1
$\begingroup$

I am new to Markov Chains and using this concept in statistics.

For a Markov Chain, may I say that $𝑃(𝑞_𝑡|𝑞_{𝑡+1},…,𝑞_𝑇)$ equals to $𝑃(𝑞_𝑡| 𝑞_{𝑡+1})$?

If yes, how can I prove that?

$\endgroup$
3
$\begingroup$

The Markov property applies both forwards and backwards

The Markov property holds that $p(q_{t+k}|q_t,...,q_{t+k-1}) = p(q_{t+k}|q_{t+k-1})$ for all $t \in \mathbb{Z}$ and $k \in \mathbb{N}^+$. This property is usually stated in its backward sense (i.e., the probability density of an outcome conditional on past values), but it also implies the result in your question, which is the same essential property in the forward sense (i.e., the probability density of an outcome conditional on future values). This is shown as follows:

$$\begin{equation} \begin{aligned} p(q_t| q_{t+1},...,q_T) &\overset{q_t}{\propto} p(q_t, q_{t+1},...,q_T) \\[12pt] &= p(q_t) \prod_{i=1}^{T-t} p(q_{t+i}| q_{t},...,q_{t+i-1}) \\[6pt] &= p(q_t) \prod_{i=1}^{T-t} p(q_{t+i}| q_{t+i-1}) \\[6pt] &\overset{q_t}{\propto} p(q_t) p(q_{t+1}|q_t) \\[12pt] &= p(q_t,q_{t+1}) \\[12pt] &\overset{q_t}{\propto} p(q_t|q_{t+1}). \\[6pt] \end{aligned} \end{equation}$$

(The third step in this working uses the backwards version of the Markov property.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.