First Returns of the Symmetric Random Walk when p=q
As we know that a related probability is the one for the event in which the first visit to position x occurs at the nth step given that the walk starts at the origin, i.e. the first passage through x. For case x=0 we will find the probability of the first return to the origin. Let A be the event that $X_n=0$, let $B_k$ be the event that the first visit to the origin occurs at the kth step. By the law of total probability
\[P(A)=\sum_{k=1}^n P(A|B_k)P(B_k) \tag{1}\label{eqn2.341}\]
Let $f_k=P(B_k)$. The conditional probability $P(A|B_k)$ is the probability that the walk returns to the origin after n–k steps i.e. $P(A|B_k)=p_{n-k}$. Note that $p_n$ is the probability that the walk is at the origin at step n and $p_n=0$ if n is odd. Hence (\ref{eqn2.341}) can be written as
\[p_n=\sum_{k=1}^n p_{n-k}f_k \tag{2}\label{eqn2.342}\]
Multiplying both sides of (\ref{eqn2.342}) by $s^n$ and summing for $n\geq1$, we have
\[\sum_{n=1}^\infty p_ns^n =\sum_{n=1}^\infty \sum_{k=1}^n p_{n-k}f_k s^n \tag{3}\label{eqn2.343}\]
as
\[H(s)=\sum_{n=0}^\infty p_ns^n=1+\sum_{n=1}^\infty p_ns^n\]
\[H(s)-1=\sum_{n=1}^\infty p_ns^n\qquad \]
Therefore from (\ref{eqn2.343})
\begin{align*}
H(s)-1&=\sum_{n=1}^\infty p_n s^n\\
&=\sum_{n=1}^\infty \sum_{k=1}^n p_{n-k}f_ks^n\tag{4}\label{eqn2.344}
\end{align*}
Since $p_0=1$ and $f_0=0$, we can replace (\ref{eqn2.344}) by
\begin{align*}
H(s)-1&=\sum_{n=1}^\infty p_ns^n\\
&=\sum_{n=0}^\infty \sum_{k=0}^n p_{n-k}f_ks^n = \sum_{n=0}^\infty p_ns^n \sum_{k=0}^\infty f_k s^k \tag{5}\label{eqn2.345}
\end{align*}
the product of two power series, also known as Convolution of the distribution.
Now if $Q(s)=\sum_{n=0}^\infty f_ns^n$ is the probability generating function of the first return distribution, then
$H(s)-1=H(s)Q(s)$.
From equation (8) of lecture “Probability Distribution after n steps” we have
\begin{align*}
Q(s)&=\frac{H(s)-1}{H(s)}=1-\frac{1}{H(s)}\\
&=1-\frac{1}{(1-s^2)^{-\frac{1}{2}}}\qquad \because H(s)=(1-s^2)^{-\frac{1}{2}}\\
&=1-(1-s^2)^{\frac{1}{2}} \tag{6}\label{eqn2.346}
\end{align*}
The probability that the walk will, at some step, return to the origin is $\sum_{n=1}^\infty f_n=Q(1)=1$;
i.e. the return is certain. In this walk the origin or any point by translation is persistent. However, the mean number of steps until this return occurs is
\begin{align*}
\sum_{n=1}^\infty nf_n &=\lim_{s \rightarrow 1-} Q'(s)\\
&=\lim_{s \rightarrow 1-} \frac{s}{(1-s^2)^{\frac{1}{2}}}=\infty
\end{align*}
In this walk which starts at the origin is certain to return there in the future, but on average, it will take an infinite number of steps.
Example: Find the probability that a symmetric random walk starting from the origin returns there for the first time after 6 steps.
solution: We require the coefficient of $s^6$ in the power series expansion of the pgf Q(s), which is
\begin{align*}
Q(s)&=1-(1-s^2)^{\frac{1}{2}}\\
&=1-\left[\frac{1^{\frac{1}{2}}}{0!}-\frac{\frac{1}{2}{(-s^2)}^1}{1!}-\frac{\frac{1}{2}(\frac{1}{2}-1){(-s^2)}^2}{2!}-\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2){(-s^2)}^3}{3!}\right]\\
&=1-1+\frac{1}{2}s^2+\frac{1}{8}s^4+\frac{1}{16}s^6+0s^8
\end{align*}
Hence the probability of a first return at step 6 in symmetric random walk example is $\frac{1}{16}$.
Read about Simple Random Walk: Unrestricted Random Walks
