Recursion for the Smallest Eigenvalue Density of $\beta$-Wishart–Laguerre Ensemble

Abstract

The statistics of the smallest eigenvalue of Wishart–Laguerre ensemble is important from several perspectives. The smallest eigenvalue density is typically expressible in terms of determinants or Pfaffians. These results are of utmost significance in understanding the spectral behavior of Wishart–Laguerre ensembles and, among other things, unveil the underlying universality aspects in the asymptotic limits. However, obtaining exact and explicit expressions by expanding determinants or Pfaffians becomes impractical if large dimension matrices are involved. For the real matrices ($\beta =1$) Edelman has provided an efficient recurrence scheme to work out exact and explicit results for the smallest eigenvalue density which does not involve determinants or matrices. Very recently, an analogous recurrence scheme has been obtained for the complex matrices ($\beta =2$). In the present work we extend this to $\beta$-Wishart–Laguerre ensembles for the case when exponent $\alpha$ in the associated Laguerre weight function, $\lambda ^\alpha e^{-\beta \lambda /2}$, is a non-negative integer, while $\beta$ is positive real. This also gives access to the smallest eigenvalue density of fixed trace $\beta$-Wishart–Laguerre ensemble, as well as moments for both cases. Moreover, comparison with earlier results for the smallest eigenvalue density in terms of certain hypergeometric function of matrix argument results in an effective way of evaluating these explicitly. Exact evaluations for large values of n (the matrix dimension) and $\alpha$ also enable us to compare with Tracy–Widom density and large deviation results of Katzav and Castillo. We also use our result to obtain the density of the largest of the proper delay times which are eigenvalues of the Wigner–Smith matrix and are relevant to the problem of quantum chaotic scattering.

Appendices

Appendix A: Proof of Recurrence Scheme

The proof of the recurrence scheme for general $\beta$ is similar to the one given by Edelman for $\beta =1$ case [22, 23]. It was also extended recently for $\beta =2$ case [12]. The smallest eigenvalue density can be obtained from the joint probability density of eigenvalues as

$$\begin{aligned} f(x)=n\int _x^\infty d\lambda _2\cdots \int _x^\infty d\lambda _n \,P(x,\lambda _2,\ldots ,\lambda _n). \end{aligned}$$

(A.1)

We now consider $\lambda _i\rightarrow \lambda _i+x$, followed by shift of indices of the integration variables as $\lambda _i\rightarrow \lambda _{i-1}$. This gives

$$\begin{aligned} f(x)= & {} nC_{n,\alpha ,\beta }\, x^\alpha e^{-\beta n x/2} \int _0^\infty d\lambda _1\cdots \int _0^\infty d\lambda _{n-1}\prod _{1\le k<j\le n-1}|\lambda _j-\lambda _k|^\beta \nonumber \\&\times \prod _{i=1}^{n-1}\lambda _i^\beta (\lambda _i+x)^\alpha e^{-\beta \lambda _i/2}. \end{aligned}$$

(A.2)

We now introduce the measure $d\Omega _i=\lambda _i^\beta \,e^{-\beta \lambda _i/2}\,d\lambda _i$ and write the above as

$$\begin{aligned} f(x)=nC_{n,\alpha ,\beta }\, x^\alpha e^{-\beta n x/2} \int _0^\infty d\Omega _1\cdots \int _0^\infty d\Omega _{n-1}|\Delta _{n-1}(\{\lambda \})|^\beta \prod _{i=1}^{n-1}(\lambda _i+x)^\alpha .\nonumber \\ \end{aligned}$$

(A.3)

Following [22, 23], we now define

$$\begin{aligned} I_{i,j}^\alpha =\int _0^\infty d\Omega _1\cdots \int _0^\infty d\Omega _{n-1}\, |\Delta _{n-1}(\{\lambda \})|^\beta u(x), \end{aligned}$$

