# 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

$$displaystyle G(z) = sum_{n=0}^{infty}Eleft(frac{X(X-1)cdots (X-n+1)}{n!}right)(z-1)^n$$

Substituting $$z=0$$ yields

$$displaystyle P(X=0) = sum_{n=0}^{infty}Eleft(frac{X(X-1)cdots (X-n+1)}{n!}right)(-1)^n$$

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}Eleft(frac{X(X-1)cdots (X-n+1)}{n!}right)(-1)^n \ &=1-1+frac{1}{2} -frac{1}{6} +sum_{n=3}^{infty}Eleft(frac{X(X-1)cdots (X-n+1)}{n!}right)(-1)^n \ &= frac{1}{3} + sum_{n=3}^{infty}Eleft(frac{X(X-1)cdots (X-n+1)}{n!}right)(-1)^nend{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?