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R/testdata.R
In ivmte: Instrumental Variables: Extrapolation by Marginal Treatment Effects
Defines functions
gendistBasic
gendistSplines
gendistCovariates
gendistMosquito
gendist6e
gendist5e
gendist4
gendist3e
gendist3
gendist2
gendist1e
gendist1
Documented in
gendist1
gendist1e
gendist2
gendist3
gendist3e
gendist4
gendist5e
gendist6e
gendistBasic
gendistCovariates
gendistMosquito
gendistSplines
#' Generate test distribution 1
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1 or 2, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 0 + u and m1 ~ 1 + u. All unobservables
#' u are integrated out.
#' @param subN integer, default set to 5. This is the number of
#' individuals possessing each value of the instrument. So the
#' total number of observations is subN * 2.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @return a data.frame.
gendist1 <- function(subN = 5, p1 = 0.4, p2 = 0.6) {
dt <- data.frame(i = rep(seq(1, subN), 2),
z = rep(c(1, 2), each = subN),
p = rep(c(p1, p2), each = subN))
## Assign treatment
dt$d <- 0
dt[dt$z == 1 & dt$i <= (p1 * subN), "d"] <- 1
dt[dt$z == 2 & dt$i <= (p2 * subN), "d"] <- 1
## Now construct Y as E[Y | X, D, Z]
## m0 = 1 + u
## m1 = 1 + u
plist <- polyparse(~ u, data = dt)
plist0 <- genGamma(plist,
lb = dt$p,
ub = 1,
multiplier = 1 / (1 - dt$p),
means = FALSE)
plist1 <- genGamma(plist,
lb = 0,
ub = dt$p,
multiplier = 1 / dt$p,
means = FALSE)
m0coef <- c(0, 6)
m1coef <- c(7, 8)
dt$ey0 <- plist0 %*% m0coef
dt$ey1 <- plist1 %*% m1coef
dt$ey <- dt$d * dt$ey1 + (1 - dt$d) * dt$ey0
## Convert data back into data.frame
return(dt)
}
#' Generate test distribution 1 with errors
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1 or 2, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 0 + u and m1 ~ 1 + u.
#'@param N integer, default set to 100. Total number of observations
#' in the data.
#' @param subN , default set to 0.5. This is the probability the agent
#' will have Z = 1.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @param v0.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 0}
#' @param v1.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 1}
#' @return a data.frame.
gendist1e <- function(N = 100, subN = 0.5, p1 = 0.4, p2 = 0.6, v0.sd = 0.5,
v1.sd = 0.75) {
dt <- data.frame(z = runif(N),
u = runif(N))
## First construct the instruments independently from U
dt$z <- as.integer(dt$z > subN) + 1
dt$d <- 0
dt[dt$z == 1 & dt$u <= p1, "d"] <- 1
dt[dt$z == 2 & dt$u <= p2, "d"] <- 1
## Now construct the error terms. The error terms are normally
## distributed, centered at the draw of u if D = 0, and -u if D =
## 1.
## dt$v0 <- sapply(dt[, "u"], rnorm, n = 1, sd = v0.sd)
## dt$v1 <- sapply(-dt[, "u"], rnorm, n = 1, sd = v1.sd)
dt$v0 <- rnorm(n = N, mean = 0, sd = v0.sd)
dt$v1 <- rnorm(n = N, mean = 0, sd = v1.sd)
## Now construct the Y values
## m0 = 0 + u
## m1 = 1 + u
dt[dt$d == 0, "y"] <- 0 + 6 * dt[dt$d == 0, "u"] + dt[dt$d == 0, "v0"]
dt[dt$d == 1, "y"] <- 7 + 8 * dt[dt$d == 1, "u"] + dt[dt$d == 1, "v1"]
## Label the types
dt[dt$d == 1 & dt$z == 1, "type"] <- 1
dt[dt$d == 0 & dt$z == 1, "type"] <- 2
dt[dt$d == 1 & dt$z == 2, "type"] <- 3
dt[dt$d == 0 & dt$z == 2, "type"] <- 4
return(dt)
}
#' Generate test distribution 2
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1, 2, or 3, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 1 + u and m1 ~ 1 + u. All unobservables
#' u are integrated out.
#' @param subN integer, default set to 5. This is the number of
#' individuals possessing each value of the instrument. So the
#' total number of observations is subN * 2.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @param p3 the probability of treatment for those with the
#' instrument Z = 3.
#' @return a data.frame.
gendist2 <- function(subN = 5, p1 = 0.4, p2 = 0.6, p3 = 0.8) {
dt <- data.frame(i = rep(seq(1, subN), 3),
z = rep(c(1, 2, 3), each = subN),
p = rep(c(p1, p2, p3), each = subN))
## Assign treatment
dt$d <- 0
dt[dt$z == 1 & dt$i <= p1 * subN, "d"] <- 1
dt[dt$z == 2 & dt$i <= p2 * subN, "d"] <- 1
dt[dt$z == 3 & dt$i <= p3 * subN, "d"] <- 1
## Now construct Y as E[Y | X, D, Z]
## m0 = 1 + u
## m1 = 1 + u
plist <- polyparse(~ u, data = dt)
plist0 <- genGamma(plist,
lb = dt$p,
ub = 1,
multiplier = 1 / (1 - dt$p),
means = FALSE)
plist1 <- genGamma(plist,
lb = 0,
ub = dt$p,
multiplier = 1 / dt$p,
means = FALSE)
m0coef <- c(0, 6)
m1coef <- c(7, 8)
dt$ey0 <- plist0 %*% m0coef
dt$ey1 <- plist1 %*% m1coef
dt$ey <- dt$d * dt$ey1 + (1 - dt$d) * dt$ey0
## Convert data back into data.frame
return(dt)
}
#' Generate test distribution 3
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1 and 2, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 1 and m1 ~ 1. All unobservables u are
#' integrated out.
#' @param subN integer, default set to 5. This is the number of
#' individuals possessing each value of the instrument. So the
#' total number of observations is subN * 2.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @return a data.frame.
gendist3 <- function(subN = 5, p1 = 0.4, p2 = 0.6) {
dt <- data.frame(i = rep(seq(1, subN), 2),
z = rep(c(1, 2), each = subN),
p = rep(c(p1, p2), each = subN))
## Assign treatment
dt$d <- 0
dt[dt$z == 1 & dt$i <= (p1 * subN), "d"] <- 1
dt[dt$z == 2 & dt$i <= (p2 * subN), "d"] <- 1
## Now construct Y as E[Y | X, D, Z]
plist <- polyparse(~ 0 + 1, data = dt)
plist0 <- genGamma(plist,
lb = dt$p,
ub = 1,
multiplier = 1 / (1 - dt$p),
means = FALSE)
plist1 <- genGamma(plist,
lb = 0,
ub = dt$p,
multiplier = 1 / dt$p,
means = FALSE)
m0coef <- c(3)
m1coef <- c(9)
dt$ey0 <- plist0 %*% m0coef
dt$ey1 <- plist1 %*% m1coef
dt$ey <- dt$d * dt$ey1 + (1 - dt$d) * dt$ey0
## Convert data back into data.frame
return(dt)
}
#' Generate test distribution 3 with errors
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1 or 2, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 0 + u and m1 ~ 1 + u.
#'@param N integer, default set to 100. Total number of observations
#' in the data.
#' @param subN , default set to 0.5. This is the probability the agent
#' will have Z = 1.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @param v0.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 0}
#' @param v1.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 1}
#' @return a data.frame.
gendist3e <- function(N = 100, subN = 0.5, p1 = 0.4, p2 = 0.6, v0.sd = 0.5,
v1.sd = 0.75) {
dt <- data.frame(z = runif(N),
u = runif(N))
## First construct the instruments independently from U
dt$z <- as.integer(dt$z > subN) + 1
dt$d <- 0
dt[dt$z == 1 & dt$u <= p1, "d"] <- 1
dt[dt$z == 2 & dt$u <= p2, "d"] <- 1
## Now construct the error terms. The error terms are normally
## distributed, centered at the draw of u if D = 0, and -u if D =
## 1.
dt$v0 <- sapply(dt[, "u"], rnorm, n = 1, sd = v0.sd)
dt$v1 <- sapply(-dt[, "u"], rnorm, n = 1, sd = v1.sd)
## Now construct the Y values
## m0 = 3
## m1 = 9
dt[dt$d == 0, "y"] <- 3 + dt[dt$d == 0, "v0"]
dt[dt$d == 1, "y"] <- 9 + dt[dt$d == 1, "v1"]
## Label the types
dt[dt$d == 1 & dt$z == 1, "type"] <- 1
dt[dt$d == 0 & dt$z == 1, "type"] <- 2
dt[dt$d == 1 & dt$z == 2, "type"] <- 3
dt[dt$d == 0 & dt$z == 2, "type"] <- 4
return(dt)
}
#' Generate test distribution 4
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1, 2, and 3, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 1 and m1 ~ 1. All unobservables u are
#' integrated out.
