investr: An R Package for Inverse Estimation
Brandon M. Greenwell
Air Force Institute of TechnologyChristine M. Schubert Kabban
Air Force Institute of Technology2014-05-27
Abstract
Inverse estimation is a classical and well-known problem in regression. In simple terms, it involves the use of an observed value of the response to make inference on the corresponding unknown value of the explanatory variable. To our knowledge, however, statistical software is somewhat lacking the capabilities for analyzing these types of problems. In this paper, we introduce investr (which stands for inverse estimation in R), a package for solving inverse estimation problems in both linear and nonlinear regression models.1Introduction
Consider the regression model \(\mathcal{Y}_i = f\left( x_i; \boldsymbol{\beta} \right) + \epsilon_i\) \((i = 1, \dotsc, n)\), where \(f\) is a known expectation function (called a calibration curve) that is monotonic over the range of interest and \(\epsilon_i \stackrel{iid}{\sim} \mathcal{N}\left( 0, \sigma^2 \right)\). A common problem in regression is to predict a future response \(\mathcal{Y}_0\) from a known value of the explanatory variable \(x_0\). Often, however, there is a need to do the reverse; that is, given an observed value of the response \(\mathcal{Y} = y_0\), estimate the unknown value of the explanatory variable \(x_0\). This is known as the calibration problem, though we refer to it more generally as inverse estimation. In this paper, we consider only controlled calibration, where the values of the explanatory variables are fixed by design. A more thorough overview of the calibration problem, including Bayesian approaches and multivariate calibration, is given in Osborne (1991).
There are three main functions available in the investr package (Greenwell 2014):
calibrate;invest;plotFit.
calibrate operates on objects of class "lm"
and can only be used when the expectation function has the form \(f(x_i; \boldsymbol{\beta}) = \beta_0 + \beta_1
x_i\) (i.e., the simple linear regression model), where
closed-form solutions are available for calculating confidence intervals
for \(x_0\). For more complicated
models (e.g., polynomial and nonlinear regression), closed-form
expressions are usually not available and iterative techniques will be
required. This is the purpose of the function invest, which
calculates the point estimate and confidence intervals for \(x_0\) by calling the function
uniroot from the stats package. The function
plotFit produces a scatterplot of the data with fitted
regression curve and the option to add confidence/prediction bands for
the response (pointwise or adjusted). It can be used with
single-predictor objects of class "lm" or
"nls"; however, for objects of class "nls",
confidence/prediction bands are based on the linear approximation and
can be misleading (Bates and Watts 1988,
65). The development version of investr can be found on
GitHub at https://github.com/w108bmg/investr. To report bugs or
issues, contact the main author or submit them to https://github.com/w108bmg/investr/issues.
Calibration for straight line regression
Consider the most common calibration model, \(\mathcal{Y}_i = \beta_0 + \beta_1 x_i + \epsilon_i\) \((i = 1, \dotsc, n)\), where \(x_i\) is fixed and \(\epsilon_i \stackrel{iid}{\sim} \mathcal{N}\left( 0, \sigma^2 \right)\). Suppose we obtain a series of \(m\) observations \(y_{01}, \dotsc, y_{0m}\) which are associated with the same (unknown) \(x_0\). It can be shown (F. A. Graybill 1976) that the maximum likelihood (ML) estimator of \(x_0\), also known as the classical estimator, is
\[\label{eq:x0-mle} \widehat{x}_0 = \frac{\bar{y}_0 - \widehat{\beta}_0}{\widehat{\beta}_1}, (\#eq:x0-mle) \]
where \(\widehat{\beta}_0\) and \(\widehat{\beta}_1\) are the usual ML estimators of \(\beta_0\) and \(\beta_1\), respectively, and \(\bar{y}_0 = \sum_{j = 1}^m y_{0j}/m\). The classical estimator in general is computed as \(\widehat{x}_0 = f^{-1}\left( \bar{y}_0; \widehat{\boldsymbol{\beta}} \right)\); that is, by inverting the fitted calibration curve at \(\bar{y}_0\) (Eisenhart 1939). Since \(\widehat{x}_0\) is a ratio of jointly normal random variables, it is not surprising to learn that it does not have any finite moments (think of a standard Cauchy distribution). The sampling distribution of \(\widehat{x}_0\) is complicated, but fortunately, not required for setting an exact \(100(1 - \alpha)\%\) confidence interval on \(x_0\). This is discussed in the next section.
