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I was reading Probability Theory by Borovkov when I was stuck on this expansion in the Proof of the Law of Large numbers for the Bernoulli Scheme. Note that the $\xi_j$'s are independent Bernoulli random variables with $p$ probability of success. $$\mathbb{E}\left(\sum_{j=1}^k(\xi_j-p)\right)^4 = \mathbb{E}\left(\sum_{j=1}^k(\xi_j-p)^4 + 6\sum_{i<j}(\xi_i-p)^2(\xi_j-p)^2\right)$$ I'm not sure how to go about showing this. I tried to look at the multinomial theorem but it doesn't seem to help.

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    $\begingroup$ Note that any term that winds up with a factor like $(\xi-p)$ has expectation $0$ since $E[\xi]=p$. $\endgroup$ – lulu Feb 23 at 0:41
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    $\begingroup$ Second hint: the only terms in $(a+b+c+\cdots)^4$ that don't have a factor with exponent $1$ are terms like $a^4$ or like $a^2b^2$. Should say: I am reading the left hand as $E\left[ \left(\sum (\cdots)\right)^4\right]$. I think the way you wrote it is ambiguous. $\endgroup$ – lulu Feb 23 at 0:43
  • $\begingroup$ @lulu I see it now, thanks $\endgroup$ – varpi Feb 23 at 0:44

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