PROBABILITY TO RETURN TO THE ORIGIN IN A SYMMETRIC RANDOM WALK:


To return to the origin, an even number of steps are needed. The probability of being at the origin at a particular step \(2n\) is:

\[\color{turquoise}{\large p_{2n}\overset{\Delta}{=}P(X_{2n}=0, n=0,1,2,\cdots)= {2n \choose n}\,\Big(pq\Big)^n} \tag 1\]

\(p\) and \(q\) are the Berouilli probabilities to turn right or left. They are equal in a symmetric random walk.

The first return happening at step \(2n\) will have a probability:

\[\color{turquoise}{\large f_{2n}\overset{\Delta}{=}P(x_{2n}=0 \cap X_{2k}\neq 0 \, \forall k<n)} \tag 2\]

The initial state is \(f_0=0\) and \(p_0=1\).

There are \(\large p_{2n}\,2^{2n}\) paths of length \(2n\) between point \((0,0)\) and \((2n,0)\). In the case of a symmetric random walk all paths have the same probability: \(\large (\frac{1}{2})^{2n}\). Therefore:

\[\large p_{2n}={2n \choose n}\,2^{-2n}.\]

We can partition these paths depending on the time of the first pass. If the first return is on step \(2k\), there is a segment from \((0,0)\) to \((2k,0)\) with no returns, and from \((2k,0)\) to \((2n,0)\) without restrictions.

The number of paths with first pass at \(2k\) are:

\[\large f_{2k}\,2^{2k} \times p_{2n-2k}\,2^{2n-2k}= f_{2k}\,p_{2n-2k}\,2^{2n}\]

Summing over \(k\):

\[\large p_{2n}\, 2^{2n}=f_0\,p_{2n}\,2^{2n}+f_2\,p_{2n-2}\,2^{2n}+\cdots+f_{2n}p_0\, 2^{2n}\]

Dividing both sides by \(2^{2n}\):

\[\color{blue}{\large p_{2n}=f_0\,p_{2n}+f_2\,p_{2n-2}+\cdots+f_{2n}p_0}\tag 2\]

This relates the probability of return to origin at time \(2n\) to first and other returns.


Based on this theorem we can set up the following convolution:

\[\large p_{n}=\displaystyle\sum_{k=1}^n\,f_k\times p_{n-k}\]

and set up the following generating functions from the power series (credit to this post):

\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] &=&\Large \color{orange}{1+F(x) P(x)}\tag 3 \end{eqnarray*}

Now,

\[\large \color{blue}{\displaystyle\sum_{n=0}^\infty\,{2n\choose n}\,x^n=\frac{1}{\sqrt{1-4x}}=(1-4x)^{-1/2}}\tag 4\]

PROOF (thanks this post)

To see this we get the Taylor series, \(\large f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\), of this expression evaluated at \(0\):

\(f(x)=(1-4x)^{-1/2}\); evaluated at \(f(0)=1.\)

\(f^1(x)=\frac{2}{(1-4x)^{3/2}}\); evaluated at \(f^1(0)=2\); coefficient \(2/1!\).

\(f^2(x)=\frac{12}{(1-4x)^{5/2}}\); evaluated at \(f^2(0)=12\); coefficient \(12/2!=6\).

\(f^3(x)=\frac{120}{(1-4x)^{7/2}}\); evaluated at \(f^3(0)=120\); coefficient \(120/3!=20\)

\(f^4(x)=\frac{1680}{(1-4x)^{9/2}}\); evaluated at \(f^4(0)=1680\); coefficient \(1680/4!=70\).

The Taylor series is:

\(1+2 x+6 x^2+20 x^3+70 x^4+O\left(x^5\right)\)

and the coefficients are \(\Large \frac{2n!}{n!n!}={2n\choose n}.\)

\(\square\)


Consequently, from equation \((1)\):

\[\color{purple}{\large P(x)= \displaystyle\sum_{n=0}^\infty=\,p_n\,x^n=\binom{2n} {n}\,(pqx)^n=(1-4pqx)^{-1/2}}\tag 5\]

and going back to equation 3:

\[F(x)=1-\frac{1}{P(x)}\]

and

\[\color{purple}{\large F(x)=1 - (1-4pqx)^{1/2}}\tag 6\]


Next:

\[\large \color{red}{\sqrt{1-x}=\displaystyle\sum_{n=0}^\infty\frac{-1}{2n-1}\,\binom{2n}{n}\,\frac{x^n}{4^n}}\tag 7\]


PROOF (thanks to this post):

Let \(f(x)=(1-x)^{1/2}\). Differentiating:

\begin{align} f^{(1)}(x)&=(-1)\left(\frac12\right) (1-x)^{-1/2}\\\\ f^{(2)}(x)&=(-1)^2\left(\frac12\right) \left(-\frac12\right) (1-x)^{-3/2} \\\\ f^{(3)}(x)&=(-1)^3\left(\frac12\right) \left(-\frac12\right) \left(-\frac32\right) (1-x)^{-5/2}\\\\ f^{(4)}(x)&=(-1)^4\left(\frac12\right) \left(-\frac12\right) \left(-\frac32\right)\left(-\frac52\right) (1-x)^{-7/2}\\\\ \vdots\\\\ f^{(n)}(x)&=-\frac{(2n-3)!!}{2^n}(1-x)^{-(2n-1)/2} \\\\ \end{align}

Therefore \(f(x)\) has the series expansion:

\(\large f(x)=1-\sum_{n=1}^\infty \frac{(2n-3)!!}{2^n\,n!}\,x^n \tag {*}\)

with

\begin{align} (2n-3)!!&=(2n-3)(2n-5)(2n-7)\cdots(5)(3)(1)\\\\ &=\frac{(2n-3)!}{(2n-4)(2n-6)(2n-8)\cdots (6)(4)(2)}\\\\ &=\frac{(2n-3)!}{2^{n-2}(n-2)!}\\\\ &=\frac{(2n)!}{(2n-1)2^n\,n!}\\\\ &=\frac{n!}{(2n-1)2^n}\binom{2n}{n} \tag {**} \end{align}

Substituting (**) into (*):

\begin{align} f(x)&=1-\sum_{n=1}^\infty \frac{1}{4^n(2n-1)}\binom{2n}{n}\,x^n\\\\ &=-\sum_{n=1}^\infty \frac{1}{4^n(2n-1)}\binom{2n}{n}\,x^n \end{align}

\(\square\)


Now substituting into \((6)\) into \((5)\):

\[\large F(x)=\displaystyle \sum_{n=0}^\infty f_n x^n= \sum_{n=1}^\infty \frac{1}{2n-1}{2m\choose n}(pq)^n\,x^n\tag 8\]

Therefore,

\[\color{LimeGreen}{\large f_n=\frac{1}{2n-1} \binom{2n}{n} (pq)^n} \tag 9\]


A probability generating function for a random variable \(X\) taking values in \(\{0,\infty\}\) will converge at \(x=1\). So at \(x=1\) the sum of all the \(f_n\) is simply the probability that the random walk ever returning to the origin (by equation \((5)\)):


\[\color{red}{\large F(1)=\displaystyle \sum_{n=1}^\infty f_n= 1 - \sqrt{1-4pq}=1-}\color{red} {|p-q|}\tag {10}\]


Here I resort to this post:

\((p-q)^2=p^2-2pq+q^2=p(1-q)-2pq+q(1-p)=p+q-4pq=1-4pq\)

With probability \(|p-q|\) the walk will never return to the origin. If \(p=q\), the probability of not returning is zero.


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