Let me summarize the basic idea of the Metropolis algorithm here. The idea is from the famous Japanese text book on Bayesian modeling, so called
"Green book".
Preliminary
Let \(q\) be a parameter to be sampled, \(L(q)\) be the likelihood of the parameter, and \(Y\) be the data observed. Suppose \(q\) follows a distribution \(p(q|Y)\), and \(p(q|Y)\) said to be stationary distribution if \(p^* \sim p(q|Y)\) where \(p^*\) is another sample from the Metropolis algorithm, and if \(p(q|Y)\) is stationary distribution, the equality \(\displaystyle p(q|Y)=\frac{L(q)}{\sum_q L(q)} \propto L(q)\) holds from the definition.
Derivation
Since \(q^*\) is selected randomly, then the probability where \(L(q^*)>L(q)\) is \(0.5\), and the probability that \(p^*\) is accepted is:
\begin{align}
p(q\to q^*) = 0.5 \times 1,
\nonumber
\end{align}
and consider the opposite situation where we already have \(q^*\) and \(q\) is proposed. The probability \(q\) is accepted is:
\begin{align}
p(q^*\to q) = 0.5 \times \frac{L(q)}{L(q^*)}.
\nonumber
\end{align}
Bind together both equations above, we have:
\begin{align}
\frac{p(q\to q^*)}{p(q^*\to q)} &= \frac{0.5 \times 1}{\displaystyle 0.5 \times \frac{L(q)}{L(q^*)}}.
\nonumber \\
L(q^*)p(q^*\to q) &= L(q)p(q\to q^*)
\nonumber \\
p(q^*|Y)p(q^*\to q) &= p(q|Y)p(q\to q^*)
\nonumber
\end{align}
Take the sum with respect to \(q\):
\begin{align}
\sum_q p(q^*|Y)p(q^*\to q) &= \sum_q p(q|Y)p(q\to q^*)
\nonumber \\
p(q^*|Y) \sum_q p(q^*\to q) &= \sum_q p(q|Y)p(q\to q^*)
\nonumber \\
p(q^*|Y) &= \sum_q p(q|Y)p(q\to q^*)
\nonumber
\end{align}
The last expression describes the fact that \(q^*\) also follows the stationary distribution \(p(q|Y)\).
Detailed Balance Condition
More general expression is:
\begin{align}
f(\theta|\theta')f(\theta'|x) = f(\theta'|\theta)f(\theta|x).
\end{align}
Comments
Post a Comment