The purpose of this vignette is to present the calculations of the costs for various univariate distributions where for each time step there are multiple independent observations.
In the follow variables identified by Greek letters are considered known a priori.
Data belongs to group k whose time stamps are the set Tk can have additive mean anomaly mk and multiplicative variance anomaly sk which are common for t ∈ Tk. For t ∈ Tk. At time step t the vector of iid observations yt = {yt, 1, …, tt, nt} is made. The probability of of yt, i is $$ P\left(y_t \left| \mu_t,m_k,\sigma_k,s_k\right.\right) = \frac{1}{\sqrt{2\pi\sigma_{t}s_{k}}}\exp\left(-\frac{1}{2\sigma_{t}s_{k}}\left(y_{t} - \mu_t - m_{k}\right)^2\right) $$
with the likelihood of of the nt observations in yt being $$ L\left(\mathbf{y}_{t} \left| \mu_t,m_k,\sigma_k,s_k\right.\right) = \left(2\pi s_{k}\right)^{-n_{t}/2} \sigma_{t}^{-n_{t}/2} \exp\left(-\frac{1}{2s_{k}\sigma_{t}}\sum\limits_{i=1}^{n_{t}} \left(y_{t,i} - \mu_t - m_{k}\right)^{2}\right) $$ or as a log likelihood $$ l\left(\mathbf{y}_{t} \left| \mu_t,m_k,\sigma_k,s_k\right.\right) = -\frac{n_{t}}{2}\log\left(2\pi s_{k}\right) -\frac{n_{t}}{2}\log\left(\sigma_{t}\right) -\frac{1}{2s_{k}\sigma_{t}}\sum\limits_{i=1}^{n_{t}} \left(y_{t,i} - \mu_t - m_{k}\right)^{2} $$
The log-likelihood of yt ∈ Tk is with $n_{k}=\sum\limits_{t\in T_{k}} n_{t}$ $$ l\left(\mathbf{y}_{t \in T_{k}} \left| \mu_t,m_k,\sigma_k,s_k\right.\right) = -\frac{n_{k}}{2} \log\left(2\pi s_{k}\right) -\frac{1}{2}\sum\limits_{t \in T_{k}} n_{t}\log\left(\sigma_{t}\right) -\frac{1}{2s_{k}}\sum\limits_{t \in T_{k}} \frac{\sum_{i=1}^{n_{t}}\left(y_{t,i} - \mu_t - m_{k}\right)^2}{\sigma_{t}} $$
with the cost being twice the negative log likelihood plus a penalty β giving
$$ C\left(\mathbf{y}_{t \in T_{k}} \left| \mu_t,m_k,\sigma_k,s_k\right.\right) = n_{k} \log\left(2\pi s_{k}\right) +\sum\limits_{t \in T_{k}} n_{t}\log\left(\sigma_{t}\right) +\frac{1}{s_{k}}\sum\limits_{t \in T_{k}} \frac{\sum_{i=1}^{n_{t}}\left(y_{t,i} - \mu_t - m_{k}\right)^2}{\sigma_{t}} +\beta $$
Estimates m̂ of m and σ̂ of σ can be selected to minimise the cost by taking $$ \hat{m}_{k} = \left( \sum\limits_{t \in T_k} \frac{\sum\limits_{i=1}^{n_t} \left(y_t-\mu_t\right)}{\sigma_t} \right)\left( \sum\limits_{t \in T_k} \frac{n_{t}}{\sigma_t}\right)^{-1} $$ and $$ \hat{s}_{k} = \frac{1}{n_{k}} \sum\limits_{t \in T_{k}} \frac{\sum_{i=1}^{n_{t}}\left(y_{t,i} - \mu_t - \hat{m}_{k}\right)^2}{\sigma_{t}} $$
There is no change in variance so sk = 1. Estimate of m̂k is unchanged from that for an anomaly in mean and variance.
These is no mean anomaly so mk = 0. Estimate of ŝk therfore changes to
$$ \hat{s}_{k} = \frac{1}{n_{k}} \sum\limits_{t \in T_{k}} \frac{\sum_{i=1}^{n_{t}}\left(y_{t,i} - \mu_t\right)^2}{\sigma_{t}} $$
Here mk = 0 and sk = 1 and there is no penalty so β = 0
Assuming at for all t there are at least 2 unique values there is no need to represent a point in time differently [Check].