# (For c-FastICA) On covariance and pseudocovariance matrix of a complex random vector

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.

You have

$$E[s_ks_k^*]=E[|s_k|^2]=1$$

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

\begin{align}E[s_k^2]&=E[x_k^2-y_k^2+2jx_ky_k]\\&=E[x_k^2]-E[y_k^2]+2jE[x_ky_k]\\&=0\end{align}

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