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Suppose There is a random walk starting in origin while probability to move right is 1/3 and probability to move left 2/3.What is the probability to return to the origin.

Thank you

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    If you let the random walk continue ad infinitum? One, I would think.2011-01-18
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    @Noldorin: No, an asymmetric walk like this one goes off to leftward infinity with probability 1. Thus it almost surely returns to the origin only finitely many times, and has a positive probability of never returning at all.2011-01-18
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    @Chris: Ah right. The analysis seems very complex indeed for the asymmetric case.2011-01-18
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    @Noldorin: proving what I've just said isn't very hard. See e.g. sections 1.5 and 1.6 of [this book](http://www.statslab.cam.ac.uk/~james/Markov/). I can't see immediately how to adapt that kind of approach to get a precise probability rather than an estimate though.2011-01-18
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    @Chris: Given I have zero formal training in Markov chains, I would consider it very difficult. :)2011-01-18

3 Answers 3

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Let $P_{i\ge 0}$ be the probability of ever reaching position $x+i$ when starting from position $x$ (this is independent of $x$, since the transition probabilities are). Clearly $P_0=1$, and $P_i\rightarrow 0$ as $i\rightarrow\infty$ provided that the right-hop probability $q < 1/2$ (in this case $q = 1/3$). Otherwise, $$ P_i = qP_{i-1} + (1-q)P_{i+1}, $$ You can guess the solution to be of the form $P_i = \alpha^i$ for some $0 < \alpha < 1$. This turns out to satisfy the conditions if $$ \alpha = q + (1-q) \alpha^2, $$ which has the solution $\alpha = q/(1-q)$ in this case.

For the problem specified, you want to know the total probability of returning to the origin after the first step. If the first step is to the right (which happens with probability $q$), then you must return to the origin; if it is to the left (with probability $1-q$), then you will return to the origin with probability $P_1 = \alpha = q/(1-q)$. So the solution is $$ P_{\text{return}} = q + (1-q)\alpha = q + (1-q)\frac{q}{1-q} = 2q $$ for general $q<1/2$, and $P_{\text{return}} = 2/3$ in this case.

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    I am not sure. I must miss something, but with your computation: $P_{1}=\frac{\frac{1}{3}}{\frac{2}{3}}=0.5$. But I think it should be $\frac{1}{3}\cdot\frac{2}{3}+\frac{2}{3}\cdot\frac{1}{3}=0.4444$. What is the problem with my computation?2011-01-18
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    "i" could be negative so we could get probability greater then one which is not possible2011-01-18
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    @daroczig: Your $P_n$ is the probability of returning to the origin from the origin in exactly $2n$ steps. My $P_n$ is the probability of _ever_ returning to the origin starting from position $(-n)$. And my calculation is specifically for non-negative $n$... the probability of returning to the origin starting from any position to the right of the origin is 1.2011-01-18
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The problem you refer to is the study of return probability for a random walk on a lattice. In your case, the lattice is the 1-D lattice of integers. This problem is very well studied on general lattices. A famous theorem by Polya says that while the probability of return is $1$ for symmetric random walks (i.e., all moves are equally likely unlike in this question) in $1$ and $2$ dimensional lattices, it is strictly less than $1$ in all higher dimensions. See here for more details.

The solution posted by mjqxxxx is very clever and is perfectly valid. If you are interested in other solutions, a very systematic way of studying this problem is through the use of generating functions. See this lecture notes for more information (In fact, these notes have the solution to the problem you posed).

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You can find a detailed answer for your question on WolframMathWorld.

To give an exact answer, you should define the number of steps you are interested in returning to the origin. Of course the number of steps should be even.

I have made a little "simulation" (rather: test run) in R:

results <- NULL for (i in 1:50) {     N=i*2     n1=N/2     p=1/3     q=1-p     results <- c(results, ((factorial(N)/(factorial(n1)*factorial(N-n1)))*(p^n1)*(q^(N-n1)))) } 

And got the following results (sorry for ugly table):

   N| probability ---+--------------   1 0.4444444444   2 0.2962962963   3 0.2194787380   4 0.1707056851   5 0.1365645481   6 0.1112748170   7 0.0918458807   8 0.0765382339   9 0.0642543198  10 0.0542592034  11 0.0460381120  12 0.0392176509  13 0.0335193598  14 0.0287308798  15 0.0246872745  16 0.0212584864  17 0.0183406549  18 0.0158499487  19 0.0137180842  20 0.0118890063  21 0.0103163865  22 0.0089617094  23 0.0077927908  24 0.0067826142  25 0.0059084106  26 0.0051509221  27 0.0044938086  28 0.0039231662  29 0.0034271337  30 0.0029955687  31 0.0026197805  32 0.0022923080  33 0.0020067342  34 0.0017575319  35 0.0015399328  36 0.0013498176  37 0.0011836238  38 0.0010382665  39 0.0009110715  40 0.0007997183  41 0.0007021917  42 0.0006167398  43 0.0005418386  44 0.0004761612  45 0.0004185515  46 0.0003680018  47 0.0003236328  48 0.0002846770  49 0.0002504641  50 0.0002204084 

As you can see here, I computed the probabilities of getting back to the original point for all possible steps (N), where N<=100. That is done by computing the probability for all even N, where the number of steps to the right is equal to the number of steps to the left with the following formula: $$P(n_{1}|N)=\frac{N!}{n_{1}!^2}p^{n_{1}}(1-p)^{n_{1}}$$

where N is the number of steps, $n_{1}=\frac{N}{2}$ and $p=\frac{1}{3}$.

Note: the sum of all above is 1.998366. That is higher than 1, as the probability of arriving back to the original point could also include a lower N's probability.

That's all I could do, I like to play with simulations, but not good at all with Markov chains extrapolated to infinite :)

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    The number of steps doesn't play a role.I am not limited by it2011-01-18
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    @Yakov: that sounds tough :) I am trying to work out a solution, but see no easy way. Till then, please see my edited answer above, that may help finding the way.2011-01-18