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Math Help - Expected value of a complex rv.

  1. #1
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    Expected value of a complex rv.

    Let Z = X + iY be complex random variable. Prove that:

    |EZ| \le E|Z|
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  2. #2
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    Quote Originally Posted by albi View Post
    Let Z = X + iY be complex random variable. Prove that:

    |EZ| \le E|Z|
    I'd start from the defintion:

    E(Z) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} z \, f(x, y) \,dx \, dy.
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  3. #3
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    First of all we do not know if joint density for the variables exists. However I know (somebody told me) that it can be done in more abstract way using:
    1. X \le Y \Rightarrow EX \le EY (for real variables)
    2. Cauchy-Schwartz inequality

    But I don't know how....
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  4. #4
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    Solution

    Let \phi = \arg EZ

    Than: |EZ| = EZe^{-i\phi} = E(X+iY)(\cos\phi - i \sin \phi) =
    =E(X\cos\phi + Y \sin \phi) + i E(Y\cos\phi - X \sin \phi)

    Because |EZ| \in \mathbb{R} we have E(Y\cos\phi - X \sin \phi) = 0

    Hence by Cauchy-Schwartz inequality (and expected value monotonicity)
    <br />
|EZ| = E(X\cos\phi + Y \sin \phi) \le E\left(\sqrt{X^2+Y^2}\sqrt{\cos^2\phi+\sin^2\phi}\  right) = E|Z|
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