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If one considers a complex random variable as the joint random variable of the real and complex part, the covariance matrix of two complex random variables $Z_{1}=X_{1}+iY_{1}$ and $Z_{2}=X_{2}+iY_{2}$ becomes a $4\times 4$ matrix $$\begin{pmatrix}C^{(rr)}&C^{(ri)}\\C^{(ir)}&C^{(ii)}\end{pmatrix}$$ $$C^{(rr)}_{ij}=E((X_{i}-E(X_{i}))(X_{j}-E(X_{j})))$$ $$C^{(ri)}_{ij}=E((X_{i}-E(X_{i}))(Y_{j}-E(Y_{j})))$$ $$C^{(ir)}_{ij}=E((Y_{i}-E(Y_{i}))(X_{j}-E(X_{j})))$$ $$C^{(ii)}_{ij}=E((Y_{i}-E(Y_{i}))(Y_{j}-E(Y_{j})))$$ However the covariance matrix of two complex random variables is often defined as a $2\times 2$ matrix $$C_{ij}=E((Z_{i}-E(Z_{i}))(Z_{j}-E(Z_{j}))^{\ast})$$ How are these two concepts related?

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$$C_{k\ell}=C_{k\ell}^{rr}+C_{k\ell}^{ii}+\mathrm i\cdot(C_{k\ell}^{ir}-C_{k\ell}^{ri})=C_{k\ell}^{rr}+C_{k\ell}^{ii}+\mathrm i\cdot(C_{k\ell}^{ir}-C_{\ell k}^{ir})$$

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Thanks! My problem however lies more in the fact that the "real covariance matrix" has 16 parameters and the "complex covariance matrix" has 8 parameters. So what is neglected? I tried deducing that from your relation, but I don't see it. –  Wox Aug 20 '12 at 12:11
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Well, the answer is in my post: for example, $C^{rr}$ and $C^{ii}$ appear in $C$ only through their sum $C^{rr}+C^{ii}$ hence, given $C$, one cannot deduce $C^{rr}$ except perhaps in some degenerate cases. (Unrelated: beware that your counts of parameters are wrong, for example any $4\times4$ covariance matrix has $10$ free parameters, not $16$.) –  Did Aug 20 '12 at 12:21
    
Yes you're right and the $2\times 2$ has $4$ free parameters, hasn't it? So it commes down to identifying the degenerate cases for which only 4 parameters of the 10 are free. –  Wox Aug 20 '12 at 12:29
    
Would this determine these degenerated cases? $$\begin{array}{l}E(X_{1}Y_{1})=0\\ E(X_{2}Y_{2})=0\\E(X_{1}^{2})= E(X_{2}^{2})\\ E(Y_{1}^{2})= E(Y_{2}^{2})\\E(X_{1}X_{2})= E(Y_{1}Y_{2})\\E(X_{1}Y_{2})= E(Y_{1}X_{2})\end{array}$$ –  Wox Aug 20 '12 at 13:01

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