Then $E(X_1+X_2+X_3+X_4)^4$ equals?

Question

Let $X_1, X_2, X_3, X_4$ are i.i.d random variable taking values $1$ and $-1$ with probability $1/2$ each. Then $E(X_1+X_2+X_3+X_4)^4$ equals?

I see that each $X_i$ is standard normal and so $X_1+X_2+X_3+X_4$ is a normal variable with mean $0$ and variance $4.$ I find the answer $48$ using MGF. But the answer is $76$. Where I gone wrong? thanks.

Answer 1

This seems one of those problems where just enumerating the possibilities is quick (even if dirty):
X1 X2 X3 X4 Sum Sum^4
-1 -1 -1 -1 -4 256
-1 -1 -1 1 -2 16
-1 -1 1 -1 -2 16
-1 -1 1 1 0 0
-1 1 -1 -1 -2 16
-1 1 -1 1 0 0
-1 1 1 -1 0 0
-1 1 1 1 2 16
1 -1 -1 -1 -2 16
1 -1 -1 1 0 0
1 -1 1 -1 0 0
1 -1 1 1 2 16
1 1 -1 -1 0 0
1 1 -1 1 2 16
1 1 1 -1 2 16
1 1 1 1 4 256
Now sum the last column and divide by $16$ as each row is equally probable to get an answer of $40$.

So either you haven't copied the question right or the answer your book has is wrong...

To double check, alternately, $$(X_1+X_2+X_3+X_4)^4 = \sum X_1^4 + 4\sum X_1^3X_2 + 6\sum X_1^2X_2^2+12\sum X_1X_2^2X_3 + 24X_1X_2X_3X_4$$ (where $\sum$ is used to denote sum of similar symmetric terms - however we dont need to count most of them, as $E(X_i^n) \in \{0, 1\}$ as $n$ is odd or even, and by independence we can multiply the expectations). $$\implies E[(X_1+X_2+X_3+X_4)^4] = 4 + 0 + 6\times \binom42+0 + 0=40$$