Extending the identity $ x^n=\sum\limits_{j=0}^n\left\{\begin{array} {}n\\ j \end{array}\right\}\prod\limits_{i=0}^{j - 1}(x-i)$ to $x\in\mathbb C$

Question

For $x \in \mathbb{R}$ and $n \in \mathbb{N}_{0}$ we have the identity $$ x^n = \sum_{j=0}^{n} \begin{Bmatrix} n\\ j \end{Bmatrix} \prod_{i=0}^{j - 1}(x-i), $$ where $\begin{Bmatrix} n\\ j \end{Bmatrix}$ denotes the Stirling number of the second kind. Now let $X$ be a complex-valued random variable. Since the identity above seems true also for $x \in \mathbb{C}$, I would expect that $$ \mathbb{E}[X^n] = \sum_{j=0}^{n} \begin{Bmatrix} n\\ j \end{Bmatrix} \mathbb{E}\bigg[\prod_{i=0}^{j - 1}(X-i)\bigg]. $$ How can this formula for the expectations be rigorously justified? Is there really no issue with going from $\mathbb{R}$ to $\mathbb{C}$?

Answer 1

Since what you have is a polynomial identity, plugging in a random variable is fine, and you get a valid identity. Now expectation is linear, and you obtain what you want. Note, also, that a polynomial identity valid over an infinite subset of $\mathbb C$ is valid over $\mathbb C$.