Exercise 17.1 - Derivation of $Q$ function for HMM

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Recall that for the auxiliay function, we are to calculate the log-likelihood w.r.t. $𝜃$ with $𝐳_{1 : T}$ replaced by their expectation w.r.t. $𝜃^{old}$ :

\begin{align} Q (𝜃, 𝜃^{old}) = & 𝔼_{p (𝐳_{1 : T} | 𝐱_{1 : T}, 𝜃^{old})} [\log p (𝐳_{1 : T}, 𝐱_{1 : T} | 𝜃)] \\ = & 𝔼_{p} [\log {\prod_{i = 1}^{N} {p (z_{i, 1} | π) \cdot (\prod_{t = 2}^{T_{i}} p (z_{i, t} | z_{i, t - 1}, 𝐀)) \cdot (\prod_{t = 1}^{T_{i}} p (x_{i, t} | z_{i, t}, 𝐁))}}] \\ = & 𝔼_{p} [\sum_{i = 1}^{N} \sum_{k = 1}^{K} 𝕀 [z_{i, 1} = k] \cdot \log π_{k} + \sum_{i = 1}^{N} \sum_{t = 2}^{T_{i}} \sum_{j = 1}^{K} \sum_{k = 1}^{K} 𝕀 [z_{i, t} = k, z_{i, t - 1} = j] \cdot \log 𝐀 (j, k) \\ + \sum_{i = 1}^{N} \sum_{t = 1}^{T_{i}} \sum_{k = 1}^{K} 𝕀 [z_{i, t} = k] \cdot \log p (x_{i, t} | z_{i, t} = k, 𝐁)] . \end{align}

From which we obtain (17.98), (17.99) and (17.100), hence (17.97).

solour_lfq

2021-03-24 13:42

Exercise 17.1 - Derivation of $Q$ function for HMM

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