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$\newcommand{\Var}{\operatorname{Var}}$Consider the zero mean independent complex random variables $X_1,\dots,X_n$ and the complex constants $a_1,\dots,a_n$. Does the formula for real valued independent random variables carry over to complex case as: $$\Var[a_1X_1+\dots +a_nX_n] = | a_1 | ^2\Var(X_1)+\dots+ | a_n |^2\Var(X_n)$$

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  • $\begingroup$ yes it should be,because modulus of complex number is properly defined,except norm,therefore this formula should work $\endgroup$ – dato datuashvili Jan 9 '14 at 17:59
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$\newcommand{\var}{\operatorname{var}}$ The variance is often defined as $$ \var(X) = \operatorname E\left((X-\mu)\overline{(X-\mu)}\,\right) $$ where $\overline{c}$ is the complex conjugate of $c$.

It follows that $$ \var(aX) = a\overline{a}\var(X) = |a|^2\var(X). $$ With real numbers $|a|^2$ is the same as $a^2$; with complex numbers they are not equal except when the complex number involved is real.

The statement about sums is true and the proof is the same as with real numbers.

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  • $\begingroup$ Thanks. Could you please comment on the sum of complex random variables? I am not sure how independence is defined for two complex random variables as there four real random variables involved (real and imaginary of each). $\endgroup$ – triomphe Jan 9 '14 at 18:05
  • $\begingroup$ $X,Y,Z,\ldots$ are independent random variables if for every choice of measurable sets $A,B,C,\ldots$, we have $\Pr(X\in A\ \&\ Y\in B\ \&\ Z\in C\ \&\cdots)$ $=\Pr(X\in A)\cdot\Pr(Y\in B)\cdot\Pr(Z\in C)\cdots$. I think the latter equality is typically required to hold only for every finitely number of factors; its truth for infinitely many factors is derived rather than assumed. That's the definition of independence regardless of whether the values of the random variables involved are real or complex or something else. $\endgroup$ – Michael Hardy Jan 9 '14 at 18:09
  • $\begingroup$ Thanks. In conclusion is the formula given in the question correct? $\endgroup$ – triomphe Jan 9 '14 at 18:11

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