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What is the number of N step random walks starting from a point (x0,y0) to a point (x1,y1) assuming each direction (right,left,up,down) has equal probability. I know the expression for the one dimension case is:

$$ \binom{N}{\frac{N+(y1-y0)}{2}} $$

is there a similar expression for two dimensions? I saw this post (Probability Distribution of a 2D lattice random walk) but the answer given is clearly wrong.

Anthony Gurunian
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2 Answers2

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The linked answer has a formula, but I wanted to provide a different perspective. I am generalizing the method used by Brian M. Scott in this answer.

Suppose the vertices your walk visits are $(x_0,y_0),(x_1,y_1),\dots,(x_N,y_N)$. (I am using different notation for the final vertex). Note that

  • The sums $x_0+y_0, x_1+y_1,\dots,x_N+y_N$ are a one-dimensional random walk from $x_0+y_0$ to $x_N+y_N$. This is because the values change by $\pm1$ each time.

  • The differences $x_0-y_0,x_1-y_1,\dots,x_N-y_N$ are also a $1$D random walk from $x_0-y_0$ to $x_N-y_N$.

  • Any pair of $1$D random walks can be generated in this fashion. You just need to check that the four possible choices of north, south, east and west for the $2$D walk generate all possible pairs of $\pm1$ for the two $1$D walks described above.

It follows that the number of $2$D random walks is the product of the number of ways to complete these two $1$D random walks, which is $$ \binom{N}{\frac12({N+x_0+y_0-x_N-y_N})}\times \binom{N}{\frac12({N+x_0-y_0-x_N+y_N})}. $$

Mike Earnest
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Here's another approach, less slick than that of @MikeEarnest (+1):

Wlog we are taking $N$ steps from $(0,0)$ to $(x,y)$. Suppose $k$ of these steps are horizontal (either left or right). Use the one-dimensional formula to count the number of ways to make $k$ (horizontal) steps starting at $0$ and ending at $x$. Next, multiply by the number of ways to embed a given sequence of $k$ horizontal steps into your total $N$ steps. Then given that embedding, count the ways to make $N-k$ (vertical) steps starting at $0$ and ending at $y$. This yields the sum $$\sum_k{k\choose \frac12(k+x)}{N\choose k}{N-k\choose\frac12(N-k+y)}.$$ Change index of summation to $a:=\frac12(k-x)$ and define $m:=\frac12(N-x-y)$. This gives $$ \sum_a\frac{N!}{(x+a)!a!(m+y-a)!(m-a)!}={N\choose m}\sum_a{m\choose a}{N-m\choose m+y-a}$$ after some rearranging along with the identity $N-m=(x+a)+(m+y-a)$. The rightmost sum evaluates to ${N\choose m+y}$, yielding $${N\choose m}{N\choose m+y}={N\choose \frac12(N-x-y)}{N\choose\frac12(N-x+y)}.$$

grand_chat
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