# Markov Chains: Example to show past and future are not independent given any information about past.

The Markov property does not imply that the past and the future are independent given any information concerning the present. Find an example of a homogeneous Markov chain $\left\{X_n\right\}_{n\ge0}$ with state space $E=\left\{1,2,3,4,5,6\right\}$ such that

$P(X_2=6\mid X_1\in\{3,4\},X_0=2)\neq P(X_2=6\mid X_1\in\{3,4\})$.

I don't understand the phrasing of this question. How should I approach this?

Could this work?

$P(X_2=6\mid X_1\in\{3,4\}) = 1$

But $P(X_2=6\mid X_1\in\{3,4\},X_0=2) =0$

## Introduction: Natural language represented using Markov model

A very simple and practical example would be that of natural language.

Now first of all, natural language is just a terminology for any language you speak and have acquired or learned, such as English. When you communicate, you have a set of words {$W$} and a set of operations you can perform on these words. You then proceed further to arrange the words that you know in a certain order that produces meaningful elements of communication/semantics -- "sentences".

When you use words sentences, they are obviously not arbitrarily placed words, they have some order. In fact, sentence formation can be well modeled as a Markov chain model with order "$k$"$=1$ (reference).

## Sentence formation and conditional probability

Now establish a relation between your possible states $[1, 6]$ and certain six words. It can be experimentally determined that $P(C | B) \geq P(C|A,B)$ where $A,B$ is an ordered occurrence of the words $A$ and $B$ in a sentence. This relation always holds in case of a natural language as $P (C | B)$ already includes all possible trigrams that can be formed as "$X B C$" where $X$ is a general word of the language. In computational linguistics, this is also termed as the transition probability.

I'll give you an example of that as well:
Let let us, however, consider the backward probability $P(A|$"$A,B$"$)$ i.e., the probability that $A$ is the word processing $B$. Now let $A$ be "chocolate" and $B$ be "chip". Now there will exist a certain probability in the historical literature of the language for $P(A|$"$A,B$"$) = p_{k=1}$. However, let us now consider another word $C$, that represents "cookie". When you compute the new probability, $P(A|A,B,C")=p_{k=2}$, then you find that $p_{k=2} \leq p_{k=1}$, because there are more words that you can say after saying "chocolate chip" other than simply "cookie". I have denoted it as "$\leq$" because theoretically it is possible that only one such meaningful triplet can be formed in a natural language, although that's rarely ever the case. You cohld simply replace is with a strict inequality relation for all practical purposes.

I think this provides a sufficient as well as intuitive example for the problem thay you have stated. Let me know if something from the answer is unclear.

Suppose that $X_0$ is equally likely to be in states $1$ and $2$, each with probability $\frac{1}{2}$.

Suppose that from state $3$, there is a $100\%$ chance of going to state $6$ in the next step.

Suppose that from state $4$, there is a $100\%$ chance of going to state $5$ in the next step.

Suppose that from state $1$, there is a $100\%$ chance of going to state $3$.

Suppose that from state $2$, there is a $100\%$ chance of going to state $4$.

Now, the LHS of your equation will evaluate to $0$, because given that you were in state $2$ at $X_0$, you must go to state $4$ in $X_1$ and therefore state $5$ in $X_2$, which means there's a $0\%$ chance of being in state $6$ at $X_2$.

The RHS of your equation will evaluate to $\frac{1}{2}$, because we are not conditioning on any value of $X_0$, so we use its initial probabilities: $0.5$ chance of $X_0=1$ and $0.5$ chance of $X_0=2$. This means that there's a $0.5$ chance of $X_3=6$ and $0.5$ chance of $X_3=5$.

It will probably help you to draw the directed graph representing the Markov chain's transition probabilities. For good practice, also try to write out the transition matrix (it will be a $6\times 6$ matrix).

• I think your solution works. Do you think the example I provided in the problem description would work? Feb 23, 2017 at 18:18
• What example? edit: oh I see you added a picture, I'll look at it Feb 23, 2017 at 18:18

The key point here is that the Markov property requires you to KNOW the value of at the previous time in order to be independent of prior history; just having some information about the previous time isn't enough to break dependence.

Try thinking of transition matrices in which the following conditions hold:

1. You are very likely to reach state 3 and unlikely to reach state 4 from a randomly chosen starting position, but when you start specifically from state 2 it is reversed (very unlikely to reach state 3, very likely to reach state 4).

2. The probability of transitioning to state 6 from state 3 is very different from the probability of transitioning to state 6 from state 4.