Exercise 19.5 - Full conditional in an Ising model

Answers

We derive an expression for (19.126) straightforwardly:

p ( x k = 1 | 𝐱 − k , 𝜃 ) = p ( x k = 1 , 𝐱 − k | 𝜃 ) p ( 𝐱 − k | 𝜃 ) = p ( x k = 1 , 𝐱 − k | 𝜃 ) p ( x k = 0 , 𝐱 − k | 𝜃 ) + p ( x k = 1 , 𝐱 − k | 𝜃 ) = 1 Z ( 𝜃 ) exp ⁡ { ∑ j ≠ k x j J 𝑘𝑗 + h k + f ( 𝐱 − k ) } 1 Z ( 𝜃 ) exp ⁡ { f ( 𝐱 − k ) } + 1 Z ( 𝜃 ) exp ⁡ { ∑ j ≠ k x j J 𝑘𝑗 + h k + f ( 𝐱 − k ) } = 1 1 + exp ⁡ { − ( ∑ j ≠ k x j J 𝑘𝑗 + h k ) } = σ ( h k + ∑ j ≠ k x j J 𝑘𝑗 ) ,

where f ( 𝐱 − k ) are the sum of terms independent from x k inside the exponential.

As for p ( x k = 1 | 𝐱 nb k , 𝜃 ) , the reason is similar, since the summation ∑ ⁡ j ≠ k x j J 𝑘𝑗 is conducted on x k ’s neighbour.

In case we use the notation x i ∈ { − 1 , + 1 } , the deduction becomes:

p ( x k = 1 | 𝐱 − k , 𝜃 ) = 1 Z ( 𝜃 ) exp ⁡ { ∑ j ≠ k x j J 𝑘𝑗 + h k + f ( 𝐱 − k ) } 1 Z ( 𝜃 ) exp ⁡ { − ∑ j ≠ k x j J 𝑘𝑗 − h k + f ( 𝐱 − k ) } + 1 Z ( 𝜃 ) exp ⁡ { ∑ j ≠ k x j J 𝑘𝑗 + h k + f ( 𝐱 − k ) } = 1 1 + exp ⁡ { − 2 ⋅ ( ∑ j ≠ k x j J 𝑘𝑗 + h k ) } = σ ( 2 ⋅ ( h k + ∑ j ≠ k x j J 𝑘𝑗 ) ) .

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2021-03-24 13:42
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