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

1. Oct 7, 2011

### alias

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 this 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
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?

2. Oct 7, 2011

### Ray Vickson

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

RGV

3. Oct 8, 2011

### alias

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. Oct 8, 2011

### Ray Vickson

No such split is needed.

RGV

5. Oct 8, 2011

### alias

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.