Let $X$ be a discrete random variable, with positive probability for non-negative integers.

If $q_k = P(X>k)$ and $Q_X(t) = \sum_{k=0}^\infty q_kt^k$, and the probability generating function defined as $G_X(t) = E(t^X)$. Then I have to show that: $$Q_X(t) = (1 - G_X(t))/(1-t). $$

What I got so far is: $$Q_X(t) = \sum_{k=0}^\infty (1 - P(X\leq k))t^k, $$ then assuming $|t|<1$:

$$Q_X(t) = \frac{1}{1-t} - \sum_{k=0}^\infty P(X\leq k)t^k = \frac{1}{1-t} - \sum_{k=0}^\infty t^k \sum_{j=0}^k P(X = j) $$

$$ = \frac{1}{1-t} - \sum_{k=0}^\infty t^k \sum_{j=0}^{k-1} P(X = j) - G_X(t), $$ taking the parts where $j=k$ on the sum to get $G_X(t)$ , but then I can't solve the sum. Is there a viable way to solve that sum? I've also tried to solve without using the complement, but it looked even harder.


1 Answer 1


Let's say $p_k = P(X=k)$, which means $q_k=\sum_{j=k+1}^\infty p_j$ and so

$Q_X(t)=\sum_{k=0}^\infty (\sum_{j=k+1}^\infty p_j)t^k$ = $\sum_{j=1}^\infty$ $(\sum_{k=0}^{j-1} t^k)p_j $

then for $0\le t<1$, this equals

$\sum_{j=1}^\infty (\frac{1-t^j}{1-t})p_j =\sum_{j=0}^\infty (\frac{1-t^j}{1-t})p_j$

$ = (1 - G_X(t))/(1-t)$


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