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From Wikipedia

In a Markov process, one assumes that $i_1 < \cdots < i_n$. Then, because of the Markov property, $$ p_{i_1,\ldots,i_n}(f_1,\ldots,f_n)=p_{i_1}(f_1)p_{i_2;i_1}(f_2\mid f_1)\cdots p_{i_n;i_{n-1}}(f_n\mid f_{n-1}), $$ where the conditional probability $p_{i;j}(f_i\mid f_j)$ is the transition probability between the times $i>j$. So, the Chapman–Kolmogorov equation takes the form $$ p_{i_3;i_1}(f_3\mid f_1)=\int_{-\infty}^\infty p_{i_3;i_2}(f_3\mid f_2)p_{i_2;i_1}(f_2\mid f_1) \, df_2. $$

I was wondering if the reverse is true. I.e., if a process satisfies the above form of the Chapman–Kolmogorov equation, will it be Markovian? Thanks and regards!

Michael Hardy
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Tim
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1 Answers1

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Yes, if the initial (unconditional) distribution holds for any subsequent propagation such that $$p_{i_2} (f_2) = \int_{-\infty}^\infty p_{i_2;i_1} (f_2|f_1) p_{i_1}(f_1) df_1 $$ then the Chapman-Kolmogorov uniquely defines a Markov process.

To see this, multiply the CK equation by the unconditional PDF of the conditioned variable, $p_{i_1} (f_1)$, to get the the joint density $$ p_{i_3,i_1} (f_1, f_3) = p_{i_1} (f_1) \cdot \int_{-\infty}^\infty p_{i_3;i_2} (f_3|f_2) \cdot p_{i_2;i_1}(f_2|f_1) \ df_2\\ $$ Differentiate by $f_2$. The lefthand side becomes a joint probability distribution of 3 variables (and the above joint probability distribution of 2 variables can now be seen as a marginal probability distribution), and the right-hand side yields the definition of a Markov process $$ p_{i_3,i_2, i_1} (f_1,f_2, f_3) = p_{i_3;i_2} (f_3|f_2) \cdot p_{i_2;i_1}(f_2|f_1) \cdot p_{i_1} (f_1) $$ This can be repeated to get the Markov definition to the $n^{th}$ order definition you gave.

Tony Beans
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