Last updated: 2018-10-13

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Suppose we observe \[ X_{t} \sim Poisson(\mu_{t}) \]

\[ L(\mu_{t}) = \frac{\mu_{t}^{x_{t}} e^{-\mu_{t}}}{x_{t}!} \\ l(\mu_{t}) = x_{t} \log \mu_{t} - \mu_{t} \]

Let \(\eta_{t} = \log(\mu_{t})\)

\[ l(\eta_{t}) = x_{t} \eta_{t} - e^{\eta_{t}} \]

Using 2nd order Taylor expansion around the maximum \(\tilde{\eta}_{t}\) \[ l(\eta_{t}) \approx l(\tilde{\eta}_{t}) + l'(\tilde{\eta}_{t})(\eta_{t} - \tilde{\eta}_{t}) + \frac{l''(\tilde{\eta}_{t})}{2}(\eta_{t} - \tilde{\eta}_{t})^{2} \]

Then \[ \eta_{t} \sim N(m, \sigma^{2}) \quad m = \tilde{\eta}_{t} - \frac{l'(\tilde{\eta}_{t})}{l''(\tilde{\eta}_{t})} \quad \sigma^2 = -\frac{1}{l''(\tilde{\eta}_{t})} \]

Therefore, \[ \eta_{t} \sim N(m, \sigma^{2}) \quad m = \tilde{\eta}_{t} - \frac{x_{t} - e^{\tilde{\eta}_{t}}}{- e^{\tilde{\eta}_{t}}} \quad \sigma^2 = -\frac{1}{-e^{\tilde{\eta}_{t}}} \]

\[ m = \tilde{\eta}_{t} - \frac{x_{t} + e^{\tilde{\eta}_{t}}}{e^{\tilde{\eta}_{t}}} = \log\tilde{\mu}_{t} - \frac{x_{t} + \tilde{\mu}_{t}}{\tilde{\mu}_{t}} \quad \sigma^2 = \frac{1}{\tilde{\mu}_{t}} \]

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