Suppose that $X\sim\text{Poisson}(\lambda)$, find $E[X(X-1)(X-2)(X-3)]$ Hint: do not use linearity, use definition of expectation.
i have a rough idea that we turn the inside into g(x), but not sure how to proceed from there.
 A: wolfies' answer requires knowing the probability generating function, that is,
the "need for sums" is hidden in previous work or something looked up in a textbook
or Wikipedia, and is replaced by the need for finding the derivative of the pgf (which fortunately is easy in this case). It also misses the whole point of the
exercise: that for a random variable that takes on integer values, 
$E[X(X-1)(X-2)(X-3)]$ can be a lot easier to compute
directly from the law of the unconscious
statistician than by expanding out $X(X-1)(X-2)(X-3)$ into a quartic polynomial
and then "exploiting" the linearity of expectation.  This is particularly
true when $P\{X=k\}$ is of the
form $\frac{g(k)}{k!}$, e.g. for Poisson, binomial, negative binomial random
variables, etc.
In this case, we have 
$$\begin{align}
E[X(X-1)(X-2)(X-3)] 
&= \sum_{n=0}^{\infty} n(n-1)(n-2)(n-3)\cdot e^{-\lambda}\frac{\lambda^n}{n!}\\
&= \sum_{n=4}^{\infty} n(n-1)(n-2)(n-3)\cdot e^{-\lambda}\frac{\lambda^n}{n!}\\
&= \sum_{n=4}^{\infty} e^{-\lambda}\frac{\lambda^n}{(n-4)!}\\
&= \lambda^4 \sum_{n=4}^{\infty} e^{-\lambda}\frac{\lambda^{n-4}}{(n-4)!}\\
&= \lambda^4\sum_{m=0}^{\infty} e^{-\lambda}\frac{\lambda^m}{m!}\\
&= \lambda^4.
\end{align}$$
As another illustration, note that for a Poisson random variable, it
is easier to calculate the variance using
$$\operatorname{var}(X) = E[X(X-1)] + E[X] - \left(E[X]\right)^2$$ rather
than the more familiar
$$\operatorname{var}(X) = E[X^2] - \left(E[X]\right)^2.$$
A: A fun way to do this is via factorial moments. In particular, the $r^{th}$ factorial moment can be obtained from the probability generating function (pgf) $\Pi(t) = E[t^X]$ via:
$$E[X(X-1) * \dots * (X-r+1) \quad = \quad \frac{d^r\Pi (t)}{dt^r}\big|_{t=1}$$
For your problem, $r = 4$, and $X \sim Poisson(\lambda)$ with pgf $\Pi(t) = E[t^X] = e^{\lambda(t-1)}$. 
The answer follows immediately, without the need for any sums.