#' @param subN integer, default set to 5. This is the number of
#' individuals possessing each value of the instrument. So the
#' total number of observations is subN * 2.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @param p3 the probability of treatment for those with the
#' instrument Z = 3.
#' @return a data.frame.
gendist4 <- function(subN = 5, p1 = 0.4, p2 = 0.6, p3 = 0.8) {
dt <- data.frame(i = rep(seq(1, subN), 3),
z = rep(c(1, 2, 3), each = subN),
p = rep(c(p1, p2, p3), each = subN))
## Assign treatment
dt$d <- 0
dt[dt$z == 1 & dt$i <= (p1 * subN), "d"] <- 1
dt[dt$z == 2 & dt$i <= (p2 * subN), "d"] <- 1
dt[dt$z == 3 & dt$i <= (p3 * subN), "d"] <- 1
## Now construct Y as E[Y | X, D, Z]
plist <- polyparse(~ 0 + 1, data = dt)
plist0 <- genGamma(plist,
lb = dt$p,
ub = 1,
multiplier = 1 / (1 - dt$p),
means = FALSE)
plist1 <- genGamma(plist,
lb = 0,
ub = dt$p,
multiplier = 1 / dt$p,
means = FALSE)
m0coef <- c(3)
m1coef <- c(9)
dt$ey0 <- plist0 %*% m0coef
dt$ey1 <- plist1 %*% m1coef
dt$ey <- dt$d * dt$ey1 + (1 - dt$d) * dt$ey0
## Convert data back into data.frame
return(dt)
}
#' Generate test distribution 5 (has errors and a covariate)
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1 or 2, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are both of the form m ~ 1 + x + u.
#'@param N integer, default set to 100. Total number of observations
#' in the data.
#' @param subN , default set to 0.5. This is the probability the agent
#' will have Z = 1.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @param v0.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 0}
#' @param v1.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 1}
#' @return a data.frame.
gendist5e <- function(N = 100, subN = 0.5, p1 = 0.4, p2 = 0.6, v0.sd = 1,
v1.sd = 1.55) {
dt <- data.frame(z = runif(N),
x = rnorm(N, mean = 1),
u = runif(N))
## First construct the instruments independently from U
dt$z <- as.integer(dt$z > subN) + 1
dt$d <- 0
dt[dt$z == 1 & dt$u <= p1, "d"] <- 1
dt[dt$z == 2 & dt$u <= p2, "d"] <- 1
## Now construct the error terms. The error terms are normally
## distributed, centered at the draw of u if D = 0, and -u if D =
## 1.
dt$v0 <- sapply(dt[, "u"], rnorm, n = 1, sd = v0.sd)
dt$v1 <- sapply(-dt[, "u"], rnorm, n = 1, sd = v1.sd)
## Now construct the Y values
dt[dt$d == 0, "y"] <- 3 + 0.4 * dt[dt$d == 0, "x"] - 1.3 *
dt[dt$d == 0, "u"] + dt[dt$d == 0, "v0"]
dt[dt$d == 1, "y"] <- 9 + 0.3 * dt[dt$d == 1, "x"] - 0.9 *
dt[dt$d == 1, "u"] + dt[dt$d == 1, "v1"]
## Label the types
dt[dt$d == 1 & dt$z == 1, "type"] <- 1
dt[dt$d == 0 & dt$z == 1, "type"] <- 2
dt[dt$d == 1 & dt$z == 2, "type"] <- 3
dt[dt$d == 0 & dt$z == 2, "type"] <- 4
return(dt)
}
#' Generate test distribution 6 (has errors and a covariate)
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that is uniformly distributed over [0, 1]. The
#' MTRs are both of the form m ~ 1 + x + x:u.
#'@param N integer, default set to 100. Total number of observations
#' in the data.
#' @param v0.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 0}
#' @param v1.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 1}
#' @return a data.frame.
gendist6e <- function(N = 100, v0.sd = 1,
v1.sd = 1.55) {
dt <- data.frame(z = runif(N),
x = rnorm(N, mean = 1),
u = runif(N))
## First construct the instruments independently from U
dt$d <- 0
dt[dt$u <= dt$z, "d"] <- 1
## Now construct the error terms.
dt$v0 <- rnorm(n = N, mean = 0, sd = v0.sd)
dt$v1 <- rnorm(n = N, mean = 0, sd = v1.sd)
## Now construct the Y values
dt[dt$d == 0, "y"] <- 3 + 0.4 * dt[dt$d == 0, "x"] - 1.3 *
dt[dt$d == 0, "x"] * dt[dt$d == 0, "u"] + dt[dt$d == 0, "v0"]
dt[dt$d == 1, "y"] <- 9 + 0.3 * dt[dt$d == 1, "x"] - 0.9 *
dt[dt$d == 1, "x"] * dt[dt$d == 1, "u"] + dt[dt$d == 1, "v1"]
return(dt)
}
#' Generate mosquito data set
#'
#' This code generates the population level data in Mogstad, Santos,
#' Torgovitsky (2018), i.e. the mosquito data set used as the running
#' example.
#'
#' @return data.frame.
gendistMosquito <- function() {
set.seed(1)
subN <- 100
p1 <- 0.12
p2 <- 0.29
p3 <- 0.48
p4 <- 0.78
dt <- data.frame(i = rep(seq(1, subN), 4),
z = rep(c(1, 2, 3, 4), each = subN),
pz = rep(c(p1, p2, p3, p4), each = subN))
## Assign treatment
dt$d <- 0
dt[dt$z == 1 & dt$i <= 12, "d"] <- 1
dt[dt$z == 2 & dt$i <= 29, "d"] <- 1
dt[dt$z == 3 & dt$i <= 48, "d"] <- 1
dt[dt$z == 4 & dt$i <= 78, "d"] <- 1
## Now construct Y as E[Y | X, D, Z]
plist <- polyparse(~ u + I(u ^ 2), data = dt)
plist0 <- genGamma(plist,
lb = dt$pz,
ub = 1,
multiplier = 1 / (1 - dt$pz),
means = FALSE)
plist1 <- genGamma(plist,
lb = 0,
ub = dt$pz,
multiplier = 1 / dt$pz,
means = FALSE)
m0coef <- c(0.9, -1.1, 0.3)
m1coef <- c(0.35, -0.3, -0.05)
dt$ey0 <- plist0 %*% m0coef
dt$ey1 <- plist1 %*% m1coef
dt$ey <- dt$d * dt$ey1 + (1 - dt$d) * dt$ey0
## Convert data back into data.frame
dtm <- data.frame(dt)
return(dtm)
}
#' Generate test data set with covariates
#'
#' This code generates population level data to test the estimation
#' function. This data includes covariates. The data generated will
#' have already integrated over the unobservable terms U, where U | X,
#' Z ~ Unif[0, 1].