Inversion interval
It can be shown (see, for example, Draper and Smith 1998, 47–51) that an exact \(100(1 - \alpha)\%\) confidence interval for \(x_0\) is given by
\[\label{eq:x0-inversion-interval} \widehat{x}_0 + \frac{\left( \widehat{x}_0 - \bar{x} \right)g \pm \left( t\widehat{\sigma}/\widehat{\beta}_1 \right)\sqrt{\left( \widehat{x}_0 - \bar{x} \right)^2 / S_{xx} + (1 - g)\left( \frac{1}{m} + \frac{1}{n} \right)}}{1 - g}, (\#eq:x0-inversion-interval) \]
where \(S_{xx} = \sum_{i=1}^n \left( x_i -
\bar{x} \right)^2\), \(g = \left( t^2
\widehat{\sigma}^2 \right) / \left( \widehat{\beta}_1^2 S_{xx}
\right)\) and \(t = t_{\alpha/2, n + m
- 3}\) is the upper \(1 -
\alpha\) percentile of a Student’s \(t\)-distribution with \(n + m - 3\) degrees of freedom. The
inversion interval @ref(eq:x0-inversion-interval) can also be obtained
using a fiducial argument, as in Fieller
(1954). For the special case \(m =
1\), the inversion interval is equivalent to inverting a \(100(1 - \alpha)\%\) prediction interval for
the response. In other words, if one draws a horizontal line through a
scatterplot of the data at \(y_0\),
then the abscissas of its intersection with the usual (pointwise)
prediction band for \(f\) correspond to
the endpoints of the inversion interval @ref(eq:x0-inversion-interval).
This interval should only be used when the usual \(\mathcal{F}\) test for testing \(\mathcal{H}_0: \beta_1 = 0\) versus \(\mathcal{H}_1: \beta_1 \ne 0\) can be
rejected at the specified \(\alpha\)
level; in other words, when the regression line is not “too” flat. If
\(\mathcal{H}_0\) is not rejected, then
such a confidence interval for \(x_0\)
may result in either the entire real line or two semi-infinite intervals
(Franklin A. Graybill and Iyer 1994,
429)—see Figure 2. The
plotFit function in the investr package can be
used for drawing scatterplots with the fitted model and
confidence/prediction bands. To calculate the inversion interval for the
linear calibration problem, we use the calibrate function
and specify the option interval = "inversion" (the default)
as in the following example.
The data frame arsenic contains the true amounts of
arsenic present in 32 water samples (Franklin A.
Graybill and Iyer 1994). Also present, is the amount of arsenic
measured by some field test, which is subject to error. A new water
sample is obtained and subjected to the field test producing a reading
of 3.0 \(\mu\)g/ml. It is desired to
infer the true amount of arsenic present in the sample. The following
code fits a simple linear regression model to the arsenic data and then
uses the calibrate function to compute the ML estimate and
corresponding 90% calibration interval based on
Equation @ref(eq:x0-inversion-interval).
