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Let $X_1 \cdots X_N$ are $N$ number of $m$ Dimensional Independent Complex Gaussian Random vectors Such that:

$ X_j \sim \mathcal{N}(\mu,\Sigma)\; \forall \;j=1 \cdots N$

Let $X=\left[\begin{array}{ccc}X_1^T \\X_2^T \\ \vdots \\ X_N^T\end{array}\right]$ be $N \times m$ matrix, Then by some algebra

$ E(X)=\left[\begin{array}{ccc}\mu^T \\\mu^T \\ \vdots \\ \mu^T\end{array}\right]$ and

$ \operatorname{Cov}(\operatorname{Vec}(X^T))=I_n \otimes \Sigma $ Then

$ X \sim \mathcal{N}(\bf{1}\mu^T, I_n \otimes \Sigma) $ Where

$ \bf{1}=\left[\begin{array}{ccc} 1 \\1 \\ \vdots \\ 1\end{array}\right]$ is Vector of all one's. Then

$ W=X^H X $ is $m \times m$ Wishart Random Matrix

$ \mathbf{f(W)=|\Sigma|^{-\frac{n}{2}}\,|W|^{\frac{n-m-1}{2}}\,e^{-Tr(\frac{1}{2}\,\Sigma^{-1}\,W)}}$

$\Pr(W \leq xI)= \int_{0 \leq W \leq xI}\,f (W) \, dW $ where $x$ is a scalar and $I$ is an $m \times m$ Identiy Matrix

Please give me any Hints..

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    What do you mean by $W\le xI$? There is a partial order on nonnegative-definite symmetric matrices that would say $A\le B$ precisely if there is some nonnegative-definite matrix $C$ such that $A+C=B$. Is that what you have in mind? I haven't thought about this for a while; I question whether it is possible that $W\le xI$ unless all off-diagonal elements of $W$ are $0$.2012-06-26

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