#'
#' @return a list of two data.frame objects. One is the distribution
#' of the simulated data, the other is the full simulated data
#' set.
gendistCovariates <- function() {
set.seed(1)
supp_x1 <- c(0, 1)
supp_x2 <- c(1, 2, 3)
supp_z1 <- c(0, 1)
supp_z2 <- c(1, 2, 3)
dtc <- data.frame(expand.grid(supp_x1, supp_x2,
supp_z1, supp_z2))
colnames(dtc) <- c("x1", "x2", "z1", "z2")
## Generate propensity scores
g0 <- -0.5
g1 <- -0.4
g2 <- 0.2
g3 <- -1
g4 <- 0.3
dtc$latent <- g0 + g1 * dtc$x1 + g2 * dtc$x2 +
g3 * dtc$z1 + g4 * dtc$z2
dtc$p <- round(1 / (1 + exp(-dtc$latent)), 2)
## Generate the counterfactual outcomes Note: the code is prepared
## such that the polynomials are all in terms of u. Thus, terms which
## have u components with the same power are all grouped together.
##
## So I use the following specifications
## m0 = 0.3 + 0.4 * x1 - 0.1 * x2 * u - 0.2 * x2 * u^2
## m1 = 0.5 + 0.2 * x1 - 0.1 * x1 * x2 - 0.02 * u +
## 0.3 * x1 * u - 0.05 * x2 * u^2
plist0 <- polyparse(~ x1 + I(x2 * u) + I(x2 * u^2),
data = dtc)
plist1 <- polyparse(~ x1 + I(x1 * x2) + u + I(x1 * u) + I(x2 * u^2),
data = dtc)
glist0 <- genGamma(plist0,
lb = dtc$p,
ub = 1,
multiplier = 1 / (1 - dtc$p),
means = FALSE)
glist1 <- genGamma(plist1,
lb = 0,
ub = dtc$p,
multiplier = 1 / dtc$p,
means = FALSE)
g0coef <- c(0.3, 0.4, -0.1, -0.2)
g1coef <- c(0.5, 0.2, -0.1, -0.02, 0.3, -0.05)
dtc$ey0 <- glist0 %*% g0coef
dtc$ey1 <- glist1 %*% g1coef
## Generate distribution: build the distributions marginally,
## allowing for correlations to occur, and then normalize the
## distribution.
dtc$f <- 0
dtc[dtc$x1 == 0, "f"] <- dtc[dtc$x1 == 0, "f"] + 0.1
dtc[dtc$x1 == 1, "f"] <- dtc[dtc$x1 == 1, "f"] + 0.13
dtc[dtc$x2 == 1, "f"] <- dtc[dtc$x2 == 1, "f"] + 0.05
dtc[dtc$x2 == 2, "f"] <- dtc[dtc$x2 == 2, "f"] + 0.1
dtc[dtc$x2 == 3, "f"] <- dtc[dtc$x2 == 3, "f"] + 0.01
dtc[dtc$x1 == 0 & dtc$z1 == 1, "f"] <-
dtc[dtc$x1 == 0 & dtc$z1 == 1, "f"] - 0.03
dtc[dtc$x1 == 0 & dtc$z2 == 2, "f"] <-
dtc[dtc$x1 == 0 & dtc$z2 == 2, "f"] - 0.01
dtc[dtc$x1 == 0 & dtc$z2 == 3, "f"] <-
dtc[dtc$x1 == 0 & dtc$z2 == 3, "f"] - 0.02
dtc[dtc$z1 == 1 & dtc$z2 == 2, "f"] <-
dtc[dtc$z1 == 1 & dtc$z2 == 2, "f"] + 0.02
dtc[dtc$z1 == 1 & dtc$z2 == 3, "f"] <-
dtc[dtc$z1 == 1 & dtc$z2 == 3, "f"] + 0.01
dtc[dtc$z2 == 2, "f"] <- dtc[dtc$z2 == 2, "f"] + 0.05
dtc[dtc$z2 == 3, "f"] <- dtc[dtc$z2 == 3, "f"] + 0.01
dtc[dtc$x2 == 2 & dtc$z2 == 2, "f"] <-
dtc[dtc$x2 == 2 & dtc$z2 == 2, "f"] + 0.01
dtc[dtc$x2 == 2 & dtc$z2 == 3, "f"] <-
dtc[dtc$x2 == 2 & dtc$z2 == 3, "f"] + 0.01
dtc[dtc$x2 == 3 & dtc$z2 == 2, "f"] <-
dtc[dtc$x2 == 3 & dtc$z2 == 2, "f"] + 0.02
dtc[dtc$x2 == 3 & dtc$z2 == 3, "f"] <-
dtc[dtc$x2 == 3 & dtc$z2 == 3, "f"] + 0.04
dtc$f <- dtc$f / sum(dtc$f)
## Check distribution (requires data.table)
## length(unique(dtc$f))
## dtc[, sum(f), by = x1]
## dtc[, sum(f), by = x2]
## dtc[, sum(f), by = z1]
## dtc[, sum(f), by = z2]
## Since distributions were arbitrarily designed, round their
## densities, and ensure they sum to 1.
dtc$f <- round(dtc$f, 2)
sum(dtc$f) ## Sum of rounded densities is 1.02. Adjust the
## densities so they sum to 1.
dtc[dtc$x1 == 1 & dtc$x2 == 1 & dtc$z1 == 0 & dtc$z2 == 1, "f"] <-
dtc[dtc$x1 == 1 & dtc$x2 == 1 & dtc$z1 == 0 & dtc$z2 == 1, "f"] - 0.01
dtc[dtc$x1 == 1 & dtc$x2 == 3 & dtc$z1 == 1 & dtc$z2 == 3, "f"] <-
dtc[dtc$x1 == 1 & dtc$x2 == 3 & dtc$z1 == 1 & dtc$z2 == 3, "f"] - 0.01
dtc$f <- round(dtc$f, 2)
## Expand data set according to distribution: We multiply each row
## by 100 * (100 * f). The first 100 is so that we assign the
## probability of treatment and control according to the variable
## p. The (100 * f) is so that we can get the distribution of
## covariates to be according to f.
dtc[, "multiplier"] <- 100 * 100 * dtc$f
sum(dtc$multiplier)
dtcf <- dtc[rep(seq_len(nrow(dtc)), dtc$multiplier), ]
## Check if the distribution of this data set matches that of what we
## generated above (requires data.table)
## round(dtcf[, .N/nrow(dtcf), by = x1], 2) == round(dtc[, sum(f), by = x1], 2)
## round(dtcf[, .N/nrow(dtcf), by = x2], 2) == round(dtc[, sum(f), by = x2], 2)
## round(dtcf[, .N/nrow(dtcf), by = z1], 2) == round(dtc[, sum(f), by = z1], 2)
## round(dtcf[, .N/nrow(dtcf), by = z2], 2) == round(dtc[, sum(f), by = z2], 2)
## Now assign treatment
dtcf$d <- 0
for (ix1 in supp_x1) {
for (ix2 in supp_x2) {
for (iz1 in supp_z1) {
for (iz2 in supp_z2) {
N <- nrow(dtcf[dtcf$x1 == ix1 &
dtcf$x2 == ix2 &
dtcf$z1 == iz1 &
dtcf$z2 == iz2, ])
dtcf[dtcf$x1 == ix1 &
dtcf$x2 == ix2 &
dtcf$z1 == iz1 &
dtcf$z2 == iz2, "i"] <- seq(1, N)
dtcf[dtcf$x1 == ix1 &
dtcf$x2 == ix2 &
dtcf$z1 == iz1 &
dtcf$z2 == iz2, "dcut"] <-
round(dtcf[dtcf$x1 == ix1 &
dtcf$x2 == ix2 &
dtcf$z1 == iz1 &
dtcf$z2 == iz2, "p"] * N)
}
}
}
}
dtcf[dtcf$i <= dtcf$dcut, "d"] <- 1
## Check if empirical distribution differs from population
## (requires data.table)
## failed <- which(dtcf[, mean(d), by = .(x1, x2, z1, z2)]$V1 !=
## dtcf[, mean(p), by = .(x1, x2, z1, z2)]$V1)
## print(failed)
## We want this list to be empty (i.e. a failure occurs when the
## empirical distribution of treatment within group differs from
## what is specified in the population).
## Assign outcomes
dtcf$ey <- dtcf$d * dtcf$ey1 + (1 - dtcf$d) * dtcf$ey0
return(list(data.full = dtcf,
data.dist = dtc))
}
#' Generate test data set with splines
#'
#' This code generates population level data to test the
#' estimation function. This data set incorporates splines in the
#' MTRs.
#'
#' The distribution of the data is as follows
#'
#' | Z
#' X/Z | 0 1
#' _______|___________
#' -1 | 0.1 0.1
#' |
#' X 0 | 0.2 0.2
#' |
#' 1 | 0.1 0.2
#'
#' The data presented below will have already integrated over the
#' unobservable terms U, and U | X, Z ~ Unif[0, 1].