library(investr)
mod <- lm(measured ~ actual, data = arsenic)
(res <- calibrate(mod, y0 = 3, interval = "inversion", level = 0.9))
## Figure 1
plotFit(mod, interval = "prediction", level = 0.9, shade = TRUE, col.pred = "skyblue")
abline(h = 3, v = c(res$lower, res$estimate, res$upper), lty = 2)Running the above block of code produces Figure 1 and the following output to the R console:
estimate lower upper
2.9314 2.6035 3.2587where estimate is the ML estimate @ref(eq:x0-mle), and
lower/upper correspond to the lower/upper
bounds of the 90% inversion interval for \(x_0\) @ref(eq:x0-inversion-interval). If
instead the new water sample was subjected to the field test three
times, thereby producing three response values corresponding to \(x_0\), say 3.17, 3.09, and 3.16 \(\mu\)g/ml, we would simply supply
calibrate with a vector of these values as in
calibrate(mod, y0 = c(3.17, 3.09, 3.16), interval = "inversion", level = 0.9)If interval = "inversion", and the slope of the model is
not ignificant at the specified \(\alpha\) level, then finite confidence
limits for \(x_0\) will not be
produced. For example, suppose badfit is an
"lm" object for which the slope is not significant at the
\(\alpha = 0.1\) level. Then, as
illustrated in Figure 2,
calibrate(badfit, y0 = 10, level = 0.9)will either produce two semi-infinite intervals, e.g.,
Error: The calibration line is not well determined. The resulting
confidence region is the union of two semi-infinite intervals:
( -Inf , -282.0006 ) U ( 393.1267 , Inf)or the entire real line, e.g.,
estimate lower upper
-97.5987 -Inf Inf
Warning message:
The calibration line is not well determined.
Wald interval
Another common approach to computing calibration intervals is to use the delta method (Dorfman 1938). It is easy to show that an approximate standard error for the ML estimator @ref(eq:x0-mle), based on a first-order Taylor series approximation, is given by
\[\label{eq:x0-se} \widehat{\mathrm{se}}\left( \widehat{x}_0 \right) = \frac{\widehat{\sigma}}{\left| \widehat{\beta}_1 \right|} \sqrt{\frac{1}{m} + \frac{1}{n} + \frac{\left( \widehat{x}_0 - \bar{x} \right)^2}{S_{xx}}}. (\#eq:x0-se) \]
Assuming large sample normality for \(\widehat{x}_0\) leads to an approximate \(100(1 - \alpha)\%\) Wald confidence interval for \(x_0\) of
\[\label{eq:x0-wald-interval} \widehat{x}_0 \pm t_{\alpha/2, n+m-3} \frac{\widehat{\sigma}}{\left| \widehat{\beta}_1 \right|} \sqrt{\frac{1}{m} + \frac{1}{n} + \frac{\left( \widehat{x}_0 - \bar{x} \right)^2}{S_{xx}}}. (\#eq:x0-wald-interval) \]
This is equivalent to taking \(g =
0\) in the inversion interval @ref(eq:x0-inversion-interval).
Unlike the inversion interval, though, the Wald interval always exists
and is symmetric about \(\widehat{x}_0\). This symmetry is
attractive, but not always realistic, such as when the calibration curve
\(f\) is nonlinear and has horizontal
asymptotes. To obtain a Wald-type interval we specify
interval = "Wald" in the call to calibrate.
For instance,
calibrate(mod, y0 = 3, interval = "Wald", level = 0.9)produces the output
estimate lower upper se
2.9314 2.6040 3.2589 0.1929The point estimate remains unchanged (as expected) but the calibration interval is slightly different. The major benefit of using the delta method to compute calibration intervals is that it always exists and provides us with an asymptotic estimate of the standard error, which in this case gives \(\widehat{\mathrm{se}} = 0.1929\). The bootstrap, as discussed in Section Bootstrap calibration intervals, also provides calibration intervals and an estimate of the standard error, but does not require large samples or specific distribution assumptions.
