Expectation of sum of product of i.i.d random variables

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 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?

• 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 $$P(k)=k!\sum_{i_1<\cdots<i_k} E(X_{i_1}\cdots X_{i_k})$$ Jun 6, 2021 at 18:37
• Also, observe the convention of where random variables are capital letters. So $X_i$ rather than $x_i.$ Makes talking probability writing clearer. Jun 6, 2021 at 18:41
• The first two formula you have for $P(1)$ and $P(2)$ do not require independence. Since they are independent, $$E[X_{i_1}X_{i_2}\cdots X_{i_k}]=E[X_{i_1}]\cdots E[X_{i_k}]=E[X]^k$$ Jun 6, 2021 at 18:46
• 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 $E[X]^2=E[X^2]$ Jun 6, 2021 at 18:55
• @ThomasAndrews You are right on the "count twice" issue. Edited the question with $i_1<i_2<i_3...,i_k$ Jun 6, 2021 at 21:22

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