# On covariance and pseudocovariance 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 $$E[\mathbf{s}\mathbf{s}^T]=\mathbf{0}_n$$ 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.

I know for sure that $$E[s_is_j^*]=E[s_i]E[s_j]=0$$ and $$E[s_is_j]=E[s_i]E[s_j]=0$$ where $$i\ne j$$ so I have
\begin{align} E[\mathbf{s}\mathbf{s}^H] &=\operatorname{diag}(E[s_is_i^*])\\ &=\operatorname{diag}(E[(a+ib)(a-ib)]) &&(\because s_i:=a+ib)\\ &=\operatorname{diag}(E[a^2+b^2])\\ &=\operatorname{diag}(E[a^2]+E[b^2])\\ &=\operatorname{diag}(\operatorname{Var}[a]+E[a]^2+\operatorname{Var}[b]+E[b]^2) && (\because \operatorname{Var}[X]=E[X^2]-E[X]^2)\\ &=\operatorname{diag}(\operatorname{Var}[s_i]+0) && (\because E[a]=E[b]=0,\\ &&& \qquad\operatorname{Var}[s]:=\operatorname{Var}[a]+\operatorname{Var}[b])\\ &=\operatorname{diag}(1) && (\because \operatorname{Var}[s]:=1)\\ &=\mathbf{I}_n \end{align} and \begin{align} E[\mathbf{s}\mathbf{s}^T] &=\operatorname{diag}(E[s_is_i])\\ &=\operatorname{diag}(E[s_i^2])\\ &=\operatorname{diag}(E[a^2+2abi-b^2])\\ &=\operatorname{diag}(E[a^2]+2E[abi]-E[b^2])\\ &=\operatorname{diag}(E[a^2]-E[b^2])\\ &=\operatorname{diag}(\operatorname{Var}[a]-\operatorname{Var}[b])\\ &=\operatorname{diag}(0)\\ &=\mathbf{0}_n \end{align}