(A.4)

where

$$\begin{aligned} u(x)\equiv & {} \underbrace{(\lambda _1+x)^\alpha \cdots (\lambda _i+x)^\alpha }_{i \mathrm {~ terms~}}\underbrace{(\lambda _{i+1}+x)^{\alpha -1}\cdots (\lambda _{i+j}+x)^{\alpha -1}}_{j \mathrm { ~terms~}}\nonumber \\&\quad \times \underbrace{(\lambda _{i+j+1}+x)^{\alpha -2}\cdots (\lambda _n+x)^{\alpha -2}}_{n-i-j-1 \mathrm {~ terms~}}. \end{aligned}$$

(A.5)

We also consider the operator

$$\begin{aligned} I_{i,j}^\alpha [v]=\int _0^\infty d\Omega _1\cdots \int _0^\infty d\Omega _{n-1} \,|\Delta _{n-1}(\{\lambda \})|^\beta u(x)\,v. \end{aligned}$$

(A.6)

Using the above, the smallest eigenvalue density can be written using (A.4) as

$$\begin{aligned} f(x)=nC_{n,\alpha ,\beta } x^\alpha e^{-\beta nx/2}I_{n-1,0}^\alpha . \end{aligned}$$

(A.7)

Moreover, Lemma 4.2 of [22] (or, Lemma 4.1 of [23]) holds:

$$\begin{aligned} I_{i,j}^\alpha [\lambda _k]=\left\{ \begin{array}{cc} I_{i+1,j-1}^\alpha -x I_{i,j}^\alpha &{} \hbox { if }i<k\le i+j,\\ I_{i,j+1}^\alpha -x I_{i,j}^\alpha &{}\hbox { if }i+j< k< n. \end{array}\right. \end{aligned}$$

(A.8)

The above result is obtained by writing $\lambda _k$ as $(\lambda _k+x)-x$ and then using the operator defined in (A.6). Now, if the terms $(\lambda _k+x)$ and $(\lambda _l+x)$ share the same exponent in the integrals (i.e., both k and l fall within one of the closed intervals $[1,i],[i+1,i+j]$, or $[i+j+1,n-1]$), then

$$\begin{aligned} I_{i,j}^\alpha \left[ \frac{\lambda _k\lambda _l}{\lambda _k-\lambda _l}\right]= & {} 0, \end{aligned}$$

(A.9)

$$\begin{aligned} I_{i,j}^\alpha \left[ \frac{\lambda _k}{\lambda _k-\lambda _l}\right]= & {} \frac{1}{2}I_{i,j}^\alpha , \end{aligned}$$

(A.10)

$$\begin{aligned} I_{i,j}^\alpha \left[ \frac{\lambda _k^2}{\lambda _k-\lambda _l}\right]= & {} I_{i,j}^{\alpha }[\lambda _k]. \end{aligned}$$

(A.11)

Equation (A.9) is a consequence of the asymmetry in $\lambda _k$ and $\lambda _l$, while  (A.10) is obtained using the identity $\lambda _k/(\lambda _k-\lambda _l)+\lambda _l/(\lambda _l-\lambda _k)=1$ and employing symmetry. Equation (A.11) follows with the aid of the identity $\lambda _k^2/(\lambda _k-\lambda _l)=\lambda _k+\lambda _k\lambda _l/(\lambda _k-\lambda _l)$ and (A.9).

The generalization of the Lemma 4.3 of [22] (or Lemma 4.2 [23]) happens to be

$$\begin{aligned} I_{i,j}^\alpha= & {} \left( x+\frac{2\alpha }{\beta }+j+2k+2\right) \! I_{i-1,j+1}^\alpha \nonumber \\&-x\left( k+\frac{2(\alpha -1)}{\beta }\right) \! I_{i-1,j}^\alpha +(i-1)x I_{i-2,j+2}^\alpha \end{aligned}$$

(A.12)

$$\begin{aligned} I_{0,j}^\alpha= & {} I_{j,n-j-1}^{\alpha -1}, \end{aligned}$$

(A.13)

with $k=n-i-j-1$. The definition (A.4) readily yields the second equation above, (A.13). The first equation of this set, (A.12), is derived using

$$\begin{aligned} I_{i,j}^\alpha =x I_{i-1,j+1}^\alpha +I_{i-1,j+1}^\alpha [\lambda _i], \end{aligned}$$

