# Expanding the power of a sum

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 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.

• Note that any term that winds up with a factor like $(\xi-p)$ has expectation $0$ since $E[\xi]=p$. – lulu Feb 23 at 0:41
• 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. – lulu Feb 23 at 0:43
• @lulu I see it now, thanks – varpi Feb 23 at 0:44