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I'm reading the following seminal paper by Besag

http://www2.stat.duke.edu/~scs/Courses/Stat376/Papers/GibbsFieldEst/BesagJRSSB1974.pdf

I'm unsure how they prove on page 10 equations 4.4 and 4.5 namely that

$$A_i(.) \equiv \alpha_i+\sum_{i=1}^n \beta_{i,j} B_j(x_j)$$ and that $$G_{i,j}(x_i,x_j) = \beta_{i,j}H_i(x_i)H_j(x_j)$$

where $H_i(x_i) = B_i(x_i) - B_i(0)$.

The proof starts on 4.3 (page 12). I follow the first part of the proof and have worked out that $$x_1x_2G(x_1,x_2) = (A_1(0,x_2,0,\ldots,0) - A_1(0))(B_1(x_1) - B_1(0))$$ and that

$$x_1x_2G(x_1,x_2) = (A_2(x_1,0,\ldots,0) - A_2(0))(B_2(x_2) - B_2(0))$$

So i guess we need to prove that $(A_1(0,x_2,0,\ldots,0) - A_1(0)) \propto (B_2(x_2) - B_2(0))$ and $(A_2(x_1,0,\ldots,0) - A_2(0)) \propto (B_1(x_1) - B_1(0))$ which i'm struggling to do.

Things i've tried are the Direct substitution we must have $$(A_1(0,x_2,0,\ldots,0) - A_1(0)) = \frac{(A_2(x_1,0,\ldots,0) - A_2(0))}{(B_1(x_1) - B_1(0))} (B_2(x_2) - B_2(0))$$ but wasn't sure how to prove that $\frac{(A_2(x_1,0,\ldots,0) - A_2(0))}{(B_1(x_1) - B_1(0))}$ was a constant. I tried directly plugging in values the conditional $$\mathrm{ln}(p_i(x_i|x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n) = A_i(.)B_i(x_i)+C_i(x_i)+D_i(.)$$ for A_i and B_i (i=1,2) and when i did things didn't cancel out nicely.

I would really appreciate any help, Thanks!

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1 Answer 1

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The answer is down to the fact that we are looking over all combinations of $x_1,x_2$

We know that $A_1(x_2)-A_1(0) = \frac{A_2(x_1)-A_2(0)}{B_1(x_1)-B_1(0)}{B_2(x_2)-B_2(0)}$ must hold. Now keep $x_2$ fixed and consider varying $x_1$ then $A_1(x_2)-A_1(0)$ and $B_2(x_2)-B_2(0)$ are constant and so the equation only holds if $\frac{A_2(x_1)-A_2(0)}{B_1(x_1)-B_1(0)}$ is the same $\forall x_1$. We can repeat for $x_2$.

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