(A.14)

which is a consequence of (A.8). For the case of general $\beta$, calculation of $I_{i-1,j+1}^\alpha [\lambda _i]$ involves the follwing result:

$$\begin{aligned}&\int _0^\infty (\lambda _i+x)^{\alpha -1}\prod _{i<l}|\lambda _l-\lambda _i|^\beta \,\lambda _i^{\beta +1}\,e^{-\beta \lambda _i/2}\,d\lambda _i\nonumber \\&\quad =\frac{2}{\beta }\int _0^\infty \frac{d}{d\lambda _i} \bigg [(\lambda _i+x)^{\alpha -1}\prod _{i<l}|\lambda _l-\lambda _i|^\beta \,\lambda _i^{\beta +1}\bigg ]e^{-\beta \lambda _i/2}\,d\lambda _i. \end{aligned}$$

(A.15)

Next, using the result

$$\begin{aligned} \frac{dI_{i-1,j+1}^\alpha }{dx}=(i-1)\alpha I_{i-2,j+2}^\alpha +(j+1)(\alpha -1)I_{i-1,j}^\alpha \end{aligned}$$

(A.16)

for $i+j=n-1$, we obtain (see Lemma 4.4, [22], or Lemma 4.3 [23])

$$\begin{aligned} I_{i,j}^\alpha= & {} \left( x+\frac{2\alpha }{\beta }+j+2\right) I_{i-1,j+1}^\alpha -\frac{2x}{\beta (j+1)}\frac{d}{dx}I_{i-1,j+1}^\alpha \nonumber \\&+x (i-1)\left( 1+\frac{2\alpha }{\beta (j+1)}\right) I_{i-2,j+2}^\alpha . \end{aligned}$$

(A.17)

For $j=n-i-1$ this yields

$$\begin{aligned} I_{i,n-i-1}^\alpha= & {} \left( x+\frac{2\alpha }{\beta }+n-i+1\right) I_{i-1,n-i}^\alpha -\frac{2x}{\beta (n-i)}\frac{d}{dx}I_{i-1,n-i}^\alpha \nonumber \\&+x (i-1)\left( 1+\frac{2\alpha }{\beta (n-i)}\right) I_{i-2,n-i+1}^\alpha . \end{aligned}$$

(A.18)

We now consider $I_{0,n-1}^\alpha$, which is same as $I_{n-1,0}^{\alpha -1}$ in view of (A.13), and use (A.18) repeatedly for $i=1$ to $n-1$ to arrive at $I_{n-1,0}^\alpha$. Therefore, we note that, starting from $I_{n-1,0}^{\alpha -1}$ we can arrive at $I_{n-1,0}^\alpha$, which is the term needed to obtain the smallest eigenvalue density expression (A.7) explicitly. This is essentially what we implement in the recursion involving $S_i:= I_{i,n-i-1}^\alpha /I_{n-1,0}^0$ for $g_{n,\alpha ,\beta }(x)$ in (10). We also observe that $I_{n-1,0}^0=1/C_{n-1,\beta ,\beta }$, which gives the constant $c_{n,\alpha ,\beta }$ of (10) as $nC_{n,\alpha ,\beta }/C_{n-1,\beta ,\beta }$.

Appendix B: Proof of Equation (15)

For a non-negative integer $\alpha$, using the Binomial theorem, we have

$$\begin{aligned} (\lambda _j+x)^\alpha =\sum _{k=0}^{\alpha }\left( {\begin{array}{c}\alpha \\ k\end{array}}\right) \lambda _j^{\alpha -k} x^k. \end{aligned}$$

(B.1)

