I'm trying to proof something but I'm getting a different answer than my textbook and I don't know where I've gone wrong.
The question concerns a random discrete variable $Y$ taking on values in $\mathbb{N}_{\geq 0}$ with a probability generating function $G_Y(s)$. If we define the sequence $U_n = \mathbb{P}(Y > n)$, it should follow that its generating function $U(s)$ satisfies:
$(1-s)U(s) = 1 - G_Y(s)$
However, I'm getting a different result, and I've gone throught it several times not getting why it's wrong (or it's just an erratum in my textbook)
My proof is as follows: By definition we have $G_Y(s) = \sum_{i=0}^{\infty} (\mathbb{P}(Y=i) \cdot s^i = \sum_{i=0}^{\infty} ((\mathbb{P}(Y > i)-\mathbb{P}(Y > i-1)) \cdot s^i = \sum_{i=0}^{\infty} (\mathbb{P}(Y > i) \cdot s^i) - \sum_{i=0}^{\infty} (\mathbb{P}(Y > i-1) \cdot s^i)=$
$ U(s) - ((\mathbb{P}(Y > -1) \cdot s^0 + \sum_{i=1}^{\infty} (\mathbb{P}(Y>i-1) \cdot s^i) =$
Now we say $i = k+1$
$U(s) - 1 - \sum_{k = 0}^{\infty} (\mathbb{P}(Y>k) \cdot s^{k+1}) = U(s) - 1 - s \cdot \sum_{k = 0}^{\infty} (\mathbb{P}(Y>k) \cdot s^{k})=$
$ U(s) - 1 - s \cdot U(s) = (1-s) \cdot U(s) -1 = G_Y(s)$
So $1- G_Y(s) = 2 - (1-s) \cdot U(s)$ Where did I go wrong?
The error is $$P(Y=i)=P(Y>i)-P(Y>i-1),$$ which should be the other way around: $$P(Y=i)=P(Y>i-1)-P(Y>i).\tag1$$ Argue this way: If $Y$ is greater than $i-1$, then either it's equal to $i$, or greater than $i$. Now rearrange that statement to get (1).