Given: Let $X$ and $Y$ be two discrete random variables with values in $\mathbb{N}_0$, such that $P(Y = k) = \frac{P(X > k)}{E[X]}$ for all $k \in \mathbb{N}$.
Show that the generating function $g_Y$ of $Y$ is given by $g_Y(z) = \frac{1}{E[X]} \frac{1 - g_X(z)}{1 - z}, \quad z \in \mathbb{C}$.
Attempt:
We start with the expression for the generating function $g_Y(z)$:$g_Y(z) = E(z^Y) = \sum_{k=0}^{\infty} z^k P(Y=k) = \sum_{k=0}^{\infty} z^k \frac{P(X > k)}{E(X)} = \frac{1}{E(X)} \sum_{k=0}^{\infty} z^k P(X > k)$
Now, using the complementary probability, we rewrite the sum:$ = \frac{1}{E(X)} \sum_{k=0}^{\infty} z^k (1 - P(X \leq k))$
We can factor out:$ = \frac{1}{E(X)} \left( \sum_{k=0}^{\infty} z^k - \sum_{k=0}^{\infty} z^k P(X \leq k) \right)$
Using the geometric series formula, we rewrite the first sum:$ = \frac{1}{E(X)} \left( \frac{1}{1 - z} - \sum_{k=0}^{\infty} z^k P(X \leq k) \right)$
At this point, I'm not sure how to proceed further. I would appreciate any hints on how to continue.