In this section, we present two exact ways to obtain the PDF and outage probability of mutual information. Moreover, the Gaussian approximation is also presented, since it leads to reasonable results and is more straightforward than the exact solutions.

### Exact results based directly on JPDF

We first calculate the PDF of mutual information. For this, we notice that one of the integrals in (11) can be easily performed because of the Dirac-delta function. To see this, we write (11) as

$$\begin{aligned} &p(\mathcal{I}_{A})=\int_{0}^{\infty} \cdots\int_{0}^{\infty} \delta\left(\log_{2}\frac{2^{\mathcal{I}_{A}}}{\prod_{j=1}^{n} (1+\lambda_{j})}\right)\\ &\qquad\qquad\qquad\qquad\times P(\lambda_{1},\ldots,\lambda_{n})\ d\lambda_{1} \cdots d\lambda_{n}\\ &=\ln 2\int_{0}^{\infty} \cdots\int_{0}^{\infty} \delta\,\left(\lambda_{1}+1-\frac{2^{\mathcal{I}_{A}}}{\prod_{j=2}^{n} \left(1+\lambda_{j}\right)}\right)\\ &\qquad\qquad\qquad\quad\times \frac{2^{\mathcal{I}_{A}}}{\prod_{j=2}^{n} \left(1+\lambda_{j}\right)} P(\lambda_{1},\ldots,\lambda_{n})\ d\lambda_{1} \cdots d\lambda_{n}. \end{aligned} $$

In the second line, we used the property *δ*(*F*(*z*))=*δ*(*z*−*z*
_{0})/|*F*
^{′}(*z*
_{0})| for *z*=*λ*
_{1}, with *z*
_{0} being the root of *F*(*z*)=0. The *λ*
_{1} integral can now be performed using the result \(\int _{0}^{\infty } \delta (z-z_{0})G(z)\ dz=G(z_{0})\Theta (z_{0})\) to yield

$$ \begin{aligned} p(\mathcal{I}_{A})&=\ln2\int_{0}^{\infty} \cdots\int_{0}^{\infty} \frac{2^{\mathcal{I}_{A}}}{\prod_{j=2}^{n} \left(1+\lambda_{j}\right)}\ P\left(\frac{2^{\mathcal{I}_{A}}}{\prod_{j=2}^{n}\left(1+\lambda_{j}\right)}-1,\lambda_{2},\ldots,\lambda_{n}\right)\\ &~~~~~\times\Theta\left(2^{\mathcal{I}_{A}}-\prod_{j=2}^{n}\left(1+\lambda_{j}\right)\right)\ d\lambda_{n} \cdots d\lambda_{2}\\ &=\ln\!2\!\!\int_{0}^{u_{2}} \!\!\cdots\!\!\int_{0}^{u_{n}}\! \!\!\frac{2^{\mathcal{I}_{A}}}{\prod_{j=2}^{n} \!(1\,+\,\lambda_{j}\!)} P\!\left(\!\frac{2^{\mathcal{I}_{A}}}{\prod_{j=2}^{n}(1\,+\,\lambda_{j})}\,-\,1,\lambda_{2},\ldots,\lambda_{n}\!\!\right)\! d\lambda_{n} \cdots d\lambda_{2}, \end{aligned} $$

(30)

where

$$ u_{\mu}=\frac{2^{\mathcal{I}_{A}}(1+\lambda_{2})}{\prod_{j=2}^{\mu} (1+\lambda_{j})}-1=\frac{2^{\mathcal{I}_{A}}}{\prod_{j=2}^{\mu-1} (1+\lambda_{j})}-1. $$

It is to be noted that \(u_{2}=2^{\mathcal {I}_{A}}-1\). In the second line of (30), we adjusted the integration limits of the variables in a particular way to take care of the theta function constraint in the first line. For instance, to take care of *Θ*(*α*−*z*
_{1}
*z*
_{2}), we can consider *z*
_{2}<*α*/*z*
_{1} for *z*
_{1},*z*
_{2},*α*>0.

