# Equivalent Definitions of the Markov Property

Assume we have a stochastic process $$\{X_n\}_\mathbb{N}$$ defined on some underlying probability space that takes values in another measurable space. One of the many definitions that I have seen of the Markov property is as follows:

The process has the Markov property iff for arbitrary $$n > s$$ and $$A$$ measurable set

$$\mathbb{P}(X_n \in A| \: \sigma(X_1,\dots,X_s)) = \mathbb{P}(X_n \in A| \: \sigma(X_s)) \tag{1}$$

Is it possible to define the Markov property as

$$\mathbb{P}(X_n \in A| \: \sigma(X_1,\dots,X_{n-1})) = \mathbb{P}(X_n \in A| \: \sigma(X_{n-1}))$$

and then deduce that $$(1)$$ holds?

Since $$\mathbb{P}(X_n \in A| \: \sigma(X_1,\dots,X_{n-1}))$$ = $$\mathbb{E}(\mathbb{1}_{X_n \in A}|\: \sigma(X_1,\dots,X_{n-1}))$$ I have been trying to use properties of conditional expectation but have not been successful.

Thanks!

Yes, they are equivalent. Let's assume that

$$\mathbb{P}(X_n \in A \mid \sigma(X_1,\ldots,X_{n-1}) = \mathbb{P}(X_n \in A \mid \sigma(X_{n-1})) \tag{1}$$

holds for all measurable sets $$A$$ and all $$n \in \mathbb{N}$$. By a standard approximation procedure, this implies $$\mathbb{E}(f(X_n) \mid \sigma(X_1,\ldots,X_{n-1}) ) = \mathbb{E}(f(X_n) \mid \sigma(X_{n-1})) \tag{1'}$$ for any bounded Borel-measurable function $$f$$.

For fixed $$n \in \mathbb{N}$$ we prove $$\mathbb{P}(X_n \in A \mid \sigma(X_1,\ldots,X_{n-k})) = \mathbb{P}(X_n \in A \mid \sigma(X_{n-k})), \qquad A \in \mathcal{A}, \tag{2}$$ by induction over $$k=1,\ldots,n$$.

Base: For $$k=1$$ this is nothing but $$(1)$$.

Inductive step: Assume that $$(2)$$ holds for some $$k=1,\ldots,j$$; we have to show that $$(2)$$ holds for $$k=j+1$$. By the tower property of conditional expectation, we have

$$\mathbb{P}(X_{n} \in A \mid \sigma(X_1,\ldots,X_{n-j-1})) = \mathbb{E} \bigg[ \mathbb{P}(X_k \in A \mid \sigma(X_1,\ldots,X_{n-j})) \mid \sigma(X_1,\ldots,X_{n-j-1}) \bigg].$$

Using our induction hypothesis, we find

$$\mathbb{P}(X_{n} \in A \mid \sigma(X_1,\ldots,X_{n-j-1})) = \mathbb{E} \bigg[ \mathbb{P}(X_n \in A \mid \sigma(X_{n-j})) \mid \sigma(X_1,\ldots,X_{n-j-1}) \bigg].$$

By the factorization lemma, there exists a measurable function $$f$$ such that

$$\mathbb{P}(X_n \in A \mid \sigma(X_{n-j})) = f(X_{n-j}),$$

and so

$$\mathbb{P}(X_{n} \in A \mid \sigma(X_1,\ldots,X_{n-j-1})) = \mathbb{E}(f(X_{n-j}) \mid \sigma(X_1,\ldots,X_{n-j-1})).$$

It follows from $$(1')$$ that

$$\mathbb{P}(X_{n} \in A \mid \sigma(X_1,\ldots,X_{n-j-1})) = \mathbb{E}(f(X_{n-j}) \mid \sigma(X_{n-j-1})). \tag{3}$$

If we take on both sides the conditional expectation with respect to $$\sigma(X_{n-j-1})$$, then we find that

$$\mathbb{P}(X_n \in A \mid \sigma(X_{n-j-1})) = \mathbb{E}(f(X_{n-j}) \mid \sigma(X_{n-j-1})). \tag{4}$$

Combining $$(3)$$ and $$(4)$$ we get

\begin{align*} \mathbb{P}(X_{n} \in A \mid \sigma(X_1,\ldots,X_{n-j-1})) &\stackrel{(3)}{=} \mathbb{E}(f(X_{n-j}) \mid \sigma(X_{n-j-1})) \\ &\stackrel{(4)}{=} \mathbb{P}(X_n \in A \mid \sigma(X_{n-j-1})), \end{align*}

i.e. $$(2)$$ holds for $$k=j+1$$.