Jtdylktuy

I am currently studying complex FastICA and the paper says that

Suppose $mathbf{s}$ is a $ntimes1$ 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}_{ntimes1}mathbf{0}_{1times 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}_{ntimes1}mathbf{0}_{1times 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]
&=Eleft{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]
&=Eleft{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.

Okay now that I ponder for a few more hours.

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 $ine j$ 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,\
&&& qquadoperatorname{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}

I will just leave my answer here in case someone were to be in my position in the future since I can't find any good reference for this.