(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}) 
&amp;= E[\mathbf{s}\mathbf{s}^H]-E[\mathbf{s}]E[\mathbf{s}^H] \\
&amp;= E[\mathbf{s}\mathbf{s}^H]-\mathbf{0}_{n\times1}\mathbf{0}_{1\times n}\\
&amp;= E[\mathbf{s}\mathbf{s}^H]\\
\end{align}
and the pseudocovariance
\begin{align}
\operatorname{pcov}(\mathbf{s}) 
&amp;= E[\mathbf{s}\mathbf{s}^T]-E[\mathbf{s}]E[\mathbf{s}^T] \\
&amp;= E[\mathbf{s}\mathbf{s}^T]-\mathbf{0}_{n\times1}\mathbf{0}_{1\times n}\\
&amp;= 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]
&amp;=E\left\{\begin{bmatrix}
s_1s_1^* &amp; s_1s_2^* &amp;\cdots &amp; s_1s_n^*\\
s_2s_1^* &amp; s_2s_2^* &amp;\cdots &amp; s_2s_n^*\\
\vdots &amp; \vdots &amp;\ddots &amp; \vdots\\
s_ns_1^* &amp; s_ns_2^* &amp;\cdots &amp; s_ns_n^*\\
\end{bmatrix}\right\}
\end{align}
and
\begin{align}
E[\mathbf{s}\mathbf{s}^T]
&amp;=E\left\{\begin{bmatrix}
s_1s_1 &amp; s_1s_2 &amp;\cdots &amp; s_1s_n\\
s_2s_1 &amp; s_2s_2 &amp;\cdots &amp; s_2s_n\\
\vdots &amp; \vdots &amp;\ddots &amp; \vdots\\
s_ns_1 &amp; s_ns_2 &amp;\cdots &amp; 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 

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

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 

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

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