Probability

I do not use mgf.
pdf of the random variable = 1 on (-0.5,0.5)
integration on the interval gives
E(Xi) = ∫xdx = 0
E(Xi)2 = ∫x2dx = 1/12
E(Xi)4 = ∫x4dx = 1/80
Expand (X1+X2+X3)4
(X1+X2+X3)4 = X14 + X24 + X34 + 6X12X22 + 6X12X32 + 6X32X22 + other terms (see note below)

E{(X1+X2+X3)4}
=E(X14) + E(X24) + E(X34) + E(6X12X22) + E(6X12X32) + E(6X32X22) + 0
=3E(Xi)4+ 18E(Xi2)E(Xi2)
=3/80+ 18/122
=13/80

Note: those other terms have zero expected value. For example, one of those other terms is 4X13X2, whose expected value is E(4X13X2) = 4E(X13)E(X2). As E(X2) = 0, this term has zero expected value.

2011-06-14 11:02:23 補充：
For your information, there are nine other terms, six of which are of the form 4 X_i^3 X_j, and the remaining three are of the form 12 X_i^2 X_j X_k.