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Now In (2.8) I have a really hard time understanding why we integrate with respect to $x_2$.

$P\left(x_3,t_3|x_1,t_1\right)=\sum _n P\left(x_3,t_3|x_n,t_2\right)P\left(x_n,t_2|x_1,t_1\right)$

This to me in a discrete case looks correct. But integrating over $x_2$ to me from the picture looks really strange. Is by integrating over $x_2$ am I integrating over the entire vertical line at $t_2$ and in that case it is in the x direction hence why not why not integrate over dx instead of $dx_2$

And most important which variable would you integrate on if there where more steps. Example $p(x_5,t_5|x_1,t_1)$

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  • $\begingroup$ Anyone, please help? $\endgroup$
    – ALEXANDER
    Jan 6, 2017 at 8:58
  • $\begingroup$ If you have $t_1 \lt t_2 \lt t_3 \lt t_4 \lt t_5$, then you could write $p(x_5,t_5|x_1,t_1)$ as $\displaystyle \int \int \int p\left(x_5,t_5\mid x_4,t_4\right)p\left(x_4,t_4\mid x_3,t_3\right)p\left(x_3,t_3\mid x_2,t_2\right)p\left(x_2,t_2\mid x_1,t_1\right)\,dx_4\,dx_3\,dx_2$ $\endgroup$
    – Henry
    Jan 10, 2017 at 10:43
  • $\begingroup$ but what does integrating over these x values mean. Given what the picture shows its to me a complete mystery. From the photo there are only one x_2 value and integrating on this would not give me the all three paths only one. If I however at t_2 integrate over x I would capture all three paths. What am I missing? $\endgroup$
    – ALEXANDER
    Jan 10, 2017 at 10:46
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    $\begingroup$ The correct formula is $$P\left(x_3,t_3|x_1,t_1\right)=\sum _{x_2} P\left(x_3,t_3|x_2,t_2\right)P\left(x_2,t_2|x_1,t_1\right)$$ It simply decomposes the set of paths $\xi$ such that $\xi(t_1)=x_1$ and $\xi(t_3)=x_3$, into the disjoint union over every possible state $x_2$ of the sets of paths $\xi$ such that $\xi(t_1)=x_1$, $\xi(t_2)=x_2$ and $\xi(t_3)=x_3$. $\endgroup$
    – Did
    Jan 10, 2017 at 11:01
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    $\begingroup$ One considers that the system is at $x_1$ at time $t_1$, at $x_3$ at time $t_3$ and at an unknown state $x_2$ at time $t_2$, hence the sum over $x_2$. "x_2 would have to be a set of multiple values" ?? No meaning I can get. "Are these part of x_2" Idem. $\endgroup$
    – Did
    Jan 10, 2017 at 12:55

1 Answer 1

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As it is said in the comments: $x_1$ at time $t_1$ and $x_3$ at time $t_3$ are fixed values. Next you fix $t_2 \in [t_1, t_3]$ and look at all possible values the process can attain at time $t_2$. For convenience this value is often called $x_2$ but you also can call it $x$, if you prefer.

However $x_2$ is not a fixed value (contrary to $x_1$ and $x_3$). So the picture might be misleading: there it seems like $x_2$ is also fixed and lies between $x_1$ and $x_3$, but this is just one (of many) possibilities. And now you sum up (or integrate) over all values $x_2$ the process can reach at time $t_2$, that's how you arrive at $$ P(x_3, t_3\mid x_1, t_1) = \sum_{x_2} P(x_2, t_2 \mid x_1, t_1) P(x_3, t_3 \mid x_2, t_2). $$ Note that the right hand side does not change if you write $x$ instead of $x_2$ $$ P(x_3, t_3\mid x_1, t_1) = \sum_{x} P(x, t_2 \mid x_1, t_1) P(x_3, t_3 \mid x, t_2). $$ It is just for convenience that the state at time $t_2$ is labeled $x_2$.

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