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When dealing with real-valued RVs, the extensions of expectation and variance are quite clear to me. For example, showing

$E(aX+b)=aE(X)+b$ and $var(aX+b)=a^2var(X)$ is relatively straightforward to me. Given my weak background in complex numbers, the analogous situation in $\mathbb{C}$ is much more elusive to me.

So, how does one prove (rigorously) the two situations below:


$\forall c_1,c_2\in\mathbb{C}, E(c_1Z+c_2)=c_1E(Z)+c_2$


What is especially confusing (and simultaneously fascinating) is that this result apparently is unaffected by the correlation between $X$ and $Y$. How is this possible?

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1 Answer 1

up vote 1 down vote accepted

If $E(\vert Z\vert)<\infty$ which is the case here since the mean and variance assumed to be well-defined, then the first relation is derived by just linearity of expectation $$ E(c_1Z) = E(c_1 (X+ j Y)) = E(c_1X + jc_1 Y) = c_1 E(X) + j c_1 E(Y) = c_1 E(Z) $$ Note that $E(\vert X\vert)< E(\vert Z\vert)<\infty$ and the similar relation is also true for $Y$.

For the second one, let $m_z=E(Z)= m_x + j m_y$: $$ var(Z) = E((Z-m_z)(Z-m_z)^*) = E((X-m_x + j(Y-m_y))(X-m_x + j (Y-m_y))^*) $$ Let $U$ be $X-m_x$ and $V$ be $Y-m_y$ the centralized random variables. Then, $$ Var(Z) = E((U+jV)(U+jV)^*) = E(UU^* -jUV^* +jVU^* + VV^*) = E(U^2) + E(V^2) = var(U) + var(V) $$

The third equality is correct since $U$ and $Y$ are real-valued RVs.

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Could you help me with your the j representing the complex portion of z? Also what are the "*" denoting? – Justin Aug 9 '12 at 15:34
Oh sorry, yes (j)^2 = -1. And * is the complex conjugate operator, i.e. (x+j y)* = x - jy for real values x and y. – rfvahid Aug 9 '12 at 16:40
where is $c_2$ addressed in the solution for part 1? – Justin Aug 11 '12 at 2:39

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