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I randomly saw a question posted here: Number of N step Random walks on 2D infinite grid or lattice. Both answers to the original post seems to agree with each other. However, I am getting confused about why the final answer is not symmetric e.g. $(x, y)$ or $(y, x)$ after $N$ steps should have the same number of paths since we can think we just change the label?

In addition to that, it is also able to derive the marginal distribution of $x$ and $y$?

DA_PA
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    Those formulae are symmetric under switching $x$ and $y$! The factor of $\binom{N}{\frac12(N-x-y)}$ is unchanged when you switch $x$ and $y$. For the other factor, using the fact that $\binom{N}k=\binom{N}{N-k}$ and letting $k=\frac12(N-x+y)$, you get $\binom{N}{\frac12(N-x+y)}=\binom N{N-\frac12(N-x+y)}=\binom{N}{\frac12(N-y+x)}$, so the other factor is unchanged as well. $\tag*{}$ Deriving the marginal distribution seems to be not so simple. – Mike Earnest Nov 14 '24 at 04:32
  • @MikeEarnest the marginal distributions are not hard and the parity complication goes away: ${2N \choose N-x}={2N \choose N+x}$ or ${2N \choose N-y}={2N \choose N+y}$, dividing by $4^N$ if you want probabilities. – Henry Nov 15 '24 at 02:19

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Those formulae are symmetric under switching $x$ and $y$! The factor of $\binom{N}{\frac12(N-x-y)}$ is unchanged when you switch $x$ and $y$. For the other factor, using the fact that $\binom{N}k=\binom{N}{N-k}$ and letting $k=\frac12(N-x+y)$, you get $\binom{N}{\frac12(N-x+y)}=\binom N{N-\frac12(N-x+y)}=\binom{N}{\frac12(N-y+x)}$, so the other factor is unchanged as well.

Here is how we derive the marginal distribution (thanks to Henry for the idea). Let $(X,Y)$ denote the coordinates of the random walk after $n$ steps. $$ \begin{align} P(X=x) &=\sum_{y} P(X=x,Y=y)=\sum_y 4^{-n}\binom{n}{\frac12(n-x-y)}\binom{n}{\frac12(n-x+y)}=4^{-n}\binom{2n}{n-x} \end{align} $$ The last step follows from Vandermonde's identity, which states $\sum_k\binom ai\binom b{k-i}=\binom{a+b}k$.

Mike Earnest
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