*Special case* (*n*=2): We have ([39], Eq. 6.40)

$$ p(\mathcal{I}_{A})=\ln2\int_{0}^{2^{\mathcal{I}_{A}}-1} \frac{2^{\mathcal{I}_{A}}}{1+\lambda_{2}}\ P\left(\frac{2^{\mathcal{I}_{A}}}{1+\lambda_{2}}-1,\lambda_{2}\right)\ d\lambda_{2}. $$

(31)

The outage probability can be written using (10) as

$$\begin{array}{*{20}l} p_{\text{out}}(R)=\int_{0}^{v_{1}} \cdots\int_{0}^{v_{n}} P\left(\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)\ d\lambda_{n} \cdots d\lambda_{1}, \end{array} $$

(32)

where

$$ v_{\mu}=\frac{2^{R}(1+\lambda_{1})}{\prod_{j=1}^{\mu} (1+\lambda_{j})}-1=\frac{2^{R}}{\prod_{j=1}^{\mu-1} (1+\lambda_{j})}-1. $$

In (32), we again adopted the strategy of modifying the integration limits to take care of the theta function in (10). Another possible expression for the outage probability, using (30), can be written as

$$ \begin{aligned} p_{\text{out}}(R) &=\int_{0}^{R} p(\mathcal{I}_{A})\ d\mathcal{I}_{A}\\ &=\ln2\int_{0}^{R} \int_{0}^{u_{2}} \!\cdots\!\int_{0}^{u_{n}} \!\!\frac{2^{\mathcal{I}_{A}}}{\prod_{j=2}^{n} (1+\lambda_{j})}\ P\left(\frac{2^{\mathcal{I}_{A}}}{\prod_{j=2}^{n}(1+\lambda_{j})}-1,\lambda_{2},\ldots,\lambda_{n}\right)\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad \times d\lambda_{n} \cdots d\lambda_{2}\ d\mathcal{I}_{A}. \end{aligned} $$

(33)

### Exact results based on Laplace transform approach

The Laplace transform of \(p(\mathcal {I}_{A})\) defined in (11) is given by

$$ \begin{aligned} \widetilde{p}(s)&=\mathcal{L}[p(\mathcal{I}_{A})](s)=\int_{0}^{\infty} \cdots\int_{0}^{\infty}\\ &e^{-s\sum_{j=1}^{n} \log_{2}(1+ \lambda_{j})} P(\lambda_{1},\ldots,\lambda_{n})\ d\lambda_{1} \cdots d\lambda_{n}. \end{aligned} $$

(34)

We note that

Thus, we see that the Laplace transform serves as the moment-generating function (MGF) for \(\mathcal {I}_{A}\), since the moments of \(\mathcal {I}_{A}\) can be obtained using the coefficients of powers of *s* in the series expansion of \(\widetilde {p}(s)\). Using the JPDF given in (14), we obtain

$$ \begin{aligned} \widetilde{p}(s)\,=\,C_{n}\int_{0}^{\infty} &\!\cdots\!\int_{0}^{\infty} \Delta(\{\lambda\})\prod_{l=1}^{n} e^{-\lambda_{l}/a}\lambda_{l}^{n_{A}-n}(1\,+\,\lambda_{l})^{-(s/\ln 2)}\ d\lambda_{1} \!\cdots\! d\lambda_{n} \\ &\times\!\det\!\left[\!U\!\left(n_{B} \,-\,j\,+\,1,n_{A}\,+\,n_{B}\,-\,j\,+\,2;\!\frac{1}{b}\,+\,\frac{\lambda_{k}}{a}\!\right)\!\right]_{j,k=1,\ldots,n}, \end{aligned} $$

(36)

where we used log2*z*= ln*z*/ ln2.

With the aid of *Andréief’s* integration formula [38], the above result can be immediately cast in the form of a determinant

$$ \widetilde{p}(s)=n!\, C_{n}\,\det[\eta_{j,k}(s)]_{j,k=1,\ldots,n}, $$

(37)

where

$$\begin{aligned} \eta_{j,k}(s)=\int_{0}^{\infty} &(1+\lambda)^{-(s/\ln2)}\lambda^{n_{A}-n+k-1} e^{-\lambda/a}\\ &\times U\left(n_{B} \,-\,j\,+\,1,n_{A}\,+\,n_{B}\,-\,j\,+\,2;~\frac{1}{b}+\frac{\lambda}{a}\right)\ d\lambda. \end{aligned} $$

The density of \(\mathcal {I}_{A}\) follows by taking the inverse Laplace of \(\widetilde {p}(s)\),

$$ p(\mathcal{I}_{A})=\mathcal{L}^{-1}\{\widetilde{p}(s)\}(\mathcal{I}_{A}). $$

(38)

The outage probability follows by taking the inverse Laplace of \(s^{-1}\widetilde {p}(s)\) [10],

$$ p_{\text{out}}(R)=\mathcal{L}^{-1}\left\{s^{-1}\widetilde{p}(s)\right\}(R). $$

(39)

With the above results, we may recover the densities by performing numerical inversion of Laplace transform as in [10]. However, we make further analytical progress below to obtain an alternative expression for the PDF of mutual information.

