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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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It's somewhat conventional to use a capital letter for a random variable, so I don't think it's a good idea to use that as the variable of integration as well. If $\Sigma$ has some non-zero off-diagonal entries, then so will the random matrix $W$. What does it mean to say that a matrix with some non-zero off-diagonal entries is $\le xI$? Throughout, I'm assuming that it's an $m\times m$ matrix and $n$ is the number of degrees of freedom, but I think you should state any such assumptions. –  Michael Hardy Jun 23 '12 at 6:20
    
ok i will be more Specific Now: –  Ekaveera Kumar Sharma Jun 23 '12 at 9:50
1  
The present version of your posting doesn't ask any question. Is the problem to find the probability $\Pr(W\le xI)$? –  Michael Hardy Jun 23 '12 at 17:26
    
yes ofcourse we need to find that.. –  Ekaveera Kumar Sharma Jun 26 '12 at 15:50
    
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 $0$. –  Michael Hardy Jun 26 '12 at 18:38
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