Let $X_{i}, i=1,2,...,n$ be i.i.d random variables who has the same distribution with variable $X$. Define $P(k)$ as the expectation of sum of product of size $k$ combination of $X_{i}$. We know that \begin{align} P(1)&=E\left[\sum_{i=1}^{n}X_i\right] = nE[X]\\ P(2)&=E\left[\sum_{i<j}X_iX_j\right] = \binom{n}{2}E[X]^2\\ P(3)&=E\left[\sum_{i<j<k}X_iX_jX_k\right]=... \end{align} Is there a way to express $P(k)$ for $k\geq3$ in the form $E[X], E[X^2], E[X^3]$ and so on?

  • $\begingroup$ Do you really want to count twice $x_1x_2$ in the formula for $P(2)?$ You can divide by $k!$ to get the expected sum for $$P(k)=k!\sum_{i_1<\cdots<i_k} E(X_{i_1}\cdots X_{i_k})$$ $\endgroup$ Jun 6, 2021 at 18:37
  • $\begingroup$ Also, observe the convention of where random variables are capital letters. So $X_i$ rather than $x_i.$ Makes talking probability writing clearer. $\endgroup$ Jun 6, 2021 at 18:41
  • $\begingroup$ The first two formula you have for $P(1)$ and $P(2)$ do not require independence. Since they are independent, $$E[X_{i_1}X_{i_2}\cdots X_{i_k}]=E[X_{i_1}]\cdots E[X_{i_k}]=E[X]^k$$ $\endgroup$ Jun 6, 2021 at 18:46
  • $\begingroup$ I’m not sure your formula for $P(2)$ is correct. The simpler formula is $P(2)=n(n-1)E[X]^2,$ which is only equal to your formula when $E[X]^2=E[X^2]$ $\endgroup$ Jun 6, 2021 at 18:55
  • $\begingroup$ @ThomasAndrews You are right on the "count twice" issue. Edited the question with $i_1<i_2<i_3...,i_k$ $\endgroup$
    – Andrew Yao
    Jun 6, 2021 at 21:22

1 Answer 1


Let us rewrite the sum $L$ inside the $P(k)$ : $P(k)= E(L) $ and count the number of term of $L$ $$L = \sum_{ \left\{ \begin{align} &a_1+...+a_n = k \\ &a_i \in \{0,1 \}, i=1,...,n \end{align} \right\} }\left(\prod_{t=1}^nx_t^{a_t}\right)$$

with $a_i$ receives only two values $0$ or $1$, and their sum is equal to $k$.

It's a combination problem and there are in total $C_k^n=\frac{n!}{k!(n-k)!}$ terms.

Hence $$P(k) = E(L) = \frac{n!}{k!(n-k)!}E(X_1X_2...X_k) = \color{red}{\frac{n!}{k!(n-k)!}E(X)^k}$$



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