#'
#' The propensity scores are generated according to the model
#'
#' p(x, z) = 0.5 - 0.1 * x + 0.2 * z
#'
#' | Z
#' p(X,Z) | 0 1
#' _______|___________
#' -1 | 0.6 0.8
#' |
#' X 0 | 0.5 0.7
#' |
#' 1 | 0.4 0.6
#'
#' The lowest common multiple of the first table is 12. The lowest
#' common multiple of the second table is 84. It turns out that 840 *
#' 5 = 4200 observations is enough to generate the population data
#' set, such that each group has a whole-number of observations.
#'
#' The MTRs are defined as follows:
#'
#' y1 ~ beta0 + beta1 * x + uSpline(degree = 2,
#' knots = c(0.3, 0.6),
#' intercept = FALSE)
#'
#' The coefficients (beta1, beta2), and the coefficients on the
#' splines, will be defined below.
#'
#' y0 = x : uSpline(degree = 0,
#' knots = c(0.2, 0.5, 0.8),
#' intercept = TRUE)
#' + uSpline(degree = 1,
#' knots = c(0.4),
#' intercept = TRUE)
#' + beta3 * I(u ^ 2)
#'
#' The coefficient beta3, and the coefficients on the splines, will be
#' defined below.
#'
#' @return a list of two data.frame objects. One is the distribution
#' of the simulated data, the other is the full simulated data
#' set.
gendistSplines <- function() {
set.seed(10L)
## Declare coefficients
pCoef <- c(0.5, -0.1, 0.2) ## Propensity score coefficients
y1Coef <- c(30, 35) ## Beta coefficients for y1
y0Coef <- c(20) ## Beta coefficients for y0
u1s1Coef <- c(-75, 45, 75, 55) ## Coefficients for spline in y1
u0s1Coef <- c(25, 60, 45, 30) ## Coefficients for first spline in y0
u0s2Coef <- c(65, 70, 40) ## Coefficients for second spline in y0
## Generate distribution
distr <- data.frame(group = seq(1, 6),
x = rep(c(-1, 0, 1), times = 2),
z = rep(c(0, 1), each = 3),
f = c(0.1, 0.2, 0.1, 0.1, 0.3, 0.2))
distr$p <- pCoef[1] + pCoef[2] * distr$x + pCoef[3] * distr$z
distr$ey1 <- y1Coef[1] * distr$p +
y1Coef[2] * distr$x * distr$p +
t(sapply(X = distr$p,
FUN = splineInt,
lb = 0,
degree = 2,
knots = c(0.3, 0.6),
intercept = FALSE)) %*% u1s1Coef
distr$ey0 <- distr$x * t(sapply(X = distr$p,
FUN = splineInt,
ub = 1,
degree = 0,
knots = c(0.2, 0.5, 0.8),
intercept = TRUE)) %*% u0s1Coef +
t(sapply(X = distr$p,
FUN = splineInt,
ub = 1,
degree = 1,
knots = c(0.4),
intercept = TRUE)) %*% u0s2Coef +
y0Coef[1] / 3 * (1 - distr$p ^ 3)
## Expand distribution into data set
N <- 4200
distr$multiplier <- N * distr$f
distr$controls <- distr$multiplier * (1 - distr$p)
dtsf <- distr[rep(seq_len(nrow(distr)), distr$multiplier), c(1:7, 9)]
rownames(dtsf) <- NULL
## Construct observed outcome
dtsf$d <- 1
dtsf$count <- 0
for (g in unique(distr$group)) {
dtsf[dtsf$group == g, "count"] <- seq(1, nrow(dtsf[dtsf$group == g, ]))
}
dtsf[round(dtsf$count) <= round(dtsf$controls), "d"] <- 0
dtsf$ey <- (1 - dtsf$d) * dtsf$ey0 + dtsf$d * dtsf$ey1
## Check if the distribution is as planned
## for (z in c(0, 1)) {
## for (x in c(-1, 0, 1)) {
## a <- nrow(dtsf[dtsf$x == x & dtsf$z == z & dtsf$d == 1, ])
## b <- nrow(dtsf[dtsf$x == x & dtsf$z == z, ])
## cat("X:", x, ", Z:", z, ", a:", a, ", b:", b,
## ", Prob of treat:", a / b, "\n")
## }
## }
## Check if treatment effects are consistent
as.numeric(t(distr$ey1 - distr$ey0) %*% distr$f)
round(as.numeric(t(distr$ey1 - distr$ey0) %*% distr$f), 7) ==
round(mean(dtsf$ey1 - dtsf$ey0), 7)
mean(dtsf[dtsf$d == 1, "ey"]) - mean(dtsf[dtsf$d == 0, "ey"])
## Drop unnecssary columns
dtsf[, c("group", "controls", "count")] <- NULL
return(list(data.full = dtsf,
data.dist = distr))
}
#' Generate basic data set for testing
#'
#' This code generates population level data to test the estimation
#' function. This is a simpler dataset, one in which we can more
#' easily estimate a correctly specified model. The data presented
#' below will have already integrated over the # unobservable terms
#' U, where U | X, Z ~ Unif[0, 1].
#'
#' @return a list of two data.frame objects. One is the distribution
#' of the simulated data, the other is the full simulated data
#' set.
gendistBasic <- function() {
set.seed(10L)
supp_x <- c(1, 2, 3)
supp_z <- c(1, 2, 3, 4)
dtb <- data.frame(expand.grid(supp_x, supp_z))
colnames(dtb) <- c("x", "z")
## Generate propensity scores
g0 <- 0.2
g1 <- -0.10
g2 <- 0.2
dtb$p <- g0 + g1 * dtb$x + g2 * dtb$z
## Generate the counterfactual outcomes.
## m0 = 2 + x + 2 * u
## m1 = 6 + 5 * u^2
plist0 <- polyparse(~ x + u,
data = dtb)
plist1 <- polyparse(~ I(u^2),
data = dtb)
## Remember that you are generating E[m | D = 0] and E[m | D = 1],
## which are analogous to E[m | u > p(X, Z)] and E[m | u < p(X,
## Z)]. This is why you include those multipliers: You're
## integrating with respect to a conditional distribution.
glist0 <- genGamma(plist0,
lb = dtb$p,
ub = 1,
multiplier = 1 / (1 - dtb$p),
means = FALSE)
glist1 <- genGamma(plist1,
lb = 0,
ub = dtb$p,
multiplier = 1 / dtb$p,
means = FALSE)
g0coef <- c(2, 1, 2)
g1coef <- c(6, 5)
dtb$ey0 <- glist0 %*% g0coef
dtb$ey1 <- glist1 %*% g1coef
## Generate distribution and expand data set
p <- matrix(runif(12), ncol = 3)
p <- p / sum(p)
pmat <- matrix(c(0.02, 0.07, 0.02, 0.19,
0.09, 0.10, 0.17, 0.15,
0.01, 0.01, 0.16, 0.01),
ncol = 4)
dtb$f <- c(pmat)
dtb$multiplier <- dtb$f * 1000
## Check treatment effect (requires data.table)
## sum(dtb$ey1 * dtb$f) - sum(dtb$ey0 * dtb$f)
## sum(dtb[dtb$ey0 > dtb$ey1, f])
## Assign treatment
dtbf <- dtb[rep(seq_len(nrow(dtb)), dtb$multiplier), ]
dtbf$d <- 0
for (ix in supp_x) {
for (iz in supp_z) {
N <- nrow(dtbf[dtbf$x == ix &
dtbf$z == iz, ])
dtbf[dtbf$x == ix &
dtbf$z == iz, "i"] <- seq(1, N)
dtbf[dtbf$x == ix &
dtbf$z == iz, "dcut"] <-
round(dtbf[dtbf$x == ix &
dtbf$z == iz, "p"] * N)
}
}
dtbf[dtbf$i <= dtbf$dcut, "d"] <- 1
dtbf$dcut <- NULL
## Check if empirical distribution differs from population
## failed <- which(round(dtbf[, mean(d), by = .(x, z)]$V1, 2) !=
## round(dtbf[, mean(p), by = .(x, z)]$V1, 2))
## print(failed)
## Assign outcomes
dtbf$ey <- dtbf$d * dtbf$ey1 + (1 - dtbf$d) * dtbf$ey0
return(list(data.full = dtbf,
data.dist = dtb))
}
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Defines functions gendistBasic gendistSplines gendistCovariates gendistMosquito gendist6e gendist5e gendist4 gendist3e gendist3 gendist2 gendist1e gendist1
Documented in gendist1 gendist1e gendist2 gendist3 gendist3e gendist4 gendist5e gendist6e gendistBasic gendistCovariates gendistMosquito gendistSplines
#' Generate test distribution 1
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1 or 2, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 0 + u and m1 ~ 1 + u. All unobservables
#' u are integrated out.
