# Applications of the Markov Property

The Markov property for a discrete state time stochastic process is defined as:

$$\mathbb{P}(X_n=x_n\mid X_{n-1}=x_{n-1}, \dots, X_0=x_0)=\mathbb{P}(X_n=x \mid X_{n-1}=x_{n-1})$$

A corollary is

$$\mathbb{P}(x_3 \mid x_0, x_1)=\mathbb{P}(x_3 \mid x_1)$$

We technically ignore state $$2$$

Now I am having an issue proving this. First I want to make sure certain steps are correct.

1. I start with $$\mathbb{P}(x_3 \mid x_0, x_1) = \sum_{x_2} \mathbb{P}(x_3, x_2 \mid x_0, x_1)$$

so $$\ \mathbb{P}(x_3 \mid x_0, x_1) = \sum_{x_2} \mathbb{P}(x_3 \mid x_0, x_1, x_2) \mathbb{P}(x_2 \mid x_0, x_1)$$

Is it true that $$\ \sum_{x_2} \mathbb{P}(x_2 \mid x_0, x_1) = 1 \ \ \ (\Psi)$$

1. If $$(\Psi)$$ is correct, then

$$\ \mathbb{P}(x_3 \mid x_0, x_1) = \mathbb{P}(x_3 \mid x_0, x_1, x_2)$$

Is it now true that:

$$\ \mathbb{P}(x_3 \mid x_0, x_1, x_2) = \mathbb{P}(x_3 \mid x_1, x_2)$$

If yes, why? It appears the M.P. only applies to the last state, and not a combination of states.

I don't see how you come to the conclusion in 2 from $$\Psi$$. Note that the event on the left hand side $$X_2=x_2$$ is not properly defined (where does $$x_2$$ come from?).
If you take another step and apply the Markov property you get: \begin{align*} P(X_3=x_3|X_0=x_0,X_1=x_1) &= \sum_{x_2}P(X_3=x_3|X_0=x_0,X_1=x_1,X_2=x_2)P(X_2=x_2|X_0=x_0,X_1=x_1) \\ &= \sum_{x_2}P(X_3=x_3|X_1=x_1,X_2=x_2)P(X_2=x_2|X_1=x_1) \\ &= \sum_{x_2}P(X_3=x_3,X_2=x_2|X_1=x_1) \\ &= P(X_3=x_3|X_1=x_1)\ . \end{align*}