Changeset - c0439e4d4206
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Tom Bannink - 8 years ago 2017-09-11 11:04:42
tom.bannink@cwi.nl
Forgot to upload graphic file, its on another pc
1 file changed with 1 insertions and 1 deletions:
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main.tex
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@@ -696,13 +696,13 @@ The following Lemma says that if a set $S$ splits the graph in two, then those t
 
        \frac{1}{z^X_2} \P(r^X_2) \cdots
 
        \frac{1}{z^X_{|\xi^{G\setminus Y}|}} \P(r^X_{|\xi^{G\setminus Y}|}) .
 
    \end{align*}
 
    and similar for $\xi^{G\setminus X}$.
 
    Instead of choosing a step in $Y,X,X,Y,\cdots$ we could have chosen other orderings. The following diagram illustrates all possible interleavings, and the red line corresponds to the particular interleaving $Y,X,X,Y$ in the example above.
 
    \begin{center}
 
        \includegraphics{diagram_paths3.pdf}
 
        \includegraphics{diagram_paths2.pdf} \todo{change to paths3.pdf}
 
    \end{center}
 
    For the labels shown within the grid, define $p_{ij} = \frac{z^X_i}{z^X_i + z^Y_j}$.
 
    The probability of this particular interleaving $\xi^G$ is given by
 
    \begin{align*}
 
        \P^{G}_S(\xi^{G})
 
        &= (1-p)^{|b^X\; b^Y|} p^{|X\cup Y|-|b^X\;b^Y|} \quad
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