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A monkey is sitting on $0$ on $\mathbb{R}$, at $t = 0$ . In every period $t\in({1,2,\dots})$ it moves one unit to the right with probability $p$ and one unit to the left with probability $1-p$. Let $P_k$ denote the probability that the monkey will reach positive integer $k$ in some period $t>0$. Find the value of $P_k$ for any positive integer $k$.

My attempt-

Let there be $R$ right and $L$ left steps, then, $$L+R = t$$ The position at $t>0$ is $$R-L = R-(t-R) = 2R-t$$

Now, $$R\sim Bin(t,p)$$ Then, if $X=k$ is the position at time $t>0$ it implies that $2R-t =k $ or $R = \frac{k+t}{2}$

So, $$P_k = {t \choose \frac{k+t}{2}}p^\frac{k+t}{2}(1-p)^\frac{t-k}{2}$$

But I don’t know how to find the answer for a specific $k$, like the question asks me to find $P_0$ when $p =\frac{1}{2}$, can somebody help me out with this.

Also if there is another way to do this question without using markov chain please let me know.

I also tried conditioning on the first step to do something similar to finding the solution to gambler’s ruin but couldn’t work that out.

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  • $\begingroup$ What do you mean by ' I don't know how to find answer to specific k'? $\endgroup$ – Tojrah Mar 25 at 15:35
  • $\begingroup$ The answer given to me says that $P_k$ is always 1 I don’t get how that is, also how do i deal with$(t/2)!$ $\endgroup$ – user601297 Mar 25 at 15:37
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Note that if $p\geq\frac12$, $P_k=1$ for all $k$. So, we are more interested in the cases when $p<\frac12$.

Now when $p<\frac12$, we set up our problem as follows: Let $P_{kj}$ be the probability that, given a random walk with $p<\frac12$, the monkey reaches $k$ before reaching $j$, where $j<0$. If we can find $P_{kj}$ for arbitrary $j$, then $P_k=\lim_{j\to-\infty}P_{kj}$.

For computing $P_{kj}$, I think that using a Markov chain to set up equations is probably the best way to go.

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  • $\begingroup$ Can you tell me why if $p>1/2$ we have $P_k=1$ for all k, I don’t understand why this is $\endgroup$ – user601297 Mar 25 at 17:42
  • $\begingroup$ Because, on average, it should be moving to the right. As such, it should eventually reach all numbers to the right of the origin. $\endgroup$ – Don Thousand Mar 25 at 17:48
  • $\begingroup$ Ok got it, thanks a lot $\endgroup$ – user601297 Mar 25 at 17:50

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