<s>[INST] Find the minimum of $P(X=0)$ when $E[X]=1,E[X^2]=2,E[X^3]=5$ using the probability generating function I was given the following exercise.

Let $X$ be a random variable that takes non-negative natural number
values such that $E[X]=1,E[X^2]=2,E[X^3]=5$. Find the minimum value of
$P(X=0)$ using the taylor expansion of the probability generating function
at $z=1$.

I know the method not using the generating function stated in this question.

My attempt:
Let $G(z)=E[z^X]$ be the probability generating function of $X$. Then by definition,
$G(z) = P(X=0)+P(X=1)z+P(X=2)z^2+\cdots $
Also, since $E[\frac{X(X-1)\cdots (X-n+1)}{n!}]=\frac{G^{(n)}(1)}{n!}$ according to Wikipedia, we have
$\displaystyle G(z) = \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](z-1)^n$
Substituting $z=0$ yields
$\displaystyle P(X=0) = \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n$
Note that for $n=0,1,2,3$, the coefficient can be calculated as follows:
$\begin{align} E[1] &= 1  \\ E[X]&= 1 \\ E[X(X-1)/2] &= E[X^2]/2 -E[X]/2 = 1-1/2 =1/2 \\ E[X(X-1)(X-2)/3!] &= E[X^3]/6 -E[X^2]/2 +E[X]/3 = 5/6-2/2+1/3 \\ &= 1/6 \end{align}$
Therefore,
$\displaystyle \begin{align} P(X=0) &= \sum_{n=0}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n \\ &=1-1+\frac{1}{2} -\frac{1}{6} +\sum_{n=4}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n \\ &= \frac{1}{3} + \sum_{n=4}^{\infty}E\left[\frac{X(X-1)\cdots (X-n+1)}{n!}\right](-1)^n\end{align}$
According to the question linked above, $1/3$ is the minimum, so I think we need to prove that the sum is non-negative to complete the proof. However, I was unable to do so.
Am I on the right path? If not, what is the correct one? If yes, how can I finish it?
 [/INST] Conditional on a given value $X=N\geq 4$ we get the tail sum
$$\sum_{n=4}^N {N\choose n} (-1)^n=- ({N\choose 0}-{N\choose 1} +{N\choose 2}-{N\choose 3})=$$
$$-(1-N+N(N-1)/2+N(N-1)(N-2)/6)$$
$$=\frac{1}{6} (N-1)(N-2)(N-3)={N-1\choose 3}>0.$$
(We re-derived a special case of a known identity for binomial coefficients.)
Of course, conditional on $X<4$ the tail sum is $0$. Thus the tail sum is non-negative (and is zero precisely if $X$ has zero probability of being above $3$).
Observe that we get that the tail sum is the expectation of $\frac{1}{6} (X-1)(X-2)(X-3)$, and basically recover the solution to which that you linked.
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