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Task
Given a random walk with a probability of going up, $p$. Find the expected time, $T^*$, when +2 or -2 is first achieved (starting from zero).

My attempt
First of all, let's denote $f_{xy}$, the expected time to get from $x$ to $y$. So, the time to get from 0 to 2 is, $f_{02}$. I assume that $T^* = p*f_{02} + (1-p)*f_{0 \ -2}$ (but I'm unsure about this assumption). Now, let's find $f_{02}$. I consider the problem as a Markov's chain, so I can write two equations: $$f_{02} = (1+f_{01})p+ f_{02}(1-p) \\ \mbox{ we can get to 2 from 1 in one go or get back to zero}$$ $$f_{01} = 1\cdot p + (1-p)f_{-1 1}$$ $$f_{-1 1} = f_{02}$$

This results in $f_{02} = 1/p + 1$. Similarly, when can find $f_{0\ -2} $. Is my approach correct?

student
  • 422

1 Answers1

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This approach is not correct, because the expected time to go from 0 to 2 is affected when -2 is also a barrier. If you write $M_k$ for the expected time to reach the boundary $\{-2,2\}$ from $k$, then you can easily find three linear equations in $M_{-1}, M_0$, and $M_1$, namely $$M_{-1}=1+pM_0$$ $$M_0=1+(1-p)M_{-1}+pM_1$$ $$M_1=1+(1-p)M_0 .$$

See also Biased random walk in 1D - expected hitting time for either edge of box where a more general question is discussed, with a reference.

Yuval Peres
  • 21,955