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

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SUMMARY

The discussion centers on proving the equality 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). Participants explore the necessity of separate proofs for discrete and continuous random variables, concluding that a single equation suffices for both cases. However, the question explicitly requests individual proofs, leading to confusion regarding the required summation and integral for each case.

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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?
 
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EX = m is a number, not a random quantity. So, expand (X-m)^4 using the binomial theorem.

RGV
 
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.
 
alias said:
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
 
Ray Vickson said:
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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