We use this within the integral in (12). Now, since (10) already contain a factor $x^\alpha$, the coefficient of $x^r$ in this equation and hence in (13) is decided by the coefficient of $x^{r-\alpha }$ in $\prod _{j=1}^{n-1}\sum _{k=0}^{\alpha }\left( {\begin{array}{c}\alpha \\ k\end{array}}\right) \lambda _j^{\alpha -k} x^k$ which, when expanded, appears as

$$\begin{aligned}&\left( \left( {\begin{array}{c}\alpha \\ 0\end{array}}\right) \lambda _1^\alpha +\left( {\begin{array}{c}\alpha \\ 1\end{array}}\right) \lambda _1^{\alpha -1}x+\left( {\begin{array}{c}\alpha \\ 2\end{array}}\right) \lambda _1^{\alpha -2}x^2+\cdots +\left( {\begin{array}{c}\alpha \\ \alpha \end{array}}\right) x^\alpha \right) \\ \times&\left( \left( {\begin{array}{c}\alpha \\ 0\end{array}}\right) \lambda _2^\alpha +\left( {\begin{array}{c}\alpha \\ 1\end{array}}\right) \lambda _2^{\alpha -1}x+\left( {\begin{array}{c}\alpha \\ 2\end{array}}\right) \lambda _2^{\alpha -2}x^2+\cdots +\left( {\begin{array}{c}\alpha \\ \alpha \end{array}}\right) x^\alpha \right) \\&~~~~~~~~~~~~~~~~~~~~~~~~\vdots ~~~~~~~~~~~~~~~~~~~~~~~~\vdots \\&\times \left( \left( {\begin{array}{c}\alpha \\ 0\end{array}}\right) \lambda _{n-1}^\alpha +\left( {\begin{array}{c}\alpha \\ 1\end{array}}\right) \lambda _{n-1}^{\alpha -1}x+\left( {\begin{array}{c}\alpha \\ 2\end{array}}\right) \lambda _{n-1}^{\alpha -2}x^2+\cdots +\left( {\begin{array}{c}\alpha \\ \alpha \end{array}}\right) x^\alpha \right) . \end{aligned}$$

Clearly, the minimum and maximum powers of x possible in the above product are 0 and $(n-1)\alpha$, respectively. Therefore, $r-\alpha$ varies from 0 to $(n-1)\alpha$, and any particular value assumed by it in this range has to be the resultant of the powers of x in the factors $(\left( {\begin{array}{c}\alpha \\ 0\end{array}}\right) \lambda _j^\alpha +\cdots +\left( {\begin{array}{c}\alpha \\ \alpha \end{array}}\right) x^\alpha )$; $j=1,\ldots ,n-1$. As a result, we look for the partitions of $r-\alpha$ using exactly $n-1$ non-negative integers which are less than or equal to $\alpha$, since the power of x varies from 0 to $\alpha$. Moreover, the different orderings of the partition constituents correspond to the exchange of different $\lambda$’s. Since the multidimensional-integral in (12) is symmetric under the exchange of eigenvalues, we may focus on a particular ordering and multiply the resultant integral by the suitable combinatorial factor, which for a partition indexed by say $\varphi$, out of 1 to L in (14), can be seen to be $(n-1)!/\prod _{k=1}^{l_\varphi }s_{\varphi ,k}!$. Furthermore, this factor appears with

$$\begin{aligned}&\left( {\begin{array}{c}\alpha \\ p_{\varphi ,k}\end{array}}\right) ^{s_{\varphi ,1}}\left( {\begin{array}{c}\alpha \\ p_{\varphi ,2}\end{array}}\right) ^{s_{\varphi ,2}}\cdots \left( {\begin{array}{c}\alpha \\ p_{\varphi ,l_\varphi }\end{array}}\right) ^{s_{\varphi ,l_\varphi }} x^{s_{\varphi ,1}p_{\varphi ,1}+s_{\varphi ,2}p_{\varphi ,2}+\cdots +s_{\varphi ,l_\varphi }p_{\varphi ,l_\varphi }}\\&\quad \times \lambda _1^{\alpha -\mathrm {p}_{j,1}}\lambda _2^{\alpha -\mathrm {p}_{j,1}}\ldots \lambda _{s_{j,1}}^{\alpha -\mathrm {p}_{j,1}} \lambda _{s_{j,1}+1}^{\alpha -\mathrm {p}_{j,2}}\ldots \lambda _{s_{j,2}}^{\alpha -\mathrm {p}_{j,2}} \ldots \lambda _{n-1}^{\alpha -\mathrm {p}_{l_j}}. \end{aligned}$$

The final result (15) therefore follows by summing the above over L distinct partitions of $r-\alpha$ and applying the multidimensional-integral appearing in (12).

