- #1
jakub jemez
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I need help with this
Consider an irreducible Markov chain with $\left|S\right|<\infty $ and transition function $p$.
Suppose that $p\left(x,x\right)=0,\ x\in S$ and that the chain has a stationary distribution $\pi .$
Let $p_x,x\ \in S,$ such that $0<\ p_x<1$ and $Q\left(x,y\right),\ x\in S,\ y\ \in S\ $where
$Q\left(x,x\right)=1-p_x$
$Q\left(x,y\right)=p_xp\left(x,y\right),\ if\ y\neq x.$
Proof that
1)The $Q$ function is a transition function and the chain becomes irreducible again with the $Q$ function.
2)The chain has a stationary distribution $\widehat{\pi }$ ($about\ Q$) given by
$\widehat{\pi }\left(x\right)=\frac{p^{-1}_x}{{\mathrm{\Sigma }}_{y\in S}p^{-1}_y\pi (y)\pi (x)},\ x\in S.$
What interpretation does the transition function $Q$ ?
Do you have any suggestions for part one?
Now, for part 2, it occurs to me that it could be as follows;
From what we already have by hypothesis, that is, that ${\mathrm{\Sigma }}_{x\in S}\pi \left(x\right)=1y$
$\sum_{x\in S}{\pi \left(x\right)}p\left(x,y\right)=\pi \left(y\right)\_\_\_\_\_\_\_\_\_(1)$
then, we look carefully (after doing several operations) that (1) (and for de definition of $Q$) it follows that
$\sum_{x\in S}{\frac{\pi \left(x\right)}{p_x}}Q\left(x,y\right)=\frac{\pi \left(y\right)}{p_y}\_\_\_\_\_\_\_\_\_(2)$And then we can see thanks to (2) that $\widehat{\pi }\left(x\right)=\frac{\widehat{\pi }\left(x\right)}{p_x}\ $it's stationary, but it's still missing that ${\mathrm{\Sigma }}_x\widehat{\pi }\left(x\right)=1$
So, I need help to complete $\widehat{\pi }\left(x\right)=\frac{\widehat{\pi }\left(x\right)}{p_x}$ in such a way that it is satisfied ${\mathrm{\Sigma }}_x\widehat{\pi }\left(x\right)=1$
Consider an irreducible Markov chain with $\left|S\right|<\infty $ and transition function $p$.
Suppose that $p\left(x,x\right)=0,\ x\in S$ and that the chain has a stationary distribution $\pi .$
Let $p_x,x\ \in S,$ such that $0<\ p_x<1$ and $Q\left(x,y\right),\ x\in S,\ y\ \in S\ $where
$Q\left(x,x\right)=1-p_x$
$Q\left(x,y\right)=p_xp\left(x,y\right),\ if\ y\neq x.$
Proof that
1)The $Q$ function is a transition function and the chain becomes irreducible again with the $Q$ function.
2)The chain has a stationary distribution $\widehat{\pi }$ ($about\ Q$) given by
$\widehat{\pi }\left(x\right)=\frac{p^{-1}_x}{{\mathrm{\Sigma }}_{y\in S}p^{-1}_y\pi (y)\pi (x)},\ x\in S.$
What interpretation does the transition function $Q$ ?
Do you have any suggestions for part one?
Now, for part 2, it occurs to me that it could be as follows;
From what we already have by hypothesis, that is, that ${\mathrm{\Sigma }}_{x\in S}\pi \left(x\right)=1y$
$\sum_{x\in S}{\pi \left(x\right)}p\left(x,y\right)=\pi \left(y\right)\_\_\_\_\_\_\_\_\_(1)$
then, we look carefully (after doing several operations) that (1) (and for de definition of $Q$) it follows that
$\sum_{x\in S}{\frac{\pi \left(x\right)}{p_x}}Q\left(x,y\right)=\frac{\pi \left(y\right)}{p_y}\_\_\_\_\_\_\_\_\_(2)$And then we can see thanks to (2) that $\widehat{\pi }\left(x\right)=\frac{\widehat{\pi }\left(x\right)}{p_x}\ $it's stationary, but it's still missing that ${\mathrm{\Sigma }}_x\widehat{\pi }\left(x\right)=1$
So, I need help to complete $\widehat{\pi }\left(x\right)=\frac{\widehat{\pi }\left(x\right)}{p_x}$ in such a way that it is satisfied ${\mathrm{\Sigma }}_x\widehat{\pi }\left(x\right)=1$