Confidence interval for a specified mean response
Instead of inferring the value of \(x_0\) that corresponds to an observed value of the response \(y_0\), suppose we want to infer the value of \(x_0\) that corresponds to a specified value of the mean response, say \(f\left( x_0; \boldsymbol{\beta} \right) = \mu_0\). The obvious ML estimate of \(x_0\) is
\[\label{eq:x0-mle-regulation} \widehat{x}_0 = \frac{\mu_0 - \widehat{\beta}_0}{\widehat{\beta}_1}, (\#eq:x0-mle-regulation) \]
which is the same as Equation @ref(eq:x0-mle) but with \(\bar{y}_0\) replaced with \(\mu_0\), a fixed population parameter. This is analogous to the difference between (i) predicting a future value of the response corresponding to a given value of the explanatory variable and (ii) estimating the mean response that corresponds to a particular value of the explanatory variable. Thus, the point estimates @ref(eq:x0-mle) and @ref(eq:x0-mle-regulation) are the same but the former has greater variability inherited from the variance of \(\overline{\mathcal{Y}}_0\). This is sometimes referred to as regulation (as opposed to calibration) and is described in more detail in Franklin A. Graybill and Iyer (1994, chap. 6). The confidence interval formulas for \(x_0\) corresponding to a specified mean response (i.e., regulation) are the same as those given in Equations @ref(eq:x0-inversion-interval) and @ref(eq:x0-wald-interval), but with \(1/m\) replaced with \(0\) and \(t_{\alpha/2, n + m - 3}\) replaced with \(t_{\alpha/2, n - 2}\). For instance, the Wald interval for \(x_0\) corresponding to \(\mu_0\) is simply \[\widehat{x}_0 \pm t_{\alpha/2, n - 2}\frac{\widehat{\sigma}}{\left| \widehat{\beta}_1 \right|} \sqrt{\frac{1}{n} + \frac{\left( \widehat{x}_0 - \bar{x} \right)^2}{S_{xx}}},\]
where \(\widehat{x}_0\) is
calculated as in Equation @ref(eq:x0-mle-regulation). To obtain
calibration intervals corresponding to a specified mean response, use
the option mean.response = TRUE.
To illustrate, consider the crystal growth data taken from Franklin A. Graybill and Iyer (1994, 119). These data are from an experiment in which the weight in grams of 14 crystals were recorded after letting the crystals grow for different (predetermined) amounts of time in hours. The weight of each crystal is plotted against time in Figure 3. Suppose we want to estimate the growth time in hours that corresponds to an average weight of 8 grams; that is, we want to estimate \(x_0\) corresponding to \(\mu_0 = 8\). The following code chunk fits a simple linear regression model to the crystal growth data and then computes the ML estimate and a 95% calibration interval for \(x_0\).
mod <- lm(weight ~ time, data = crystal)
(res <- calibrate(mod, y0 = 8, mean.response = TRUE))
## Figure 3
plotFit(mod, interval = "confidence", pch = 19, shade = TRUE, col.conf = "plum",
extend.range = TRUE)
abline(h = 8, v = c(res$lower, res$estimate, res$upper), lty = 2)The output for this code chunk should be
estimate lower upper
15.8882 14.6590 17.1596 Thus, in order to produce crystals with an average weight of 8 grams,
they should be grown for an estimated 15.8882 hours. A 95% confidence
interval for the growth time is (14.6590, 17.1596). Obviously, if
mean.response = TRUE, then y0 can only take a
single value; otherwise, an error will be displayed as in the
following:
calibrate(mod, y0 = c(8, 9), mean.response = TRUE)which displays the message
Error in calibrate.default(cbind(x, y), ...) :
Only one mean response value allowed.
This type of calibration problem is similar to computing the median
effective dose (\(ED_{0.5}\)), or more
generally \(ED_p\) where \(0 < p < 1\), in binary response
models. In R, these models are usually fit with the glm
function from the stats package. The function
dose.p from the MASS package
(Venables and Ripley 2002) can then be
used to compute the ML estimate of \(ED_p\) for specified \(p\). An estimate of the asymptotic standard
error based on the delta method is also given which can be used to
calculate a Wald-based confidence interval for \(ED_p\). A future release of
investr will likely make an inversion-type interval available
for these models as well (i.e., by inverting a confidence interval for
the mean response on the link scale). The package drc (Ritz and Streibig 2005), for fitting
dose-response curves, may also be of interest.