*Special case* (*n*=1): Let us consider the *n*=1 case. The density of mutual information can be obtained from (11) as

$$\begin{aligned} p(\mathcal{I}_{A})=C_{1} \int_{0}^{\infty} &\delta(\mathcal{I}_{A}-\log_{2}(1+\lambda)) e^{-\lambda/a}\lambda^{n_{A}-1}\\ &\times U\left(n_{B},n_{A}+n_{B}+1;\frac{1}{b}+\frac{\lambda}{a}\right)\ d\lambda. \end{aligned} $$

This one-dimensional integral can be readily performed because of the presence of Dirac-delta function and yields the exact result for *n*=1 as

$$ \begin{aligned} p(\mathcal{I}_{A})=&\frac{a^{-n_{A}}b^{-n_{B}}}{\Gamma(n_{A})} (\ln2)\, 2^{\mathcal{I}}_{A}\, (2^{\mathcal{I}}_{A}-1)^{n_{A}-1}\\ & \times \exp\left(-\frac{2^{\mathcal{I}}_{A}-1}{a}\right) \,\!U\left(n_{B},n_{A}+n_{B}+1;\frac{1}{b}+\frac{2^{\mathcal{I}}_{A}-1}{a}\right). \end{aligned} $$

(40)

Comparing (40) with (37) evaluated for *n*=1, we arrive at the following inverse Laplace transform identity:

$$ \begin{aligned} &\mathcal{L}^{-1}\left[\int_{0}^{\infty} \lambda^{\gamma} e^{-\lambda/a} (1+\lambda)^{-s/\ln2}\,U \left(\alpha,\beta,\frac{1}{b}+\frac{\lambda}{a}\right)d \lambda\right](t)=\\ &\qquad\qquad\qquad(\ln2)2^{t}\!\exp\!\left(\!-\frac{\left(2^{t}\,-\,1\right)}{a}\!\right)\!(2^{t}\,-\,1)^{\gamma}\!U\!\left(\!\alpha,\beta,\frac{1}{b}\,+\,\frac{2^{t}-1}{a}\!\right)\!. \end{aligned} $$

With this interesting result in our hands, we can use the convolution property of the Laplace transform and write an expression for the PDF of mutual information for arbitrary *n* as a (*n*−1) fold integral. To this end, we expand the determinant in (37) and afterwards use the following result for inverse Laplace transform of product of *n* functions, which follows from the result for product of two functions [40]

$$ \begin{aligned} \mathcal{L}^{-1}&\left[\widetilde{F}_{1}(s)\widetilde{F}_{2}(s)\cdots\widetilde{F}_{n}(s)\right](x_{1})=\int_{0}^{x_{1}} \int_{0}^{x_{2}} \cdots \int_{0}^{x_{n-1}} \\ &F_{1}(x_{1}\,-\,x_{2}) F_{2}(x_{2}\,-\,x_{3})\cdots F_{n\,-\,1}(x_{n\,-\,1}\,-\,x_{n})F_{n}(x_{n})\ dx_{n}\ \!\cdots\! dx_{3}\ dx_{2}, \end{aligned} $$

