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Let $p = (p_x)_{x \in Z}$ be an i.i.d collection of $(0,1)$ valued random variables with common distribution $\alpha$. For fixed $p$, let $X = (X_n)_{n \in N_0}$ be the Markov chain on $Z$, starting at $X_0 = 0$, with transition probabilities $P(X_{n+1}=y|X_n = x) = p_x$ if $y = x+1, =1-p_x$ if $y=x-1, =0$ otherwise.

What will happen if $p_x = 1/2$ for all $x$? The problem is at that time I am getting probability measure is always zero for any element of the sigma algebra of the probability space which cannot be. What is the mistake I am making?

guest
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  • Not sure what you're doing, but if $p_x$ were zero with probability $1$, then each transition would be to the left, making $X_i=-i$ with probability $1$. Note, though, that $p_x$ can't be zero: the range is $(0,1)$. – mjqxxxx Mar 07 '15 at 15:44
  • Sorry I made a mistake : p_x = 1/2 not 0...sorry again – guest Mar 07 '15 at 16:52
  • By probability measure I mean to say P(A) = ? where A is an element of the sigma algebra. It is coming always 0 if I take p_x=1/2 for all x. But that cant be the case. Right ? – guest Mar 07 '15 at 17:01
  • What do you mean by "for fixed $p$"? It seems you never meantion $p$ again (unless you meant a realization of the already-defined sequence $p= (p_{x})_{x\in Z}$). – Quinn Culver Mar 07 '15 at 17:26
  • p is a tuple (p_x)_{x \in Z}. Hence p is random since p_x are random. Now consider a particular p. I was reading "Random walk in Random environment". Here p is an environment. – guest Mar 07 '15 at 17:36
  • P(A) = \sum_{a_1,a_2,...}P(X_0=0, X_1=a_1,X_2=a_2,...)= \sum_{a_1,a_2,...}p_{0,a_1}p_{a_1,a_2}... = 0 where a=(0,a_1,a_2,...) \in A – guest Mar 07 '15 at 17:44
  • @guest Then I think the common distribution $\alpha$ is not relevant to your question. – Quinn Culver Mar 07 '15 at 17:51
  • @guest If $p_{x}=1/2$ for all $x$, then you just have the standard random walk where the probability of moving left and right is the same (i.e. $1/2$) and independent of position. – Quinn Culver Mar 07 '15 at 17:53
  • Certainly the probability measure is zero for any particular infinite random walk. But what sigma algebra are you using? It should have an element $[X_1=1]$, for instance, whose probability measure is clearly not zero. – mjqxxxx Mar 08 '15 at 19:17

1 Answers1

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Let me see, if I understand your question correctly. If $p_x=0$ for all $x$, then $P(X_{n+1}=x+1|X_n=x)=0$, while $P(X_{n+1}=x-1|X_n=x)=1$, for all $x$. This means that since your chain starts out at $X_0=0$, your next step of the chain will almost surely be $-1$, then $-2$, etc.

I don't get the part, where you write about the "measure always being zero". I don't think there is such a problem here.

Mankind
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  • Sorry I made a mistake : p_x = 1/2 not 0...sorry again – guest Mar 07 '15 at 16:50
  • By probability measure I mean to say P(A) = ? where A is an element of the sigma algebra. It is coming always 0 if I take p_x=1/2 for all x. But that cant be the case. Right ? – guest Mar 07 '15 at 17:01
  • P(A) = \sum_{a_1,a_2,...}P(X_0=0, X_1=a_1,X_2=a_2,...)= \sum_{a_1,a_2,...}p_{0,a_1}p_{a_1,a_2}... = 0 where a=(0,a_1,a_2,...) \in A – guest Mar 07 '15 at 17:45
  • Why is this equal to $0$? And have you noticed that there are uncountably many possible sequences $(a_1,a_2,\ldots)$, and that you therefore may sum over an uncountable set? As a side note, I came to think that it is weird that the $p_x$ are random variables, because it makes your probabilities random as well? Maybe it is the $p$ vector that is random, and by "fixing $p$", you choose a specific instance of the variable. – Mankind Mar 07 '15 at 18:15