#' @param subN integer, default set to 5. This is the number of
#' individuals possessing each value of the instrument. So the
#' total number of observations is subN * 2.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @return a data.frame.
gendist1 <- function(subN = 5, p1 = 0.4, p2 = 0.6) {
dt <- data.frame(i = rep(seq(1, subN), 2),
z = rep(c(1, 2), each = subN),
p = rep(c(p1, p2), each = subN))
## Assign treatment
dt$d <- 0
dt[dt$z == 1 & dt$i <= (p1 * subN), "d"] <- 1
dt[dt$z == 2 & dt$i <= (p2 * subN), "d"] <- 1
## Now construct Y as E[Y | X, D, Z]
## m0 = 1 + u
## m1 = 1 + u
plist <- polyparse(~ u, data = dt)
plist0 <- genGamma(plist,
lb = dt$p,
ub = 1,
multiplier = 1 / (1 - dt$p),
means = FALSE)
plist1 <- genGamma(plist,
lb = 0,
ub = dt$p,
multiplier = 1 / dt$p,
means = FALSE)
m0coef <- c(0, 6)
m1coef <- c(7, 8)
dt$ey0 <- plist0 %*% m0coef
dt$ey1 <- plist1 %*% m1coef
dt$ey <- dt$d * dt$ey1 + (1 - dt$d) * dt$ey0
## Convert data back into data.frame
return(dt)
}
#' Generate test distribution 1 with errors
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1 or 2, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 0 + u and m1 ~ 1 + u.
#'@param N integer, default set to 100. Total number of observations
#' in the data.
#' @param subN , default set to 0.5. This is the probability the agent
#' will have Z = 1.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @param v0.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 0}
#' @param v1.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 1}
#' @return a data.frame.
gendist1e <- function(N = 100, subN = 0.5, p1 = 0.4, p2 = 0.6, v0.sd = 0.5,
v1.sd = 0.75) {
dt <- data.frame(z = runif(N),
u = runif(N))
## First construct the instruments independently from U
dt$z <- as.integer(dt$z > subN) + 1
dt$d <- 0
dt[dt$z == 1 & dt$u <= p1, "d"] <- 1
dt[dt$z == 2 & dt$u <= p2, "d"] <- 1
## Now construct the error terms. The error terms are normally
## distributed, centered at the draw of u if D = 0, and -u if D =
## 1.
## dt$v0 <- sapply(dt[, "u"], rnorm, n = 1, sd = v0.sd)
## dt$v1 <- sapply(-dt[, "u"], rnorm, n = 1, sd = v1.sd)
dt$v0 <- rnorm(n = N, mean = 0, sd = v0.sd)
dt$v1 <- rnorm(n = N, mean = 0, sd = v1.sd)
## Now construct the Y values
## m0 = 0 + u
## m1 = 1 + u
dt[dt$d == 0, "y"] <- 0 + 6 * dt[dt$d == 0, "u"] + dt[dt$d == 0, "v0"]
dt[dt$d == 1, "y"] <- 7 + 8 * dt[dt$d == 1, "u"] + dt[dt$d == 1, "v1"]
## Label the types
dt[dt$d == 1 & dt$z == 1, "type"] <- 1
dt[dt$d == 0 & dt$z == 1, "type"] <- 2
dt[dt$d == 1 & dt$z == 2, "type"] <- 3
dt[dt$d == 0 & dt$z == 2, "type"] <- 4
return(dt)
}
#' Generate test distribution 2
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1, 2, or 3, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 1 + u and m1 ~ 1 + u. All unobservables
#' u are integrated out.
#' @param subN integer, default set to 5. This is the number of
#' individuals possessing each value of the instrument. So the
#' total number of observations is subN * 2.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @param p3 the probability of treatment for those with the
#' instrument Z = 3.
#' @return a data.frame.
gendist2 <- function(subN = 5, p1 = 0.4, p2 = 0.6, p3 = 0.8) {
dt <- data.frame(i = rep(seq(1, subN), 3),
z = rep(c(1, 2, 3), each = subN),
p = rep(c(p1, p2, p3), each = subN))
## Assign treatment
dt$d <- 0
dt[dt$z == 1 & dt$i <= p1 * subN, "d"] <- 1
dt[dt$z == 2 & dt$i <= p2 * subN, "d"] <- 1
dt[dt$z == 3 & dt$i <= p3 * subN, "d"] <- 1
## Now construct Y as E[Y | X, D, Z]
## m0 = 1 + u
## m1 = 1 + u
plist <- polyparse(~ u, data = dt)
plist0 <- genGamma(plist,
lb = dt$p,
ub = 1,
multiplier = 1 / (1 - dt$p),
means = FALSE)
plist1 <- genGamma(plist,
lb = 0,
ub = dt$p,
multiplier = 1 / dt$p,
means = FALSE)
m0coef <- c(0, 6)
m1coef <- c(7, 8)
dt$ey0 <- plist0 %*% m0coef
dt$ey1 <- plist1 %*% m1coef
dt$ey <- dt$d * dt$ey1 + (1 - dt$d) * dt$ey0
## Convert data back into data.frame
return(dt)
}
#' Generate test distribution 3
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1 and 2, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 1 and m1 ~ 1. All unobservables u are
#' integrated out.
#' @param subN integer, default set to 5. This is the number of
#' individuals possessing each value of the instrument. So the
#' total number of observations is subN * 2.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @return a data.frame.
gendist3 <- function(subN = 5, p1 = 0.4, p2 = 0.6) {
dt <- data.frame(i = rep(seq(1, subN), 2),
z = rep(c(1, 2), each = subN),
p = rep(c(p1, p2), each = subN))
## Assign treatment
dt$d <- 0
dt[dt$z == 1 & dt$i <= (p1 * subN), "d"] <- 1
dt[dt$z == 2 & dt$i <= (p2 * subN), "d"] <- 1
## Now construct Y as E[Y | X, D, Z]
plist <- polyparse(~ 0 + 1, data = dt)
plist0 <- genGamma(plist,
lb = dt$p,
ub = 1,
multiplier = 1 / (1 - dt$p),
means = FALSE)
plist1 <- genGamma(plist,
lb = 0,
ub = dt$p,
multiplier = 1 / dt$p,
means = FALSE)
m0coef <- c(3)
m1coef <- c(9)
dt$ey0 <- plist0 %*% m0coef
dt$ey1 <- plist1 %*% m1coef
dt$ey <- dt$d * dt$ey1 + (1 - dt$d) * dt$ey0
## Convert data back into data.frame
return(dt)
}
#' Generate test distribution 3 with errors
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1 or 2, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 0 + u and m1 ~ 1 + u.
#'@param N integer, default set to 100. Total number of observations
#' in the data.
#' @param subN , default set to 0.5. This is the probability the agent
#' will have Z = 1.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @param v0.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 0}
#' @param v1.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 1}
#' @return a data.frame.
gendist3e <- function(N = 100, subN = 0.5, p1 = 0.4, p2 = 0.6, v0.sd = 0.5,
v1.sd = 0.75) {
dt <- data.frame(z = runif(N),
u = runif(N))
## First construct the instruments independently from U
dt$z <- as.integer(dt$z > subN) + 1
dt$d <- 0
dt[dt$z == 1 & dt$u <= p1, "d"] <- 1
dt[dt$z == 2 & dt$u <= p2, "d"] <- 1
## Now construct the error terms. The error terms are normally
## distributed, centered at the draw of u if D = 0, and -u if D =
## 1.
dt$v0 <- sapply(dt[, "u"], rnorm, n = 1, sd = v0.sd)
dt$v1 <- sapply(-dt[, "u"], rnorm, n = 1, sd = v1.sd)
## Now construct the Y values
## m0 = 3
## m1 = 9
dt[dt$d == 0, "y"] <- 3 + dt[dt$d == 0, "v0"]
dt[dt$d == 1, "y"] <- 9 + dt[dt$d == 1, "v1"]
## Label the types
dt[dt$d == 1 & dt$z == 1, "type"] <- 1
dt[dt$d == 0 & dt$z == 1, "type"] <- 2
dt[dt$d == 1 & dt$z == 2, "type"] <- 3
dt[dt$d == 0 & dt$z == 2, "type"] <- 4
return(dt)
}
#' Generate test distribution 4
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1, 2, and 3, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are m0 ~ 1 and m1 ~ 1. All unobservables u are
#' integrated out.