Simultaneous calibration intervals
The calibration intervals discussed so far are one-at-a-time
intervals. If \(k\) new values of the
response are observed where each corresponds to a different unknown, say
\(x_{01}, \dotsc x_{0k}\), then we need
to adjust the critical value used in the inversion and Wald intervals
accordingly. The simplest approaches are of course, the Bonferroni and
Scheffé procedures. These can be computed by specifying the
adjust option which can take any of the following
arguments: "none" (the default), "bonferroni",
or "scheffe". The value \(k\) also needs to be specified. See Miller (1981, chap. 3) for more details.
Nonlinear and polynomial calibration
In application, many relationships are nonlinear (e.g., dose-response curves). The classical estimator along with the inversion and Wald intervals can easily be extended to such nonlinear calibration curves. However, classical inference in these models (such as prediction intervals) are based on large samples and linear approximations (see Bates and Watts 1988, chap. 2). Thus, for nonlinear calibration curves, the inversion interval provides only an approximate \(100(1 - \alpha)\%\) confidence interval for \(x_0\) as does the Wald interval. Calibration in nonlinear models is discussed in further detail in Schwenke and Milliken (1991), Seber and Wild (2003, 245–50), and Huet (2004, chap. 5). For calibration in polynomial models, see Brown (1993, 47–88) and Seber and Lee (2003, 172).
The invest function can be used for inverse estimation
with any univariate regression model in R that inherits from class
"lm" or "nls" (with the exception of weighted
fits). For instance, consider the regression model \(\mathcal{Y}_i = f\left( x_i; \boldsymbol{\theta}
\right) + \epsilon_i\) \((i = 1,
\dotsc, n)\), where \(f\) may or
may not be linear in \(\boldsymbol{\theta}\). If we wish to
estimate \(x_0\) given an observation
\(y_0\), then the point estimate \(\widehat{x}_0\) is given by solving \(y_0 = f\left(x;
\widehat{\boldsymbol{\theta}}\right)\) for \(x\). The solution will be unique as long as
\(f\) is monotonic in the region of
interest. The invest function computes \(\widehat{x}_0\) by calling the
stats functions predict and uniroot
to solve \[y_0 - f\left(x;
\widehat{\boldsymbol{\theta}}\right) = 0\] numerically for \(x\).
Approximate confidence intervals
Equation @ref(eq:x0-inversion-interval) gives a closed-form expression for the inversion interval for the case of simple linear regression. In more complicated cases, such an expression is not available and the interval must be found numerically. It can be shown (see, for example, Seber and Wild 2003, 245–46) that
\[\label{eq:predictive-pivot} \mathcal{T}_{x_0} = \frac{\mathcal{Y}_0 - f\left(x_0; \widehat{\boldsymbol{\theta}}\right)}{\left\{\widehat{\sigma}^2 + \widehat{S}_{x_0}^2\right\}^{1/2}} \stackrel{\cdot}{\sim} t_{n-p}, (\#eq:predictive-pivot) \]
where \(\widehat{S}_{x_0}^2\) is the estimated variance of \(f\left(x_0; \widehat{\boldsymbol{\theta}}\right)\). An approximate \(100(1-\alpha)\%\) confidence interval for \(x_0\) is then given by the set
\[\label{eq:inversion} \bigg\{ x: t_{\alpha/2, n-p} < \mathcal{T}_x < t_{1-\alpha/2, n-p} \bigg\}. (\#eq:inversion) \]
Essentially, invest finds the lower and upper limits for
this interval by solving the equations \[\mathcal{T}_x - t_{\alpha/2, n-p} = 0 \quad
\text{and} \quad \mathcal{T}_x - t_{1-\alpha/2, n-p} = 0\]
numerically for \(x_0\). For the
special case of the simple linear regression model, these limits
coincide with Equation @ref(eq:x0-inversion-interval) and the coverage
probability is exactly \(1-\alpha\).
The Wald interval for \(x_0\) is also easily extended. It has the basic form: \(\widehat{x}_0 \pm t_{\alpha/2, n-p}\widehat{\mathrm{se}}\left(\widehat{x}_0\right)\), where \(p\) is the dimension of \(\boldsymbol{\theta}\). The estimated standard error \(\widehat{\mathrm{se}}\left(\widehat{x}_0\right)\) is based on a first-order Taylor series approximation. For the special case of the simple linear regression model, this approximation results in Equation @ref(eq:x0-se).
