Timeline for Markov Chain: state reduction
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 7, 2013 at 0:59 | comment | added | Anthony Quas | I don't think this quite captures it. Firstly, you can't sum over $H_1$ because it's uncountable. Secondly, there shouldn't be a $p_2$ term in the first line. Thirdly you shouldn't be summing over all of $H_1$, but rather those paths in $H_1$ starting from $x_0$ and going through $a_1,\ldots,a_k$ in order before hitting any other element of $X_2$. There are a countable number of finite paths starting at $x_0$ and going through the $a$'s in order up to $a_k$. You should be summing over these, expanding out the $P_1$ probabilities of these, and then grouping them to make the $p_2$'s. | |
| Sep 6, 2013 at 17:36 | vote | accept | Cal | ||
| Sep 6, 2013 at 17:34 | comment | added | Cal | Thank you very much, it looks clearer to me, I was exactly confused with the notation $P_1$ and $P_2$. Just to make sure I understand it right: if I sum over all the possible paths $h$ in $H_1$ going through $a_1,...,a_k$ then is this what I have got? \begin{align} P_1(x_0,F^-1(a_1,...,a_k)) & = \sum_{h\in H_1} P_1(h)p_2(x_0,a_1,...,a_k) \\ & = \sum_h P_1(h)p_2(x_0,a_1)p_2(a_1,a_2)...p_2(a_{k-1},a_k) \\ & = p_2(x_0,a_1)p_2(a_1,a_2)...p_2(a_{k-1},a_k) \sum_h P_1(h) \\ & = p_2(x_0,a_1)p_2(a_1,a_2)...p_2(a_{k-1},a_k) = P_2(a_1,...,a_k) \end{align} Thanks again! | |
| Sep 5, 2013 at 18:36 | history | answered | Anthony Quas | CC BY-SA 3.0 |