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Separate proofs for discrete and cont. rv. cases of E(X-mu)^4

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data
    X is a random variable with moments, E[X], E[X^2], E[X^3], and so forth. Prove the following is true for i) X is discrete, ii) X is continuous

    2. Relevant equations
    E[X-mu]^4 = E(X^4) - 4[E(X)][E(X^3)] + 6[E(X)]^2[E(X^2)] - 3[E(X)]^4
    where mu=E(X)

    3. The attempt at a solution
    discrete case: summation [X-mu]^4 = E(X^4)-4[E(X)][E(X^3)]+6[E(X)]^2[E(X^2)]-3[E(X)]^4
    continuous case: integral (from -inf to inf) [X-mu]^4 = E(X^4)-4[E(X)][E(X^3)]+6[E(X)]^2[E(X^2)]-3[E(X)]^4
    Not sure how to show this as a full proof. I do know that the expanson of E[(X-mu)^4] works for both cases but I am asked to prove them separately.
     
  2. jcsd
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