Since the point estimate and confidence intervals for \(x_0\) are obtained numerically using
uniroot, invest has the additional options
lower, upper, tol, and
maxiter. See the reference manual for details.
To get an idea of how invest works, consider the data in
Figure 4 which were generated from the
model \(\mathcal{Y} = 5 + x-\sin(x) +
1.5\mathcal{Z}\), where \(\mathcal{Z}\) is a standard normal random
variable. Suppose we want to estimate \(x_0\) given a new observation, say \(y_0 = 22\). Fitting a model of the form
\(f(x; \theta) = \theta_1 +
\theta_2\left[x-\sin(x)\right]\) to the sample, we obtain \(\widehat{f}(x) = 5.490 +
0.941\left[x-\sin(x)\right]\), which is certainly invertible, but
\(\widehat{f}^{-1}\) cannot be
expressed in closed-form using a finite number of terms; thus, \(\widehat{x}_0 = \widehat{f}^{-1}(y_0 =
22)\) must be obtained numerically. The following chunk of code
generates the sample data, calculates \(\widehat{x}_0\) and an approximate 95% Wald
interval for \(x_0\) using the
invest function.
set.seed(101) # for reproducibility
x <- rep(seq(from = 0, to = 25, by = 2), each = 2)
d <- data.frame(x, y = 5 + x - sin(x) + rnorm(length(x), sd = 1.5))
mod <- lm(y ~ I(x - sin(x)), data = d)
res <- invest(mod, y0 = 22, interval = "Wald")
## Figure 4
plotFit(mod, interval = "prediction", shade = TRUE, col.pred = "seagreen1",
extend.range = TRUE)
abline(h = 22, v = res$estimate, lty = 2)The point estimate is \(\widehat{x}_0 =
16.7053\) with an estimated standard error of 0.8909. The code
used by invest for obtaining the point estimate is
essentially
uniroot(function(x) predict(mod, newdata = list("x" = x)) - 22,
lower = min(d$x), upper = max(d$x))$rootThe code for obtaining the standard error used in the Wald interval is slightly more involved:
## Write x0.hat as function of parameters (theta1.hat, theta2.hat, Y0)
x0.fun <- function(params, object = mod) {
object$coefficients <- params[1:2]
uniroot(function(x) predict(object, list("x" = x)) - params[3],
lower = 0, upper = 25, tol = 1e-10)$root
}
## Variance-covariance matrix of (theta1.hat, theta2.hat, Y0)'
covmat <- diag(3)
covmat[1:2, 1:2] <- vcov(mod)
covmat[3, 3] <- summary(mod)$sigma^2
## Numerically evaluate gradient
params <- c(coef(mod), y0 = 22)
grad <- attr(numericDeriv(quote(x0.fun(params)), "params"), "gradient")
## Calculate standard error
(se <- sqrt(grad %*% covmat %*% t(grad)))
The following example uses the nasturtium data from the drc package. These data were analyzed in Racine-Poon (1988) using an approximate Bayesian approach. Bioassays were performed on a type of garden cress called nasturtium. Six replicates of the response (plant weight in mg after the third week of growth) were measured at seven preselected concentrations of an agrochemical. The weights corresponding to three new soil samples, all sharing the same (unknown) concentration \(x_0\), were observed to be 309, 296, and 419 mg. The block of code below fits a log-logistic model \[f(x; \boldsymbol{\theta}) = \left\{ \begin{array}{l l} \theta_1, &\quad x = 0, \\ \theta_1 / \left[1 + \exp\left\{\theta_2 + \theta_3\ln(x)\right\}\right], &\quad x > 0, \end{array} \right.\] and computes an approximate 95% inversion interval for \(x_0\).
