Exercise 20.3 - Message passing on a tree

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The complete likelihood for this model is:

p (X_{1}, X_{2}, X_{3}, G_{1}, G_{2}, G_{3}) = p (G_{1}) p (G_{2} | G_{1}) p (G_{3} | G_{1}) p (X_{1} | G_{1}) p (X_{2} | G_{2}) p (X_{3} | G_{3}) .

For question (a), we have the following variable elimination:

\begin{aligned} p (X_{2} = 50) & = \int_{X_{1}} \int_{X_{3}} \sum_{G_{1}} \sum_{G_{2}} \sum_{G_{3}} p (X_{1}, X_{2} = 50, X_{3}, G_{1}, G_{2}) \\ = \sum_{G_{1}} p (G_{1}) \sum_{G_{2}} p (G_{2} | G_{1}) p (X_{2} = 50 | G_{2}) \sum_{G_{3}} p (G_{3} | G_{1}) \int_{X_{3}} p (X_{3} | G_{3}) \int_{X_{1}} p (X_{1} | G_{1}) \\ = \sum_{G_{1}} p (G_{1}) \sum_{G_{2}} p (G_{2} | G_{1}) p (X_{2} = 50 | G_{2}) \\ = \end{aligned}

On the other hand:

\begin{aligned} p (X_{2} = 50, G_{1} = 1) & = \int_{X_{1}} \int_{X_{3}} \sum_{G_{2}} \sum_{G_{3}} p (X_{1}, X_{2} = 50, X_{3}, G_{1} = 1, G_{2}) \\ = p (G_{1} = 1) \sum_{G_{2}} p (G_{2} | G_{1} = 1) p (X_{2} = 50 | G_{2}) . \end{aligned}

Hence:

\begin{aligned} p (G_{1} = 1 | X_{2} = 50) & = \frac{p (G_{1} = 1, X_{2} = 50)}{p (X_{2} = 50)} \\ = \frac{0.45 + 0.05 \times e^{- 5}}{0.5 + 0.5 \times e^{- 5}} \\ \approx 0.9 . \end{aligned}

For question (b), we have:

p (X_{2} = 50, X_{3} = 50) = \sum_{G_{1}} p (G_{1}) \sum_{G_{2}} p (G_{2} | G_{1}) p (X_{2} = 50 | G_{2}) \sum_{G_{3}} p (G_{3} | G_{1}) p (X_{3} = 50 | G_{3}) \int_{X_{1}} p (X_{1} | G_{1}),

similar for $p (G_{1} = 1, X_{2} = 50, X_{3} = 50)$ , hence:

\begin{aligned} p (G_{1} = 1 | X_{2} = 50, X_{3} = 50) & = \frac{p (G_{1} = 1, X_{2} = 50, X_{3} = 50)}{p (X_{2} = 50, X_{3} = 50)} \\ = \frac{0.405 + 0.09 \times e^{- 5} + 0.005 \times e^{- 10}}{0.41 + 0.18 \times e^{- 5} + 0.41 \times e^{- 10}} \\ \approx 0.99 . \end{aligned}

Since both $X_{2}$ , $X_{3}$ equal fifty implies that both $G_{2}$ , $G_{3}$ are approximately unity, this confirms the belief that $G_{1} = 1$ .

For question (c), the setting is symmetric to (b), hence:

p (G_{1} = 0 | X_{2} = 60, X_{3} = 60) \approx 0.01 .

For question (d), we follow the analysis as before:

p (G_{1} = 1 | X_{2} = 50, X_{3} = 60) = \frac{p (G_{1} = 1) p (X_{2} = 50, X_{3} = 60) | G_{1})}{p (X_{2} = 50, X_{3} = 60)} .

Note that:

p (X_{2} = 50, X_{3} = 60) | G_{1}) = p (X_{2} = 50, X_{3} = 60) | G_{10})

by symmetry, hence:

p (G_{1} = 1 | X_{2} = 50, X_{3} = 60) = \frac{1}{2} .

Since two contradictive evidence cancell each other.

This is a soft/fuzzy version of the ordinary logical reasoning, where we could have set $p (G_{2, 3} | G_{1})$ to then identity matrices and the variance for generating $X_{i}$ to zero.

solour_lfq

2021-03-24 13:42

Exercise 20.3 - Message passing on a tree

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