prove tail probabilities equation

Question

Let's say $p(z) = E[z^X]$, which is a probability generating function of a random variable X. Could we prove following equation?

$\sum_{k \geq 0} Pr[X \geq k] z^k = \frac{1-zp(z)}{1-z}$

in which $Pr[X \geq k]$ is the probability of $X \geq k$.

Answer 1

Note that $$ \sum_{k=0}^\infty P(X\geqslant k)z^k=\sum_{k=0}^\infty\sum_{i=k}^\infty P(X=i)z^k=\sum_{i=0}^\infty P(X=i)\sum_{k=0}^iz^k $$ hence $$ \sum_{k=0}^\infty P(X\geqslant k)z^k=\sum_{i=0}^\infty P(X=i)\frac{1-z^{i+1}}{1-z}=\frac{1-zE(z^X)}{1-z}. $$