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Regarding mixing time of a Markov chain

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
832
Let $X=X_0, X_1, X_2, \ldots$ be a Markov chain on a finite state space $S$, and let $P$ denote the transition matrix.
Assume that there is an $\varepsilon>0$ such that whenever $\mu_0$ and $\nu_o$ are point distributions on $S$ (in other words, $\mu_0$ and $\nu_0$ are Direac masses) we have
$$\|\mu_0P-\nu_0P\|_{TV}\leq \varepsilon$$

Now let $Y=Y_0, Y_1, Y_2, \ldots$ be an independent copy of $X$.

Question. What can we say about the magnitude of $P[X_1\neq Y_1|X_0=x_0, Y_0=y_0]$, where $x_0$ and $y_0$ are two different states in $S$.

I intuitively think that we should be above to bound the above quantity by $\varepsilon$ up to multiplication by an absolute constant. But I am unable to prove it.

We have that
$$P[X_1\neq Y_1|X_0=x_0, Y_0=y_0] = \sum_{x\in S}\sum_{y\in S, y\neq x}P[X_1=x, Y_1=y|X_0=x_0, Y_0=y_0]$$
which is equal to
$$\sum_{x\in S} p(x_0, x)(1-p(y_0, x))$$
but I am unable to make progress.
 

steep

Member
Dec 17, 2018
51
Really a standard exercise/result for countable state random variables (with emphasis on markov chains) is to show, where, for notational reasons it is understood in your particular case that $X_0 = 0 $ and $Y_0 = 0$

$P[X_1\neq Y_1|X_0=x_0, Y_0=y_0] = P[X_1\neq Y_1] \geq d_{TV}(X_1,Y_1)= \|\mu_0P-\nu_0P\|_{TV}$
with $\mu := \mathbf e_0$ $\nu := \mathbf e_0$

and the inequality is reached with equality in the case of a maximal coupling. This is somewhat trivial here but useful for other couplings e.g. if $\nu_0$ was the steady state vector.

(nit: there are couple different formulations of total variation distance I am using the 1/2 L1 norm of difference between probability vectors interpretation for countable state r.v.v)

the inequality you're trying to prove 'points in the wrong way' so to speak. More context / background information is needed to make sense of what you're trying to do. E.g. if you don't know what coupling, that is a big deal and my post won't make any sense.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
832
Really a standard exercise/result for countable state random variables (with emphasis on markov chains) is to show, where, for notational reasons it is understood in your particular case that $X_0 = 0 $ and $Y_0 = 0$

$P[X_1\neq Y_1|X_0=x_0, Y_0=y_0] = P[X_1\neq Y_1] \geq d_{TV}(X_1,Y_1)= \|\mu_0P-\nu_0P\|_{TV}$
with $\mu := \mathbf e_0$ $\nu := \mathbf e_0$

and the inequality is reached with equality in the case of a maximal coupling. This is somewhat trivial here but useful for other couplings e.g. if $\nu_0$ was the steady state vector.

(nit: there are couple different formulations of total variation distance I am using the 1/2 L1 norm of difference between probability vectors interpretation for countable state r.v.v)

the inequality you're trying to prove 'points in the wrong way' so to speak. More context / background information is needed to make sense of what you're trying to do. E.g. if you don't know what coupling, that is a big deal and my post won't make any sense.
I realize now that the statement I want to prove does not hold for the random walk on a big complete graph.