Univariate Poisson Cost Calculations
The purpose of this vignette is to present the calculations of the costs for the univariate Poisson distribution.
Each time step \(t\) belongs to group \(k\) whose time stamps are the set \(T_{k}\). A group can have a multiplicative rate anomaly \(\lambda_{k}\) which is common for \(t \in T_{k}\). Assuming the {} known rate \(r_{t}\) the data generating distribution gives for \(t \in T_{k}\)
\[ P\left(y_t \left| r_t, \lambda_k\right.\right) = \frac{\lambda_{k}^{y_{t}} r_{t}^{y_{t}} \exp\left(-r_{t}\lambda_{k}\right)}{y_{t}!} \]
The cost is computed as twice the negative log likelhiood plus a penalty term \(\beta\) giving
\[ C\left(y_{t \in T_{k}} \left| \lambda_k,r_{t \in T_{k}}\right.\right) = 2 \lambda_{k} \sum\limits_{t \in T_{k}} r_{t} - 2 \log\left(\lambda_{k}\right) \sum\limits_{t \in T_{k}} y_{t} - 2 \sum\limits_{t \in T_{k}} y_{t} \log\left(r_{t}\right) + 2 \sum\limits_{t \in T_{k}} \log\left( y_{t}! \right) + \beta \]
No Anomaly (Baseline)
In this case \(\lambda_{k}=1\) and there is no penalty so
\[ C_{B}\left(y_{t \in T_{k}} \left| r_{t \in T_{k}}\right.\right) = 2 \sum\limits_{t \in T_{k}} r_{t} - 2 \sum\limits_{t \in T_{k}} y_{t} \log\left(r_{t}\right) + 2 \sum\limits_{t \in T_{k}} \log\left( y_{t}! \right) \]
Anomaly in Rate
An estimate \(\hat{\lambda}_{k}\) of \(\lambda_{k}\) can be selected to minimise the cost by taking \[ \hat{\lambda}_{k} = \frac{ \sum\limits_{t \in T_{k}} y_{t} }{\sum\limits_{t \in T_{k}} r_{t}} \]
\[ C_{A}\left(y_{t \in T_{k}} \left| \hat{\lambda}_k,r_{t \in T_{k}}\right.\right) = 2 \sum\limits_{t \in T_{k}} y_{t} - 2 \log\left(\hat{\lambda}_{k}\right) \sum\limits_{t \in T_{k}} y_{t} - 2 \sum\limits_{t \in T_{k}} y_{t} \log\left(r_{t}\right) + 2 \sum\limits_{t \in T_{k}} \log\left( y_{t}! \right) + \beta \]
An anomaly will be accepted whenever
\[ C_{A}\left(y_{t \in T_{k}} \left| \hat{\lambda}_k,r_{t \in T_{k}}\right.\right) - C_{B}\left(y_{t \in T_{k}} \left| r_{t \in T_{k}}\right.\right) = 2 \sum\limits_{t \in T_{k}} y_{t} - 2 \log\left(\hat{\lambda}_{k}\right) \sum\limits_{t \in T_{k}} y_{t} + \beta - 2 \sum\limits_{t \in T_{k}} r_{t} <0 \]
Rearranging this expression in terms of \(\sum\limits_{t \in T_{k}} r_{t}\), the expected number of counts if the period was not anomalous gives
\[ C_{A}\left(y_{t \in T_{k}} \left| \hat{\lambda}_k,r_{t \in T_{k}}\right.\right) - C_{B}\left(y_{t \in T_{k}} \left| r_{t \in T_{k}}\right.\right) = \beta - 2 \left( 1 - \hat{\lambda}_{k} + \hat{\lambda}_{k}\log\left(\hat{\lambda}_{k}\right)\right) \sum\limits_{t \in T_{k}} r_{t} \]
This form suggests the selection of \(\beta\) based on a minimum change in \(\hat{\lambda}_{k}\) away from 1 causing at least a certain change from the expected number of counts. The figure below show \(\gamma = 2 \left( 1 - \hat{\lambda}_{k} + \hat{\lambda}_{k}\log\left(\hat{\lambda}_{k}\right)\right)\). Using this we could select \(gamma=0.01\) to ensure a change in in lambda of 10%, which combined with a desire to detect only changes of 50 units would result in \(\beta = 0.5\)