#' @param subN integer, default set to 5. This is the number of
#' individuals possessing each value of the instrument. So the
#' total number of observations is subN * 2.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @param p3 the probability of treatment for those with the
#' instrument Z = 3.
#' @return a data.frame.
gendist4 <- function(subN = 5, p1 = 0.4, p2 = 0.6, p3 = 0.8) {
dt <- data.frame(i = rep(seq(1, subN), 3),
z = rep(c(1, 2, 3), each = subN),
p = rep(c(p1, p2, p3), each = subN))
## Assign treatment
dt$d <- 0
dt[dt$z == 1 & dt$i <= (p1 * subN), "d"] <- 1
dt[dt$z == 2 & dt$i <= (p2 * subN), "d"] <- 1
dt[dt$z == 3 & dt$i <= (p3 * subN), "d"] <- 1
## Now construct Y as E[Y | X, D, Z]
plist <- polyparse(~ 0 + 1, data = dt)
plist0 <- genGamma(plist,
lb = dt$p,
ub = 1,
multiplier = 1 / (1 - dt$p),
means = FALSE)
plist1 <- genGamma(plist,
lb = 0,
ub = dt$p,
multiplier = 1 / dt$p,
means = FALSE)
m0coef <- c(3)
m1coef <- c(9)
dt$ey0 <- plist0 %*% m0coef
dt$ey1 <- plist1 %*% m1coef
dt$ey <- dt$d * dt$ey1 + (1 - dt$d) * dt$ey0
## Convert data back into data.frame
return(dt)
}
#' Generate test distribution 5 (has errors and a covariate)
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that takes on values of 1 or 2, and the
#' distribution of the values for the binary instrument is
#' uniform. The MTRs are both of the form m ~ 1 + x + u.
#'@param N integer, default set to 100. Total number of observations
#' in the data.
#' @param subN , default set to 0.5. This is the probability the agent
#' will have Z = 1.
#' @param p1 the probability of treatment for those with the
#' instrument Z = 1.
#' @param p2 the probability of treatment for those with the
#' instrument Z = 2.
#' @param v0.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 0}
#' @param v1.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 1}
#' @return a data.frame.
gendist5e <- function(N = 100, subN = 0.5, p1 = 0.4, p2 = 0.6, v0.sd = 1,
v1.sd = 1.55) {
dt <- data.frame(z = runif(N),
x = rnorm(N, mean = 1),
u = runif(N))
## First construct the instruments independently from U
dt$z <- as.integer(dt$z > subN) + 1
dt$d <- 0
dt[dt$z == 1 & dt$u <= p1, "d"] <- 1
dt[dt$z == 2 & dt$u <= p2, "d"] <- 1
## Now construct the error terms. The error terms are normally
## distributed, centered at the draw of u if D = 0, and -u if D =
## 1.
dt$v0 <- sapply(dt[, "u"], rnorm, n = 1, sd = v0.sd)
dt$v1 <- sapply(-dt[, "u"], rnorm, n = 1, sd = v1.sd)
## Now construct the Y values
dt[dt$d == 0, "y"] <- 3 + 0.4 * dt[dt$d == 0, "x"] - 1.3 *
dt[dt$d == 0, "u"] + dt[dt$d == 0, "v0"]
dt[dt$d == 1, "y"] <- 9 + 0.3 * dt[dt$d == 1, "x"] - 0.9 *
dt[dt$d == 1, "u"] + dt[dt$d == 1, "v1"]
## Label the types
dt[dt$d == 1 & dt$z == 1, "type"] <- 1
dt[dt$d == 0 & dt$z == 1, "type"] <- 2
dt[dt$d == 1 & dt$z == 2, "type"] <- 3
dt[dt$d == 0 & dt$z == 2, "type"] <- 4
return(dt)
}
#' Generate test distribution 6 (has errors and a covariate)
#'
#' This function generates a data set for testing purposes. There is a
#' single instrument that is uniformly distributed over [0, 1]. The
#' MTRs are both of the form m ~ 1 + x + x:u.
#'@param N integer, default set to 100. Total number of observations
#' in the data.
#' @param v0.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 0}
#' @param v1.sd numeric, standard deviation of error term for
#' counterfactual \code{D = 1}
#' @return a data.frame.
gendist6e <- function(N = 100, v0.sd = 1,
v1.sd = 1.55) {
dt <- data.frame(z = runif(N),
x = rnorm(N, mean = 1),
u = runif(N))
## First construct the instruments independently from U
dt$d <- 0
dt[dt$u <= dt$z, "d"] <- 1
## Now construct the error terms.
dt$v0 <- rnorm(n = N, mean = 0, sd = v0.sd)
dt$v1 <- rnorm(n = N, mean = 0, sd = v1.sd)
## Now construct the Y values
dt[dt$d == 0, "y"] <- 3 + 0.4 * dt[dt$d == 0, "x"] - 1.3 *
dt[dt$d == 0, "x"] * dt[dt$d == 0, "u"] + dt[dt$d == 0, "v0"]
dt[dt$d == 1, "y"] <- 9 + 0.3 * dt[dt$d == 1, "x"] - 0.9 *
dt[dt$d == 1, "x"] * dt[dt$d == 1, "u"] + dt[dt$d == 1, "v1"]
return(dt)
}
#' Generate mosquito data set
#'
#' This code generates the population level data in Mogstad, Santos,
#' Torgovitsky (2018), i.e. the mosquito data set used as the running
#' example.
#'
#' @return data.frame.
gendistMosquito <- function() {
set.seed(1)
subN <- 100
p1 <- 0.12
p2 <- 0.29
p3 <- 0.48
p4 <- 0.78
dt <- data.frame(i = rep(seq(1, subN), 4),
z = rep(c(1, 2, 3, 4), each = subN),
pz = rep(c(p1, p2, p3, p4), each = subN))
## Assign treatment
dt$d <- 0
dt[dt$z == 1 & dt$i <= 12, "d"] <- 1
dt[dt$z == 2 & dt$i <= 29, "d"] <- 1
dt[dt$z == 3 & dt$i <= 48, "d"] <- 1
dt[dt$z == 4 & dt$i <= 78, "d"] <- 1
## Now construct Y as E[Y | X, D, Z]
plist <- polyparse(~ u + I(u ^ 2), data = dt)
plist0 <- genGamma(plist,
lb = dt$pz,
ub = 1,
multiplier = 1 / (1 - dt$pz),
means = FALSE)
plist1 <- genGamma(plist,
lb = 0,
ub = dt$pz,
multiplier = 1 / dt$pz,
means = FALSE)
m0coef <- c(0.9, -1.1, 0.3)
m1coef <- c(0.35, -0.3, -0.05)
dt$ey0 <- plist0 %*% m0coef
dt$ey1 <- plist1 %*% m1coef
dt$ey <- dt$d * dt$ey1 + (1 - dt$d) * dt$ey0
## Convert data back into data.frame
dtm <- data.frame(dt)
return(dtm)
}
#' Generate test data set with covariates
#'
#' This code generates population level data to test the estimation
#' function. This data includes covariates. The data generated will
#' have already integrated over the unobservable terms U, where U | X,
#' Z ~ Unif[0, 1].
