Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 206 / 1 / e

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- Solution e

Solution

Let

I_{0} = {i : y_{i} = 0}

and

I_{+} = {i : y_{i} > 0}

. Under the proposed zero-inflated negative binomial model, independence gives the likelihood function

L (π, β; r) = i \in I_{0} \prod [π + (1 - π) (\frac{r}{r + e ^{x_{i}^{T} β}})^{r}] i \in I_{+} \prod (1 - π) f_{NB} (y_{i}; r, e^{x_{i}^{T} β}) .

The positive-count factor uses the ordinary, untruncated negative binomial distribution. With

f_{i} = f_{NB} (y_{i}; r, λ_{i})

, the log-likelihood is

ℓ = \sum_{i \in I_{0}} lo g [π + (1 - π) f_{i}] + \sum_{i \in I_{+}} [lo g (1 - π) + lo g f_{i}] .

For latent indicators

a_{i}

, the complete-data log-likelihood is

ℓ_{c} = \sum_{i} {a_{i} lo g π + (1 - a_{i}) [lo g (1 - π) + lo g f_{i}]},

where

a_{i} = 1

is allowed only when

y_{i} = 0

. This separation is what makes the expectation-maximization algorithm convenient.

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