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

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Homework Help Overview

The problem involves proving a relationship for the fourth moment of a random variable X, specifically showing that E[X-mu]^4 is equal to a given expression. The context includes both discrete and continuous cases for the random variable.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss expanding the expression using the binomial theorem and generalizing the variance. There are questions about the necessity of treating discrete and continuous cases separately, with some expressing uncertainty about how to approach the proofs for each case.

Discussion Status

Some participants have made progress in expanding the expression, while others are questioning the need for separate treatments of discrete and continuous cases. There is a recognition that one equation may suffice for both cases, but the original poster is seeking clarification on the specific requirements of the problem.

Contextual Notes

The original poster notes that the problem explicitly requests separate proofs for discrete and continuous cases, which adds to the complexity of their approach.

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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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