#'
#' @return a list of two data.frame objects. One is the distribution
#' of the simulated data, the other is the full simulated data
#' set.
gendistCovariates <- function() {
set.seed(1)
supp_x1 <- c(0, 1)
supp_x2 <- c(1, 2, 3)
supp_z1 <- c(0, 1)
supp_z2 <- c(1, 2, 3)
dtc <- data.frame(expand.grid(supp_x1, supp_x2,
supp_z1, supp_z2))
colnames(dtc) <- c("x1", "x2", "z1", "z2")
## Generate propensity scores
g0 <- -0.5
g1 <- -0.4
g2 <- 0.2
g3 <- -1
g4 <- 0.3
dtc$latent <- g0 + g1 * dtc$x1 + g2 * dtc$x2 +
g3 * dtc$z1 + g4 * dtc$z2
dtc$p <- round(1 / (1 + exp(-dtc$latent)), 2)
## Generate the counterfactual outcomes Note: the code is prepared
## such that the polynomials are all in terms of u. Thus, terms which
## have u components with the same power are all grouped together.
##
## So I use the following specifications
## m0 = 0.3 + 0.4 * x1 - 0.1 * x2 * u - 0.2 * x2 * u^2
## m1 = 0.5 + 0.2 * x1 - 0.1 * x1 * x2 - 0.02 * u +
## 0.3 * x1 * u - 0.05 * x2 * u^2
plist0 <- polyparse(~ x1 + I(x2 * u) + I(x2 * u^2),
data = dtc)
plist1 <- polyparse(~ x1 + I(x1 * x2) + u + I(x1 * u) + I(x2 * u^2),
data = dtc)
glist0 <- genGamma(plist0,
lb = dtc$p,
ub = 1,
multiplier = 1 / (1 - dtc$p),
means = FALSE)
glist1 <- genGamma(plist1,
lb = 0,
ub = dtc$p,
multiplier = 1 / dtc$p,
means = FALSE)
g0coef <- c(0.3, 0.4, -0.1, -0.2)
g1coef <- c(0.5, 0.2, -0.1, -0.02, 0.3, -0.05)
dtc$ey0 <- glist0 %*% g0coef
dtc$ey1 <- glist1 %*% g1coef
## Generate distribution: build the distributions marginally,
## allowing for correlations to occur, and then normalize the
## distribution.
dtc$f <- 0
dtc[dtc$x1 == 0, "f"] <- dtc[dtc$x1 == 0, "f"] + 0.1
dtc[dtc$x1 == 1, "f"] <- dtc[dtc$x1 == 1, "f"] + 0.13
dtc[dtc$x2 == 1, "f"] <- dtc[dtc$x2 == 1, "f"] + 0.05
dtc[dtc$x2 == 2, "f"] <- dtc[dtc$x2 == 2, "f"] + 0.1
dtc[dtc$x2 == 3, "f"] <- dtc[dtc$x2 == 3, "f"] + 0.01
dtc[dtc$x1 == 0 & dtc$z1 == 1, "f"] <-
dtc[dtc$x1 == 0 & dtc$z1 == 1, "f"] - 0.03
dtc[dtc$x1 == 0 & dtc$z2 == 2, "f"] <-
dtc[dtc$x1 == 0 & dtc$z2 == 2, "f"] - 0.01
dtc[dtc$x1 == 0 & dtc$z2 == 3, "f"] <-
dtc[dtc$x1 == 0 & dtc$z2 == 3, "f"] - 0.02
dtc[dtc$z1 == 1 & dtc$z2 == 2, "f"] <-
dtc[dtc$z1 == 1 & dtc$z2 == 2, "f"] + 0.02
dtc[dtc$z1 == 1 & dtc$z2 == 3, "f"] <-
dtc[dtc$z1 == 1 & dtc$z2 == 3, "f"] + 0.01
dtc[dtc$z2 == 2, "f"] <- dtc[dtc$z2 == 2, "f"] + 0.05
dtc[dtc$z2 == 3, "f"] <- dtc[dtc$z2 == 3, "f"] + 0.01
dtc[dtc$x2 == 2 & dtc$z2 == 2, "f"] <-
dtc[dtc$x2 == 2 & dtc$z2 == 2, "f"] + 0.01
dtc[dtc$x2 == 2 & dtc$z2 == 3, "f"] <-
dtc[dtc$x2 == 2 & dtc$z2 == 3, "f"] + 0.01
dtc[dtc$x2 == 3 & dtc$z2 == 2, "f"] <-
dtc[dtc$x2 == 3 & dtc$z2 == 2, "f"] + 0.02
dtc[dtc$x2 == 3 & dtc$z2 == 3, "f"] <-
dtc[dtc$x2 == 3 & dtc$z2 == 3, "f"] + 0.04
dtc$f <- dtc$f / sum(dtc$f)
## Check distribution (requires data.table)
## length(unique(dtc$f))
## dtc[, sum(f), by = x1]
## dtc[, sum(f), by = x2]
## dtc[, sum(f), by = z1]
## dtc[, sum(f), by = z2]
## Since distributions were arbitrarily designed, round their
## densities, and ensure they sum to 1.
dtc$f <- round(dtc$f, 2)
sum(dtc$f) ## Sum of rounded densities is 1.02. Adjust the
## densities so they sum to 1.
dtc[dtc$x1 == 1 & dtc$x2 == 1 & dtc$z1 == 0 & dtc$z2 == 1, "f"] <-
dtc[dtc$x1 == 1 & dtc$x2 == 1 & dtc$z1 == 0 & dtc$z2 == 1, "f"] - 0.01
dtc[dtc$x1 == 1 & dtc$x2 == 3 & dtc$z1 == 1 & dtc$z2 == 3, "f"] <-
dtc[dtc$x1 == 1 & dtc$x2 == 3 & dtc$z1 == 1 & dtc$z2 == 3, "f"] - 0.01
dtc$f <- round(dtc$f, 2)
## Expand data set according to distribution: We multiply each row
## by 100 * (100 * f). The first 100 is so that we assign the
## probability of treatment and control according to the variable
## p. The (100 * f) is so that we can get the distribution of
## covariates to be according to f.
dtc[, "multiplier"] <- 100 * 100 * dtc$f
sum(dtc$multiplier)
dtcf <- dtc[rep(seq_len(nrow(dtc)), dtc$multiplier), ]
## Check if the distribution of this data set matches that of what we
## generated above (requires data.table)
## round(dtcf[, .N/nrow(dtcf), by = x1], 2) == round(dtc[, sum(f), by = x1], 2)
## round(dtcf[, .N/nrow(dtcf), by = x2], 2) == round(dtc[, sum(f), by = x2], 2)
## round(dtcf[, .N/nrow(dtcf), by = z1], 2) == round(dtc[, sum(f), by = z1], 2)
## round(dtcf[, .N/nrow(dtcf), by = z2], 2) == round(dtc[, sum(f), by = z2], 2)
## Now assign treatment
dtcf$d <- 0
for (ix1 in supp_x1) {
for (ix2 in supp_x2) {
for (iz1 in supp_z1) {
for (iz2 in supp_z2) {
N <- nrow(dtcf[dtcf$x1 == ix1 &
dtcf$x2 == ix2 &
dtcf$z1 == iz1 &
dtcf$z2 == iz2, ])
dtcf[dtcf$x1 == ix1 &
dtcf$x2 == ix2 &
dtcf$z1 == iz1 &
dtcf$z2 == iz2, "i"] <- seq(1, N)
dtcf[dtcf$x1 == ix1 &
dtcf$x2 == ix2 &
dtcf$z1 == iz1 &
dtcf$z2 == iz2, "dcut"] <-
round(dtcf[dtcf$x1 == ix1 &
dtcf$x2 == ix2 &
dtcf$z1 == iz1 &
dtcf$z2 == iz2, "p"] * N)
}
}
}
}
dtcf[dtcf$i <= dtcf$dcut, "d"] <- 1
## Check if empirical distribution differs from population
## (requires data.table)
## failed <- which(dtcf[, mean(d), by = .(x1, x2, z1, z2)]$V1 !=
## dtcf[, mean(p), by = .(x1, x2, z1, z2)]$V1)
## print(failed)
## We want this list to be empty (i.e. a failure occurs when the
## empirical distribution of treatment within group differs from
## what is specified in the population).
## Assign outcomes
dtcf$ey <- dtcf$d * dtcf$ey1 + (1 - dtcf$d) * dtcf$ey0
return(list(data.full = dtcf,
data.dist = dtc))
}
#' Generate test data set with splines
#'
#' This code generates population level data to test the
#' estimation function. This data set incorporates splines in the
#' MTRs.
#'
#' The distribution of the data is as follows
#'
#' | Z
#' X/Z | 0 1
#' _______|___________
#' -1 | 0.1 0.1
#' |
#' X 0 | 0.2 0.2
#' |
#' 1 | 0.1 0.2
#'
#' The data presented below will have already integrated over the
#' unobservable terms U, and U | X, Z ~ Unif[0, 1].