We consider an example to enunciate the above. Suppose $n=5, \alpha =3, \beta =2$, and we are interested in finding the coefficient of $x^r$ with $r=7$. Then we look for the partition of $7-3=4$ and find the unique partitions $\{3,1,0,0\}, \{2,2,0,0\}, \{2,1,1,0\}, \{1,1,1,1\}$ up to ordering. Therefore, we have $L=4$, and the following parameters:

$$\begin{aligned}&l_1=3: p_{1,1}=3, s_{1,1}=1; p_{1,2}=1, s_{1,2}=1; p_{1,3}=0, s_{1,3}=2,\\&l_2=2: p_{2,1}=2, s_{2,1}=2; p_{2,2}=0, s_{2,2}=2,\\&l_3=3: p_{3,1}=2, s_{3,1}=1; p_{3,2}=1, s_{1,2}=2; p_{3,3}=0, s_{3,3}=1,\\&l_4=1: p_{4,1}=1, s_{4,1}=4. \end{aligned}$$

Equation (15) then tells that the coefficient of $x^7$ would be

$$\begin{aligned} \kappa _7= & {} 4!\,c_{5,3,2}\Bigg [\frac{\left( {\begin{array}{c}3\\ 3\end{array}}\right) ^1}{1!}\frac{\left( {\begin{array}{c}3\\ 1\end{array}}\right) ^1}{1!}\frac{\left( {\begin{array}{c}3\\ 0\end{array}}\right) ^2}{2!}\langle \lambda _1^0\lambda _2^2\lambda _3^3\lambda _4^3 \rangle _\Lambda +\frac{\left( {\begin{array}{c}3\\ 2\end{array}}\right) ^2}{2!}\frac{\left( {\begin{array}{c}3\\ 0\end{array}}\right) ^2}{2!}\langle \lambda _1^1\lambda _2^1\lambda _3^3\lambda _4^3 \rangle _\Lambda \\&+\frac{\left( {\begin{array}{c}3\\ 2\end{array}}\right) ^1}{1!}\frac{\left( {\begin{array}{c}3\\ 1\end{array}}\right) ^2}{2!}\frac{\left( {\begin{array}{c}3\\ 0\end{array}}\right) ^1}{1!}\langle \lambda _1^1\lambda _2^2\lambda _3^2\lambda _4^3 \rangle _\Lambda +\frac{\left( {\begin{array}{c}3\\ 1\end{array}}\right) ^4}{4!}\langle \lambda _1^2\lambda _2^2\lambda _3^2\lambda _4^2 \rangle _\Lambda \Bigg ]. \end{aligned}$$

We find that $\langle \lambda _1^0\lambda _2^2\lambda _3^3\lambda _4^3 \rangle _\Lambda =3175200$, $\langle \lambda _1^1\lambda _2^1\lambda _3^3\lambda _4^3 \rangle _\Lambda =1360800$, $\langle \lambda _1^1\lambda _2^2\lambda _3^2\lambda _4^3 \rangle _\Lambda =680400$, $\langle \lambda _1^2\lambda _2^2\lambda _3^2\lambda _4^2 \rangle _\Lambda =302400,$ which gives $\kappa _7=159/16$. This agrees with the coefficient of $x^7$ extracted after applying the recursion, as it should.

Appendix C: Mathematica Codes

The following code can be implemented in Mathematica [72] to obtain exact expressions for the smallest eigenvalue density for the unrestricted $\beta$-Wishart–Laguerre ensemble:

For generating the smallest eigenvalue density for unit-trace $\beta$-Wishart–Laguerre ensemble, the following code can be used along with the above.

Subsequently, the following codes can be used to obtain the moments: