4th moment, show that E[X-mu]^4 is equal to

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Homework Statement


X is a random variable with moments, E[X], E[X^2], E[X^3], and so forth. Prove this is true for i) X is discrete, ii) X is continuous


Homework 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)

The Attempt at a Solution


I've been trying to generalize expanding the variance, E[(X-mu)^2], into the above result with no success. Not sure about the discrete and continuous proofs either. Anyone have any ideas?
 

Answers and Replies

  • #2
Ray Vickson
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EX = m is a number, not a random quantity. So, expand (X-m)^4 using the binomial theorem.

RGV
 
  • #3
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I got the expansion, thanks. I'm not sure why I need to split the cases for continuous/discrete either. It's asked in the question and I don't know how to prove with continuous/discrete separately.
 
  • #4
Ray Vickson
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I got the expansion, thanks. I'm not sure why I need to split the cases for continuous/discrete either. It's asked in the question and I don't know how to prove with continuous/discrete separately.

No such split is needed.

RGV
 
  • #5
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No such split is needed.

RGV

I know that the one equation will cover both the discrete and continuous cases, but the question specifically asks to show each case individually and I'm not sure what the specific summation and integral that I have to work out is.
 

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