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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.

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    $\begingroup$ How did you infer they are std normal? The distribution is given explicitly as something else. Anyway the space has size only $2^4=16$ so just enumerating should not be too tedious. $\endgroup$
    – Macavity
    Jan 31, 2018 at 16:55
  • $\begingroup$ Sorry, actually I misunderstood as because mean and variance of $X_i$ are $0$ and $1$ resp. So I thought.....Please help on solving $\endgroup$
    – ram ram
    Jan 31, 2018 at 17:01
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    $\begingroup$ One way is to directly expand it with multinomial theorem and note that $E[X^n] = 0$ when $n$ is odd and $X^n = 1$ when $n$ is even. Another way is to relate it to Binomial, and see if you are more familiar with the moment of Binomial or not. $\endgroup$
    – BGM
    Jan 31, 2018 at 17:02

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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). $$\implies E[(X_1+X_2+X_3+X_4)^4] = 4 + 0 + 6\times \binom42+0 + 0=40$$

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  • $\begingroup$ Looks good (+1). Nice approach. Checked (approx) by simulation in R: s = replicate(10^6, sum(sample(c(-1,1), 4, repl=T))); mean(s^4) returns 39.89854. Second run: 40.01542. [Also, by simulation, seems $SD(S^4) \approx 82.$] Very hard to believe $E(S^4) = 76.$ $\endgroup$
    – BruceET
    Jan 31, 2018 at 19:33

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