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The move is uniformly random in four directions (u/d/l/r), at 1 unit. What is the expected number of moves? What about from (0,0) to (2, 2)?

I have also an easier version: what is the expected number of moves from 0 to 2 on number axis, if the move is uniformly random in both directions?

I have tried to create equations but it seems I need an infinite number of equations.

Curious George
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    @PeterForeman: The two-dimensional random walk (like the one-dimensional and unlike the three-dimensional) returns to the origin with probability $1$. – joriki Feb 09 '20 at 18:19

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The expected return time to the origin is infinite already for a one-dimensional random walk. Since going from $(0,0)$ to $(1,1)$ includes going from a nearest neighbour of $(1,1)$ (say, $(0,1)$) to $(1,1)$ and the horizontal component of this walk is a one-dimensional walk that tries to return to the origin from its nearest neighbour, the expected time to get from $(0,0)$ to $(1,1)$ is also infinite.

joriki
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  • Yes, that makes sense intuitively. But, is there a way to prove "The expected return time to the origin is infinite already for a one-dimensional random walk"? – Curious George Feb 09 '20 at 19:25
  • @ThePianist: That's a standard fact that you can easily find proofs of by searching this site or the net. E.g. on this site: https://math.stackexchange.com/questions/536, https://math.stackexchange.com/questions/3486291, https://math.stackexchange.com/questions/2109655, – joriki Feb 09 '20 at 19:44
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    @ThePianist - the easy proof comes from using reversibility of the markov chain -- solving the detailed balance equations implies $\pi_i = \pi_j$ for all $i,j$ so there is no way $\sum_{j}\pi_j \lt \infty$ and hence no steady state distribution and hence the chain cannot be positive recurrent -- i.e. expected return times cannot be finite. – user8675309 Feb 09 '20 at 21:57