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Given \begin{equation}\label{eq:definition_of_z} \begin{split} \textbf{Z} = \left[\begin{array}{cccc} {z}_{11} & {z}_{12} & \cdots & {z}_{1P} \\ {z}_{21} & {z}_{22} & \cdots & {z}_{2P} \\ {z}_{31} & {z}_{32} & \cdots & {z}_{3P} \\ \vdots & \vdots & \ddots & \vdots \\ {z}_{M1} & {z}_{M2} & \cdots & {z}_{MP} \\ \end{array} \right] = \left[\begin{array}{cccc} \textbf{z}_{1} & \textbf{z}_{2} & \cdots & \textbf{z}_{P} \\ \end{array} \right] \end{split} \end{equation}

where $\textbf{z}_{i}$ represents each of the columns of $\textbf{Z} \sim \mathcal{CN}(\textbf{0}_{M \times P},\text{cM}\textbf{I}_{P})$ and $z_{ij} \sim \mathcal{CN}(0,c)$.

Other necessary definition is given by

\begin{equation}\label{eq:definition_of_z_z} \textbf{Z}^{H} \textbf{Z} = \left[\begin{array}{c} \textbf{z}_{1}^{H} \\ \textbf{z}_{2}^{H} \\ \vdots \\ \textbf{z}_{P}^{H} \\ \end{array} \right] \left[\begin{array}{cccc} \textbf{z}_{1} & \textbf{z}_{2} & \cdots & \textbf{z}_{P} \\ \end{array} \right] = \left[\begin{array}{cccc} \textbf{z}_{1}^{H}\textbf{z}_{1} & \textbf{z}_{1}^{H} \textbf{z}_{2} & \cdots & \textbf{z}_{1}^{H} \textbf{z}_{P} \\ \textbf{z}_{2}^{H} \textbf{z}_{1} & \textbf{z}_{2}^{H} \textbf{z}_{2} & \cdots & \textbf{z}_{2}^{H} \textbf{z}_{P} \\ \vdots & \vdots & \ddots & \vdots \\ \textbf{z}_{P}^{H} \textbf{z}_{1} & \textbf{z}_{P}^{H} \textbf{z}_{2} & \cdots & \textbf{z}_{P}^{H} \textbf{z}_{P} \\ \end{array} \right] \end{equation}

I'd like to know the following expectation:

\begin{equation}\label{eq:channel_matrix} \mathbb{E} \left\lbrace \frac{ \textbf{Z}^{H} \textbf{Z} }{ \text{Tr} \left( \textbf{Z}^{H} \textbf{Z} \right)^{2} } \right\rbrace = \mathbb{E} \left\lbrace \frac{ \textbf{Z}^{H} \textbf{Z} }{ \left( \textbf{z}_{1}^{H}\textbf{z}_{1} + \textbf{z}_{2}^{H}\textbf{z}_{2} + \cdots + \textbf{z}_{P}^{H}\textbf{z}_{P} \right)^{2} } \right\rbrace \end{equation}

where $\text{Tr}$ is the Trace operator.

It is important to note that the elements of the main diagonal of $\textbf{Z}^{H} \textbf{Z}$, namely $\textbf{z}_{1}^{H}\textbf{z}_{1}, \ \textbf{z}_{2}^{H}\textbf{z}_{2}, \ \cdots, \ \textbf{z}_{P}^{H}\textbf{z}_{P} \sim \Gamma(M,2c)$.

$\textbf{EDIT: 30/12/2016}$: So far we now from What is the distribution of the ratio between independent Beta and Gamma random variables? the distribution of the ratio between independent Beta and Gamma random variables, $i.e.$,

\begin{equation}\label{eq:3} y_1 = \frac{\textbf{z}_{i}^{H}\textbf{z}_{j}}{\left( \textbf{z}_{1}^{H}\textbf{z}_{1} + \textbf{z}_{2}^{H}\textbf{z}_{2} + \cdots + \textbf{z}_{P}^{H}\textbf{z}_{P} \right)^{2} }, \ \text{when} \ i = j. \end{equation}

What is still missing is the distribution of the following ratio:

\begin{equation}\label{eq:4} y_2 = \frac{\textbf{z}_{i}^{H}\textbf{z}_{j}}{\left( \textbf{z}_{1}^{H}\textbf{z}_{1} + \textbf{z}_{2}^{H}\textbf{z}_{2} + \cdots + \textbf{z}_{P}^{H}\textbf{z}_{P} \right)^{2} }, \ \text{when} \ i \neq j. \end{equation}

Would the above function also have the same distribution as when $i = j$?

