Changeset - 7b239b9b9891
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Tom Bannink - 8 years ago 2017-09-10 22:42:04
tombannink@gmail.com
Fix typo in equation
1 file changed with 1 insertions and 1 deletions:
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@@ -718,13 +718,13 @@ The following Lemma says that if a set $S$ splits the graph in two, then those t
 
        \frac{z^Y_1}{z^X_1+z^Y_1} 
 
        \frac{z^X_1}{z^X_1+z^Y_2} 
 
        \frac{z^X_2}{z^X_2+z^Y_2} 
 
        \cdots
 
        \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \; \P^{G\setminus X}_S(\xi^{G\setminus X})
 
        \tag{definition} \\
 
        &= (1-p_{1,1}) \; p_{1,2} \; p_{2,2} \; (1-p_{3,2}) \; \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \; \P^{G\setminus X}_S(\xi^{G\setminus X})
 
        &= (1-p_{1,1}) \; p_{1,2} \; p_{2,2} \; (1-p_{3,2}) \cdots \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \; \P^{G\setminus X}_S(\xi^{G\setminus X})
 
        \tag{definition of $p_{i,j}$} \\
 
        &= \P(\text{path in grid}) \; \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \; \P^{G\setminus X}_S(\xi^{G\setminus X})
 
    \end{align*}
 
    In the grid we see that at every point the probabilities sum to 1, and we always reach the end, so we know the sum of all paths in the grid is 1. This proves the required equality.
 
    We obtain
 
    \begin{align*}
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