How to obtain probability function from its probability generating function For a discrete random variable $X$, we have known the exact expression of its PGF $G(z)=E[z^X]$.
The question is how can I get $Pr\{X>k\}$ from this PGF. 
I want to have an explicit expression of the probability in terms of $G(z)$.
The Z-Transform is too complex to use. And an explicit expression is preferred rather than a upper bound. Is there some other way?
Thanks!
 A: 
Using complex integration over the direct unit circle $S^1$, one gets
  $$\mathbb P(X\gt k)=\frac1{2\mathrm i\pi}\oint_{S^1}\frac{1-G(z)}{1-z}\cdot\frac{\mathrm dz}{z^{k+1}}.
$$

To prove this, one can rely on two ingredients:
(1) A distribution transform
For every nonnegative integer valued random variable $X$ integrable and not almost surely zero, there exists a random variable $Y$ such that, for every $k\geqslant0$,
$$
\mathbb P(Y=k)=\frac{\mathbb P(X\gt k)}{\mathbb E(X)}.
$$
This is because each RHS is nonnegative and their sum is $1$ thanks to the well-known formula
$$
\sum_{k\geqslant0}\mathbb P(X\gt k)=\mathbb E(X).
$$
If the PGF of $X$ is the function $G$ defined by $G(z)=\mathbb E(z^X)$, then $\mathbb E(X)=G'(1)$ and the PGF of $Y$ is the function $F$ defined by
$$
F(z)=\mathbb E(z^Y)=\frac{1-G(z)}{\mathbb E(X)\cdot(1-z)}.
$$
(2) The "extraction" of coefficients of an entire series through Cauchy formula
For every nonnegative integer valued random variable $Y$ with PGF $F$ and every $k\geqslant0$,
$$
\mathbb P(Y=k)=\frac1{2\mathrm i\pi}\oint_{S^1}F(z)\cdot\frac{\mathrm dz}{z^{k+1}}.
$$
To see this, expand $F$ as $F(z)=\sum\limits_{n\geqslant0}\mathbb P(Y=n)z^n$ and use the fact that, for every $n\geqslant0$,
$$
\oint_{S^1}z^n\cdot\frac{\mathrm dz}{z^{k+1}}=\left\{\begin{array}{ccc}2\mathrm i\pi & \text{if} & n=k,\\ 0 & \text{if} & n\ne k.\end{array}\right.
$$
Using (1) and (2) together, one gets
$$
\frac{\mathbb P(X\gt k)}{\mathbb E(X)}=\frac1{2\mathrm i\pi}\oint_{S^1}\frac{1-G(z)}{\mathbb E(X)\cdot(1-z)}\cdot\frac{\mathrm dz}{z^{k+1}},
$$
which is equivalent to the desired formula.
A: If $$G(z) = \operatorname{E} (z^X) = \sum_{x=0}^{\infty}\Pr(X=x)z^x$$ then to identify $\Pr(X=x)$ for a particular (non-negative integer) $x$, you take the $x$th derivative of $G(z)$ at $0$ and divide by $x!$. 
Or use standard generating function techniques, such as those described in generatingfunctionology.
So for $\Pr(X \gt k)=1 -\Pr(X \le k)$ you can use $$1-\sum_{x=0}^{k}\Pr(X=x) =1- \sum_{x=0}^{k}\frac{G^{(x)}(0)}{x!}$$
A: This solution is for the positive RV, i.e. RV that takes only positive values. In this case one can show that 
$$
\sum_{k \geq 0}\mathbf{P}(X \geq k)=\mathbf{E}X=G'_{X}(0)
$$
