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?
Expectation of sum of product of i.i.d random variables
expected valueprobabilityrandom variables
Best Answer
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 $$P(k) = E(L) = \frac{n!}{k!(n-k)!}E(X_1X_2...X_k) = \color{red}{\frac{n!}{k!(n-k)!}E(X)^k}$$
Q.E.D