#'
#' The propensity scores are generated according to the model
#'
#' p(x, z) = 0.5 - 0.1 * x + 0.2 * z
#'
#' | Z
#' p(X,Z) | 0 1
#' _______|___________
#' -1 | 0.6 0.8
#' |
#' X 0 | 0.5 0.7
#' |
#' 1 | 0.4 0.6
#'
#' The lowest common multiple of the first table is 12. The lowest
#' common multiple of the second table is 84. It turns out that 840 *
#' 5 = 4200 observations is enough to generate the population data
#' set, such that each group has a whole-number of observations.
#'
#' The MTRs are defined as follows:
#'
#' y1 ~ beta0 + beta1 * x + uSpline(degree = 2,
#' knots = c(0.3, 0.6),
#' intercept = FALSE)
#'
#' The coefficients (beta1, beta2), and the coefficients on the
#' splines, will be defined below.
#'
#' y0 = x : uSpline(degree = 0,
#' knots = c(0.2, 0.5, 0.8),
#' intercept = TRUE)
#' + uSpline(degree = 1,
#' knots = c(0.4),
#' intercept = TRUE)
#' + beta3 * I(u ^ 2)
#'
#' The coefficient beta3, and the coefficients on the splines, will be
#' defined below.
#'
#' @return a list of two data.frame objects. One is the distribution
#' of the simulated data, the other is the full simulated data
#' set.
gendistSplines <- function() {
set.seed(10L)
## Declare coefficients
pCoef <- c(0.5, -0.1, 0.2) ## Propensity score coefficients
y1Coef <- c(30, 35) ## Beta coefficients for y1
y0Coef <- c(20) ## Beta coefficients for y0
u1s1Coef <- c(-75, 45, 75, 55) ## Coefficients for spline in y1
u0s1Coef <- c(25, 60, 45, 30) ## Coefficients for first spline in y0
u0s2Coef <- c(65, 70, 40) ## Coefficients for second spline in y0
## Generate distribution
distr <- data.frame(group = seq(1, 6),
x = rep(c(-1, 0, 1), times = 2),
z = rep(c(0, 1), each = 3),
f = c(0.1, 0.2, 0.1, 0.1, 0.3, 0.2))
distr$p <- pCoef[1] + pCoef[2] * distr$x + pCoef[3] * distr$z
distr$ey1 <- y1Coef[1] * distr$p +
y1Coef[2] * distr$x * distr$p +
t(sapply(X = distr$p,
FUN = splineInt,
lb = 0,
degree = 2,
knots = c(0.3, 0.6),
intercept = FALSE)) %*% u1s1Coef
distr$ey0 <- distr$x * t(sapply(X = distr$p,
FUN = splineInt,
ub = 1,
degree = 0,
knots = c(0.2, 0.5, 0.8),
intercept = TRUE)) %*% u0s1Coef +
t(sapply(X = distr$p,
FUN = splineInt,
ub = 1,
degree = 1,
knots = c(0.4),
intercept = TRUE)) %*% u0s2Coef +
y0Coef[1] / 3 * (1 - distr$p ^ 3)
## Expand distribution into data set
N <- 4200
distr$multiplier <- N * distr$f
distr$controls <- distr$multiplier * (1 - distr$p)
dtsf <- distr[rep(seq_len(nrow(distr)), distr$multiplier), c(1:7, 9)]
rownames(dtsf) <- NULL
## Construct observed outcome
dtsf$d <- 1
dtsf$count <- 0
for (g in unique(distr$group)) {
dtsf[dtsf$group == g, "count"] <- seq(1, nrow(dtsf[dtsf$group == g, ]))
}
dtsf[round(dtsf$count) <= round(dtsf$controls), "d"] <- 0
dtsf$ey <- (1 - dtsf$d) * dtsf$ey0 + dtsf$d * dtsf$ey1
## Check if the distribution is as planned
## for (z in c(0, 1)) {
## for (x in c(-1, 0, 1)) {
## a <- nrow(dtsf[dtsf$x == x & dtsf$z == z & dtsf$d == 1, ])
## b <- nrow(dtsf[dtsf$x == x & dtsf$z == z, ])
## cat("X:", x, ", Z:", z, ", a:", a, ", b:", b,
## ", Prob of treat:", a / b, "\n")
## }
## }
## Check if treatment effects are consistent
as.numeric(t(distr$ey1 - distr$ey0) %*% distr$f)
round(as.numeric(t(distr$ey1 - distr$ey0) %*% distr$f), 7) ==
round(mean(dtsf$ey1 - dtsf$ey0), 7)
mean(dtsf[dtsf$d == 1, "ey"]) - mean(dtsf[dtsf$d == 0, "ey"])
## Drop unnecssary columns
dtsf[, c("group", "controls", "count")] <- NULL
return(list(data.full = dtsf,
data.dist = distr))
}
#' Generate basic data set for testing
#'
#' This code generates population level data to test the estimation
#' function. This is a simpler dataset, one in which we can more
#' easily estimate a correctly specified model. The data presented
#' below will have already integrated over the # unobservable terms
#' U, where U | X, Z ~ Unif[0, 1].
#'
#' @return a list of two data.frame objects. One is the distribution
#' of the simulated data, the other is the full simulated data
#' set.
gendistBasic <- function() {
set.seed(10L)
supp_x <- c(1, 2, 3)
supp_z <- c(1, 2, 3, 4)
dtb <- data.frame(expand.grid(supp_x, supp_z))
colnames(dtb) <- c("x", "z")
## Generate propensity scores
g0 <- 0.2
g1 <- -0.10
g2 <- 0.2
dtb$p <- g0 + g1 * dtb$x + g2 * dtb$z
## Generate the counterfactual outcomes.
## m0 = 2 + x + 2 * u
## m1 = 6 + 5 * u^2
plist0 <- polyparse(~ x + u,
data = dtb)
plist1 <- polyparse(~ I(u^2),
data = dtb)
## Remember that you are generating E[m | D = 0] and E[m | D = 1],
## which are analogous to E[m | u > p(X, Z)] and E[m | u < p(X,
## Z)]. This is why you include those multipliers: You're
## integrating with respect to a conditional distribution.
glist0 <- genGamma(plist0,
lb = dtb$p,
ub = 1,
multiplier = 1 / (1 - dtb$p),
means = FALSE)
glist1 <- genGamma(plist1,
lb = 0,
ub = dtb$p,
multiplier = 1 / dtb$p,
means = FALSE)
g0coef <- c(2, 1, 2)
g1coef <- c(6, 5)
dtb$ey0 <- glist0 %*% g0coef
dtb$ey1 <- glist1 %*% g1coef
## Generate distribution and expand data set
p <- matrix(runif(12), ncol = 3)
p <- p / sum(p)
pmat <- matrix(c(0.02, 0.07, 0.02, 0.19,
0.09, 0.10, 0.17, 0.15,
0.01, 0.01, 0.16, 0.01),
ncol = 4)
dtb$f <- c(pmat)
dtb$multiplier <- dtb$f * 1000
## Check treatment effect (requires data.table)
## sum(dtb$ey1 * dtb$f) - sum(dtb$ey0 * dtb$f)
## sum(dtb[dtb$ey0 > dtb$ey1, f])
## Assign treatment
dtbf <- dtb[rep(seq_len(nrow(dtb)), dtb$multiplier), ]
dtbf$d <- 0
for (ix in supp_x) {
for (iz in supp_z) {
N <- nrow(dtbf[dtbf$x == ix &
dtbf$z == iz, ])
dtbf[dtbf$x == ix &
dtbf$z == iz, "i"] <- seq(1, N)
dtbf[dtbf$x == ix &
dtbf$z == iz, "dcut"] <-
round(dtbf[dtbf$x == ix &
dtbf$z == iz, "p"] * N)
}
}
dtbf[dtbf$i <= dtbf$dcut, "d"] <- 1
dtbf$dcut <- NULL
## Check if empirical distribution differs from population
## failed <- which(round(dtbf[, mean(d), by = .(x, z)]$V1, 2) !=
## round(dtbf[, mean(p), by = .(x, z)]$V1, 2))
## print(failed)
## Assign outcomes
dtbf$ey <- dtbf$d * dtbf$ey1 + (1 - dtbf$d) * dtbf$ey0
return(list(data.full = dtbf,
data.dist = dtb))
}
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