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% Generated by roxygen2: do not edit by hand
% Please edit documentation in R/ID.R
\name{ID}
\alias{ID}
\title{ID method}
\usage{
ID(NT, mu)
}
\arguments{
\item{NT}{Vector of case counts.}

\item{mu}{Mean of the serial distribution. This needs to match case counts in time units. For example, if case counts
are weekly and the serial distribution has a mean of seven days, then \code{mu} should be set to one If case
counts are daily and the serial distribution has a mean of seven days, then \code{mu} should be set to seven.}
}
\value{
\code{ID} returns a single value, the estimate of R0.
}
\description{
This function implements a least squares estimation method of R0 due to Fisman et al. (PloS One, 2013).
See details for implementation notes.
}
\details{
The method is based on a straightforward incidence decay model. The estimate of R0 is the value which
minimizes the sum of squares between observed case counts and cases counts 'expected' under the model.

This method is based on an approximation of the SIR model, which is most valid at the beginning of an epidemic.
The method assumes that the mean of the serial distribution (sometimes called the serial interval) is known.
The final estimate can be quite sensitive to this value, so sensitivity testing is strongly recommended.
Users should be careful about units of time (e.g., are counts observed daily or weekly?) when implementing.
}
\examples{
## ===================================================== ##
## Illustrate on weekly data                             ##
## ===================================================== ##

NT <- c(1, 4, 10, 5, 3, 4, 19, 3, 3, 14, 4)
## obtain Rhat when serial distribution has mean of five days
ID(NT=NT, mu=5/7)
## obtain Rhat when serial distribution has mean of three days
ID(NT=NT, mu=3/7)

## ========================================================= ##
## Compute Rhat using only the first five weeks of data      ##
## ========================================================= ##

ID(NT=NT[1:5], mu=5/7) # serial distribution has mean of five days

}