I am currently studying complex FastICA and the paper says that

Suppose $\mathbf{s}$ is a $n\times1$ complex random vector. If $\mathbf{s}$ has zero mean, unit variance, and uncorrelated real and imaginary part of equal variances, then $E[\mathbf{s}\mathbf{s}^H]=\mathbf{I}_n$ and $E[\mathbf{s}\mathbf{s}^T]=\mathbf{0}_n$.

I don't quite get how $E[\mathbf{s}\mathbf{s}^H]=\mathbf{I}_n$ and $E[\mathbf{s}\mathbf{s}^T]=\mathbf{0}_n$ come about from the conditions.

We have the covariance matrix as \begin{align} \operatorname{cov}(\mathbf{s}) &= E[\mathbf{s}\mathbf{s}^H]-E[\mathbf{s}]E[\mathbf{s}^H] \\ &= E[\mathbf{s}\mathbf{s}^H]-\mathbf{0}_{n\times1}\mathbf{0}_{1\times n}\\ &= E[\mathbf{s}\mathbf{s}^H]\\ \end{align} and the pseudocovariance \begin{align} \operatorname{pcov}(\mathbf{s}) &= E[\mathbf{s}\mathbf{s}^T]-E[\mathbf{s}]E[\mathbf{s}^T] \\ &= E[\mathbf{s}\mathbf{s}^T]-\mathbf{0}_{n\times1}\mathbf{0}_{1\times n}\\ &= E[\mathbf{s}\mathbf{s}^T]\\ \end{align} I don't quite get how to equate the last line of covariance matrix to identity and the pseucovariance to zero.

If I were to write out the matrix, \begin{align} E[\mathbf{s}\mathbf{s}^H] &=E\left\{\begin{bmatrix} s_1s_1^* & s_1s_2^* &\cdots & s_1s_n^*\\ s_2s_1^* & s_2s_2^* &\cdots & s_2s_n^*\\ \vdots & \vdots &\ddots & \vdots\\ s_ns_1^* & s_ns_2^* &\cdots & s_ns_n^*\\ \end{bmatrix}\right\} \end{align} and \begin{align} E[\mathbf{s}\mathbf{s}^T] &=E\left\{\begin{bmatrix} s_1s_1 & s_1s_2 &\cdots & s_1s_n\\ s_2s_1 & s_2s_2 &\cdots & s_2s_n\\ \vdots & \vdots &\ddots & \vdots\\ s_ns_1 & s_ns_2 &\cdots & s_ns_n\\ \end{bmatrix}\right\} \end{align} I still can't quite figure how all of these eventually becomes identity and zeros.


1 Answer 1


You have


because the complex random variables $s_k$ have zero mean and unit variance. That means that all elements of the main diagonal of $E[\mathbf{s}\mathbf{s}^H]$ equal $1$. Furthermore, with $s_k=x_k+jy_k$we have

$$\begin{align}E[s_ks^*_l]&=E[x_kx_l+y_ky_l+j(x_ly_k-x_ky_l)]\\&=E[x_kx_l]+E[y_ky_l]+jE[x_ly_k]-jE[x_ky_l]\\&=0,\qquad k\neq l\end{align}$$

because real and imaginary parts are uncorrelated. Consequently, all off-diagonal elements of $E[\mathbf{s}\mathbf{s}^H]$ are zero.

The off-diagonal elements of $E[\mathbf{s}\mathbf{s}^T]$ are zero for the same reason. (There's just a sign difference in the sum of the expectations, but since each of them is zero the result is the same). The main diagonal elements are


because real and imaginary parts are uncorrelated, and they have equal variance, i.e., $E[x_k^2]=E[y_k^2]$.


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