Transition from convolution of PMF's to convolution of power series in a random walk

Question

In the proof that symmetric random walks end up regressing to the origin with probability $1$, I have found this didactic post on-line. In it the following two definitions are given:

1. Probability of returning to the origin at time $2n$:

   $\large p_{2n}\overset{\Delta}{=}P(X_{2n}=0), n=0,1,2,\cdots\tag 1$

2. Probability that the first return happens at time $2n$: $\large f_{2n}\overset{\Delta}{=}P(X_{2n}=0 \cap X_{2k}\neq 0 \, \forall k<n) \tag 2$

The initial conditions are $f_0=0$ and $p_0=0$.

Then it proceeds:

Since a walk that returns to the origin at time $2n$ should have visited the origin at some time between $2$ and $2n$ for the first time, we have the relation:

$$\Large p_n=\displaystyle \sum_{k=1}^n f_k\,p_{n-k}\tag 3$$

and proceeds:

The convolution-like product suggests that the next step is to go to the generating functions of $p_n$ and $f_n$. Define the formal power series $F(x) = \sum_{n=0}^\infty f_n\,x^n$ and similarly for $P(x)$. Then, the above convolution can be written in this domain as:

$$\large P(x) − 1 = P(x)F(x)\tag 4$$

with the $-1$ being the consequence of the initial conditions.

I understand every point up to equation $(3)$ but I would like to get some background context or explanation about the transition from the convolution of pmf's in $(3)$ to the convolution of power series in $(4)$, as well as the role of the $-1$ in $(4)$.

Are there any assumptions or requisites allowing such seamless transition? Is it always possible? What are the mathematical circumstances where this move is advisable?

Answer 1

The correct initial conditions are $p_0=1$ and $f_0=0$. The "1" is the first term $p_0 x^0$ which has to be separated out since the convolution formula (3) is only valid for $n\geq 1$. The rest is simple algebra:

\begin{eqnarray*} P(x)&=&\sum_{n=0}^\infty p_n x^n\\[5pt] &=&1+\sum_{n=1}^\infty p_n x^n\\[5pt] &=&1+\sum_{n=1}^\infty \sum_{k=1}^n f_k p_{n-k} x^n\\[5pt] &=&1+\sum_{k=1}^\infty \sum_{n=k}^\infty f_k p_{n-k} x^n\\[5pt] &=&1+\sum_{k=1}^\infty f_k x^k \sum_{n=k}^\infty p_{n-k} x^{n-k}\\[5pt] &=&1+\sum_{k=0}^\infty f_k x^k \sum_{j=0}^\infty p_{j} x^{j}\\[5pt] &=&1+F(x) P(x). \end{eqnarray*}