(41)

where

$$ \mathcal{L}^{-1}[\widetilde{F}_{j}(s)](t)=F_{j}(t),~~~~ j=1,\ldots,n. $$

Therefore, with the help of (41) in (38), we obtain the following expression:

$$ \begin{aligned} p(\mathcal{I}_{A})=n!\, C_{n}\,(\ln 2)^{n}\,2^{x_{1}} &\int_{0}^{x_{1}}\int_{0}^{x_{2}} \cdots \int_{0}^{x_{n-1}}\prod_{j>k}(2^{x_{j}-x_{j+1}}-2^{x_{k}-x_{k+1}})\\ &\times \prod_{j=1}^{n}(2^{x_{j}-x_{j+1}}-1)^{n_{A}-n}\exp\left(\!-\frac{\left(2^{x_{j}-x_{j+1}}\,-\,1\right)}{a} \right)\\ &\times\! U\!\left(\!n_{B}\,-\,j\,+\,1,n_{A}\,+\,n_{B}\,-\,j\,+\,2;\!\frac{1}{b}\,+\,\frac{2^{x_{j}-x_{j+1}}\,-\,1}{a}\!\right)\\ &\times dx_{n}\cdots dx_{3}\ dx_{2}, \end{aligned} $$

(42)

where \(x_{1}\equiv \mathcal {I}_{A}\) and *x*
_{
n+1}≡0.

*Special case* (*n*=2): We have the following explicit result

$$ \begin{aligned} p(\mathcal{I}_{A}) &=\frac{a^{1\,-\,2n_{A} }b^{\,-\,2n_{B}}}{\Gamma\!(n_{A})\!\Gamma\!(n_{A}\!\,-\,1)}(\ln \!2)^{2}2^{\mathcal{I}_{A}}\! \int_{0}^{\mathcal{I}_{A}}\!(2^{x}\!\,-\,2^{\mathcal{I}_{A}-x})\!(2^{\mathcal{I}_{A}-x}\,-\,1)^{n_{A}\,-\,2}(\!2^{x}\!\,-\,\!1\!)^{n_{A}\!-2}\\ &\times \!\exp\!\left(\,-\,\frac{(2^{\mathcal{I}_{A}\,-\,x}\,+\,2^{x}\,-\,2)}{a} \!\right) \!U\!\left(n_{B},n_{A}\,+\,n_{B}\,+\,1;\frac{1}{b}\,+\,\frac{2^{\mathcal{I}_{A}-x}-1}{a}\right)\\ &\times U\left(n_{B}-1,n_{A}+n_{B};\frac{1}{b}+\frac{2^{x}-1}{a}\right)\ dx. \end{aligned} $$

(43)

The outage probability, which is the CDF of the mutual information, can be written as the integral of (42) from 0 to *R* as

$$ \begin{aligned} p_{out}(R)&=\int_{0}^{R} p(x_{1})\ dx_{1}\\ &=n!\! C_{n}\,\!(\ln 2)^{n}\!\! \!\int_{0}^{R}\! \int_{0}^{x_{1}}\!\int_{0}^{x_{2}} \!\!\cdots\! \int_{0}^{x_{n-1}} \!2^{x_{1}}\!\!\prod_{j>k}\!(2^{x_{j}-x_{j+1}}\,-\,2^{x_{k}-x_{k+1}})\\ &\times \prod_{j=1}^{n}(2^{x_{j}-x_{j+1}}-1)^{n_{A}-n}\exp\left(-\frac{(2^{x_{j}-x_{j+1}}-1)}{a} \right)\\ &\times U\left(n_{B}-j+1,n_{A}+n_{B}-j+2;\frac{1}{b}+\frac{2^{x_{j}-x_{j+1}}-1}{a}\right)\\ &\qquad \times dx_{n}\cdots dx_{3}\ dx_{2}\ dx_{1}. \end{aligned} $$

(44)

We should remark at this point that for the evaluation of PDF and outage probability of \(\mathcal {I}_{A}\), as far as number of integrals is concerned, we have not gained anything. However, the above exact expressions provide an alternative route to calculate these quantities compared to the expressions derived in the last subsection, where we adopted the strategy of integrating the JPDF directly.

### Gaussian approximation

The expressions for PDF and CDF presented above use the JPDF of the eigenvalues or Laplace transform, which involves the calculation of multiple integrals. A more straightforward method is to use the Gaussian approximation that depends only on integrals involving up to two eigenvalue densities.

The PDF of mutual information can be approximated by a Gaussian distribution as [10]

$$ p(\mathcal{I}_{A})\approx \frac{1}{\sqrt{2\pi\sigma_{\mathcal{I}_{A}}^{2}}}\exp\left(-\frac{\left(\mathcal{I}_{A}-\mu_{\mathcal{I}_{A}}\right)^{2}}{2\sigma_{\mathcal{I}_{A}}^{2}}\right). $$

(45)

Correspondingly, the outage probability is given by

$$ p_{\text{out}}(R)\approx \frac{1}{2}\text{erfc}\left(\frac{\mu_{\mathcal{I}_{A}}-R}{\sqrt{2\sigma_{\mathcal{I}_{A}}^{2}}}\right), $$

(46)

where erfc (·) represents the complementary error function.

