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Math Help - Finding E[|b|^4] = E[(bb*)^2]

  1. #1
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    Finding E[|b|^4] = E[(bb*)^2]

    I need to work out E((bb*)^2) for b~N(0,1/2)+iN(0,1/2) b* is complex conjugate.

    Using prior working:
    E[bb*]= E[ Re{b}^2 + Im{b}^2] = E[Re{b}^2] + E[Im{b}^2]
    using the variance formula
    E[Re{b}^2] = var[Re{b}] + E[Re{b}]^2 = 1/2
    E[Im{b}^2] = var[Im{b}] + E[Im{b}]^2 = 1/2
    so E[bb*] = 1

    So far I have that:

    E[(bb*)^2] = E[Re{b}^4 + Im{b}^4 + 2*Re{b}^2*Im{b}^2]
    = E[Re{b}^4] + E[Im{b}^4] + 2*E[Re{b}^2]*E[Im{b}^2]

    which tells me that 2*E[Re{b}^2]*E[Im{b}^2] = 2*1/2*1/2 = 1/2
    But I am unsure of how to compute E[Re{b}^4] and E[Im{b}^4] easily as I do not know the variance of bb*

    Please can someone verify my working so far, show me how to compute the rest of the expectation and tell me what answer I should arrive at (so I can check working etc)?

    Thankyou

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  2. #2
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    Quote Originally Posted by ash89 View Post
    I need to work out E((bb*)^2) for b~N(0,1/2)+iN(0,1/2) b* is complex conjugate.

    Using prior working:
    E[bb*]= E[ Re{b}^2 + Im{b}^2] = E[Re{b}^2] + E[Im{b}^2]
    using the variance formula
    E[Re{b}^2] = var[Re{b}] + E[Re{b}]^2 = 1/2
    E[Im{b}^2] = var[Im{b}] + E[Im{b}]^2 = 1/2
    so E[bb*] = 1

    So far I have that:

    E[(bb*)^2] = E[Re{b}^4 + Im{b}^4 + 2*Re{b}^2*Im{b}^2]
    = E[Re{b}^4] + E[Im{b}^4] + 2*E[Re{b}^2]*E[Im{b}^2]

    which tells me that 2*E[Re{b}^2]*E[Im{b}^2] = 2*1/2*1/2 = 1/2
    But I am unsure of how to compute E[Re{b}^4] and E[Im{b}^4] easily as I do not know the variance of bb*
    Your work seems correct so far. What you would need is E[X^4] when X\sim\mathcal{N}(0,\sigma^2), i.e. compute

    E[X^4]=\int x^4 e^{-x^2/(2\sigma^2)}\frac{dx}{\sqrt{2\pi\sigma^2}}

    You can integrate by parts by writing the integrand as (x^3) (xe^{-x^2/(2\sigma^2)}), deriving the first factor and integrating the second one. Thus you will resume to something like C x^2 e^{-x^2/(2\sigma^2)} for some constant C, and you can relate this to

    \sigma^2=E[X^2]=\int x^2 e^{-x^2/(2\sigma^2)}\frac{dx}{\sqrt{2\pi\sigma^2}}.

    You should find E[X^4]=3\sigma^4, I think. Another way consists in differentiating the characteristic (or moment generating) function four times and evaluating it at 0, if you know that technique.
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