It's also worth mentioning that $\textbf{z}_{1}^{H}\textbf{z}_{1} + \textbf{z}_{2}^{H}\textbf{z}_{2} + \cdots + \textbf{z}_{P}^{H}\textbf{z}_{P}$ results in a scalar random variable.

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  • $\begingroup$ Does someone know how to find the pdf of $y_2$? That would be a good starting point to find out the matrix's expectation. If I happen to know $\textbf{z}_{i}^{H}\textbf{z}_{i}$'s pdf when $i \neq j$, I could use prudentAI's solution. $\endgroup$ – Felipe Augusto de Figueiredo Jan 4 '17 at 8:36
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I don’t see an obvious way of finding an exact analytical solution, but you could fall back on Taylor expansions. As explained on the Wikipedia article on Taylor expansions for the moments of functions of random variables, the approximate expected value of the division of the random variable X by the random variable Y is

$\operatorname{E}\left[\frac{X}{Y}\right]\approx\frac{\operatorname{E}\left[X\right]}{\operatorname{E}\left[Y\right]} -\frac{\operatorname{cov}\left[X,Y\right]}{\operatorname{E}\left[Y\right]^2}+\frac{\operatorname{E}\left[X\right]\operatorname{var}\left[Y\right]}{\operatorname{E}\left[Y\right]^3}$

which you may be able to adapt to your matrix expressions. This approximation will likely be poor if there is an appreciable probability that the denominator is close to 0.

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For the following case I have found the following equation for its expectation

$$ \mathbb{E} \left\lbrace y_1 \right\rbrace = \mathbb{E} \left\lbrace \frac{\textbf{z}_{i}^{H}\textbf{z}_{j}}{\left( \textbf{z}_{1}^{H}\textbf{z}_{1} + \textbf{z}_{2}^{H}\textbf{z}_{2} + \cdots + \textbf{z}_{P}^{H}\textbf{z}_{P} \right)^{2} } \right\rbrace = \frac{1}{2cP(MP-1)}, \ \text{when} \ i = j \ \text{and} \ MP \gt 1. $$

$\textbf{Proof}$:

Writing $y_1$ as

$$y_1 = \frac{V}{U}$$

where $V = \frac{\textbf{z}_{i}^{H}\textbf{z}_{j}}{ \textbf{z}_{1}^{H}\textbf{z}_{1} + \textbf{z}_{2}^{H}\textbf{z}_{2} + \cdots + \textbf{z}_{P}^{H}\textbf{z}_{P} }$ and $U = \textbf{z}_{1}^{H}\textbf{z}_{1} + \textbf{z}_{2}^{H}\textbf{z}_{2} + \cdots + \textbf{z}_{P}^{H}\textbf{z}_{P}$ where $U \sim \Gamma(MP,2c) $ and $V \sim \beta(M,M(MP-1))$ are independent random variables.

Then taking into account that $V$ and $U$ are independent we have:

$$\mathbb{E} \left\lbrace y_1 \right\rbrace = \mathbb{E} \left\lbrace \frac{V}{U} \right\rbrace = \mathbb{E} \left\lbrace V \right\rbrace \mathbb{E} \left\lbrace \frac{1}{U} \right\rbrace $$

Note that $\frac{1}{U}$ follows the inverse-gamma distribution and then if $U \sim \Gamma(MP,2c)$ thus $\frac{1}{U} \sim Inv-Gamma(MP,2c)$.

Next, the following expectations $\mathbb{E} \left\lbrace \frac{1}{U} \right\rbrace = \frac{1}{2c(MP-1)}, MP \gt 1$ and $\mathbb{E} \left\lbrace V \right\rbrace = \frac{1}{P}$ give us the final result.

Finally, I'm supposing (but I'm not sure as I haven't figured out the distribution of $\textbf{z}_{i}^{H}\textbf{z}_{j}$ when $ i \neq j$):

$$ \mathbb{E} \left\lbrace y_2 \right\rbrace = \mathbb{E} \left\lbrace \frac{\textbf{z}_{i}^{H}\textbf{z}_{j}}{\left( \textbf{z}_{1}^{H}\textbf{z}_{1} + \textbf{z}_{2}^{H}\textbf{z}_{2} + \cdots + \textbf{z}_{P}^{H}\textbf{z}_{P} \right)^{2} } \right\rbrace = 0, \ \text{when} \ i \neq j. $$

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  • $\begingroup$ Does someone know how to prove the assumption for the second expectation above? I've run some simulations on matlab and confirmed it is the right result but I'd like a proof for that. $\endgroup$ – Felipe Augusto de Figueiredo Jan 27 '17 at 22:50

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