}(1+\gamma ^2)^{m-1-2l} (4\gamma ^2)^l \int _{-1}^1 u^{2l}\sqrt {1-u^2}{\mathop{}\!\mathrm{d}} u\\ &= \beta ^m \sum _{l=0}^{[(m-1)/2]} \binom {m-1}{2l} \frac {1}{l+1}\binom {2l}{l} (1+\gamma ^2)^{m-1-2l} \gamma ^{2l}\\ &= \beta ^m\sum _{l=0}^{[(m-1)/2]}\sum _{s=0}^{m-1-l} \frac {(m-1)!}{l!(l+1)!s!(m-1-2l-s)!}\gamma ^{2(l+s)}\\ &=\beta ^m \frac {1}{m}\sum _{r=0}^{m-1} \gamma ^{2r} \binom {m}{r} \sum _{l=0}^{r\wedge (m-1-r)}\binom {r}{l} \binom {m-r}{m-r-l-1}\\ &=\beta ^m \frac {1}{m}\sum _{r=0}^{m-1}\binom {m}{r}\binom {m}{r+1}\gamma ^{2r}\\ &=\beta ^m \sum _{r=0}^{m-1}\frac {1}{r+1}\binom {m}{r}\binom {m-1}{r}\gamma ^{2r}. \end{align*}

Next, we notice that \[ {\mathbb{E}} _{\mu _N} x^m =\frac {1}{N} \sum _{i=1}^N \lambda _i^N = \frac {1}{N}{\operatorname{tr}} \left [\left (\frac {XX^*}{S}\right )^m\right ] \] and \[ \lim _{N\to \infty } \frac {1}{N}{\operatorname{tr}} \left [\left (\frac {XX^*}{S}\right )^m\right ] = \beta ^m \sum _{r=0}^{m-1}\frac {1}{r+1}\binom {m}{r}\binom {m-1}{r}\gamma ^{2r}. \] So we have \[ \lim _{N\to \infty }{\mathbb{E}} _{\mu _N} x^m = {\mathbb{E}} _{\mu } x^m. \] Similar to the proof of semi-circle law, we have \(\rho ^N\) converges (weakly, in probability) to the Marchenko-Pastur law. □