We propose below two approaches to calculate the mean and variance of the mutual information \(\mathcal {I}_{A}\).

#### Laplace transform based

We can obtain an arbitrary moments of the mutual information using the Laplace transform results (34) and (35) as

In particular, the first moment (mean) and the second moment are obtained by differentiating the determinant-based expression in (34), and then setting *s*=0. We find

Here,

$$ \eta^{(\mu)}_{j,k}\,=\, \left\{\begin{array}{ll} -\int_{0}^{\infty} \!\log_{2}(1\,+\,\lambda)\,\!\lambda^{n_{A}-n+k-1} e^{-\lambda/a}\,\! f_{j}(\lambda)\, d\lambda &\quad j=\mu,\\ h_{j,k} &\quad j\neq \mu, \end{array}\right. $$

(50)

$$ \xi^{(\mu)}_{j,k}\,=\, \left\{\!\begin{array}{ll} \int_{0}^{\infty} (\log_{2}(1\,+\,\lambda))^{2}\,\!\lambda^{n_{A}-n+k-1} e^{-\lambda/a}\, \!f_{j}(\lambda)\, d\lambda &\quad j=\mu,\\ h_{j,k} &\quad j\neq \mu, \end{array}\right. $$

(51)

$$ \xi^{(\mu,\nu)}_{j,k}\!\,=\, \left\{\!\begin{array}{ll} \,-\,\int_{0}^{\infty} \!\log_{2}\!(1\,+\,\lambda)\lambda^{n_{A}\!-n+k\,-\,1} e^{\,-\,\lambda/a} f_{j}(\lambda) d\lambda &\quad\!\! j=\!\mu\, \text{or}\, \nu\!,\\ \!h_{j,k} &\quad\!\! j\neq \!\mu. \end{array}\right. $$

(52)

Here, *f*
_{
j
}(*λ*) and *h*
_{
j,k
} are as in (16) and (18), respectively. The variance of \(\mathcal {I}_{A}\) can be calculated as . Using the mean and variance of the mutual information in (45) and (46), we obtain the approximations for the PDFs of the mutual information and outage probability, respectively.

#### Correlation function based

Under this approach, we use the correlation function expression (19) to obtain the mean and variance of the mutual information. The \(\mu _{\mathcal {I}_{A}}\) can be obtained by averaging over the ensemble of **W** as [9]

and using the one-point function *R*
_{1}(*λ*
_{1}), this can be written as

$$ \mu_{\mathcal{I}_{A}}=\int_{0}^{\infty} R_{1}(\lambda_{1})\log_{2}(1+\lambda_{1})\ d\lambda_{1}. $$

(53)

Similarly, the \(\sigma _{\mathcal {I}_{A}}^{2}\) can be obtained as [10]

Therefore, with the aid of one-point and two-point correlation functions (19), this can be written as

$$ \begin{aligned} \sigma_{\mathcal{I}_{A}}^{2} &=\int_{0}^{\infty} R_{1}(\lambda_{1})\log_{2}^{2}(1+\lambda_{1})\ d\lambda_{1}\\ &+\int_{0}^{\infty} \int_{0}^{\infty} R_{2}(\lambda_{1},\lambda_{2})\log_{2}(1\,+\,\lambda_{1})\log_{2}(1\,+\,\lambda_{2})\ d\lambda_{1}\ d\lambda_{2}\,-\,\mu_{\mathcal{I}_{A}}^{2}. \end{aligned} $$

(55)

We note that while obtaining the exact density requires evaluation of *n*−1 or *n*-fold integrals, for Gaussian approximation, we need to perform up to a twofold integral. Both the approaches mentioned above can be easily implemented in symbolic manipulation computational packages, such as Mathematica. The first approach based on the Laplace transform is preferable, as it gives the result in terms of determinants where only some of the elements involve onefold integrals. The second approach, on the other hand, involves up to twofold integrals of determinantal expressions.