margMLE: Compute the marginal MLE of phi

In dkahle/poisDoubleSamp: Confidence Intervals with Poisson Double Sampling

Description

Compute the marginal MLE of the ratio of two Poisson rates in a two-sample Poisson rate problem with misclassified data given fallible and infallible datasets.

Usage

margMLE(data, N1, N2, N01, N02, l = 0.001, u = 1000, out = c("par",
  "all"))

Arguments

data	the vector of counts of the fallible data (z11, z12, z21, z22) followed by the infallible data (m011, m012, m021, m022, y01, y02)
N1	the opportunity size of group 1 for the fallible data
N2	the opportunity size of group 2 for the fallible data
N01	the opportunity size of group 1 for the infallible data
N02	the opportunity size of group 2 for the infallible data
l	the lower end of the range of possible phi's (for optim)
u	the upper end of the range of possible phi's (for optim)
out	"par" or "all" (for the output of optim)

Value

a named vector containing the marginal mle of phi

Examples

## Not run: 

# small example
z11 <- 34; z12 <- 35; N1 <- 10;
z21 <- 22; z22 <- 31; N2 <- 10;
m011 <- 9; m012 <- 1; y01 <- 3; N01 <- 3;
m021 <- 8; m022 <- 8; y02 <- 2; N02 <- 3;
data <- c(z11, z12, z21, z22, m011, m012, m021, m022, y01, y02)

fullMLE(data, N1, N2, N01, N02)
margMLE(data, N1, N2, N01, N02)



# big example :
z11 <- 477; z12 <- 1025; N1 <- 16186;
z21 <- 255; z22 <- 1450; N2 <- 18811;
m011 <- 38;  m012 <- 90; y01 <- 15; N01 <- 1500;
m021 <- 41; m022 <- 200; y02 <-  9; N02 <- 2500;
data <- c(z11, z12, z21, z22, m011, m012, m021, m022, y01, y02)

fullMLE(data, N1, N2, N01, N02)
margMLE(data, N1, N2, N01, N02)





## End(Not run)

dkahle/poisDoubleSamp documentation built on May 15, 2019, 9:09 a.m.