## Load package containing nasturtium data
library(drc)
## Fit log-logistic model
mod <- nls(weight ~ theta1/(1 + exp(theta2 + theta3 * log(conc))),
start = list(theta1 = 1000, theta2 = -1, theta3 = 1),
data = nasturtium)
plotFit(mod, interval = "prediction") # figure not shown
## Compute approximate 95% inversion interval
invest(mod, y0 = c(309, 296, 419), interval = "inversion")The interval obtained is (1.7722, 2.9694) with a point estimate of 2.2639. We can check the point estimate manually. Some algebra gives \[\widehat{x}_0 = \exp\left\{\frac{\log\left(\widehat{\theta}_1/\bar{y}_0 - 1\right) - \widehat{\theta}_2}{\widehat{\theta}_3}\right\} = 2.2639.\]
A Wald interval can also be obtained by running the following line of code:
invest(mod, y0 = c(309, 296, 419), interval = "Wald")the output for which is
estimate lower upper se
2.2639 1.6889 2.8388 0.2847In certain cases (i.e., when \(\widehat{x}_0\) can be written in
closed-form), one can use the very useful deltaMethod
function from the car package
(Fox and Weisberg 2011) to compute the
approximate standard error used in the Wald interval. For the nasturtium
example, the minimal code to obtain \(\widehat{\mathrm{se}}\left(\widehat{x}_0\right)\)
is
## Using the deltaMethod function in the car package
library(car)
covmat <- diag(4)
covmat[1:3, 1:3] <- vcov(mod)
covmat[4, 4] <- summary(mod)$sigma^2 / 3 # since length(y0) = 3
(se <- deltaMethod(c(coef(mod), y0.bar = mean(c(309, 296, 419))),
g = "exp((log(theta1 / y0.bar - 1) - theta2) / theta3)",
vcov. = covmat)$SE)which produces an estimated standard error of 0.2847019, the same as
that obtained automatically by invest.
As indicated by the bootstrap distribution obtained in the next section, the symmetric Wald interval obtained here seems unrealistic for this problem. In the next section, we show how to obtain a bias-corrected and accelerated (\(BC_a\)) bootstrap confidence interval for \(x_0\) using the boot package (Canty and Ripley 2014).
Bootstrap calibration intervals
The bootstrap (Efron 1979) provides an
alternative means for computing calibration intervals. This is useful in
nonlinear calibration problems where inference traditionally relies on
large samples, approximate normality, and linear approximations. A
practical guide to the bootstrap is provided by Davison and Hinkley (1997) and the companion R
package boot, and bootstrap resampling for controlled
calibration is discussed in Jones and Rocke
(1999). Although it is likely for a "bootstrap"
option to appear in a future release of investr, it is quite
simple to set up using the recommended boot package. First, we
discuss a naive approach to calculating bootstrap calibration
intervals.
A simple, but ultimately wrong approach to resampling in controlled calibration is demonstrated in the following example for the nasturtium data:
library(boot)
## Function to compute estimate of x0
x0.fun <- function(object, y) {
theta <- unname(coef(object))
exp((log(theta[1] / mean(y) - 1) - theta[2]) / theta[3])
}
## Bootstrap setup
y0 <- c(309, 296, 419)
res <- resid(mod) - mean(resid(mod)) # center the residuals
n <- length(res)
boot.data <- data.frame(nasturtium, res = res, fit = fitted(mod))
boot.fun <- function(data, i) {
boot.mod <- nls(fit + res[i] ~ theta1 / (1 + exp(theta2 + theta3 * log(conc))),
start = list(theta1 = 1000, theta2 = -1, theta3 = 1), data = data)
## Make sure the original estimate also gets returned
if (all(i == 1:n)) x0.fun(mod, y0) else x0.fun(boot.mod, y0)
}
## Run bootstrap simulation (takes about 50s on a standard laptop)
set.seed(123) # for reproducibility
res <- boot(boot.data, boot.fun, R = 9999) # collect 9,999 bootstrap samples
boot.ci(res, type = "bca") # obtain BCa confidence interval for x0This produces an approximate 95% \(BC_a\) calibration interval of (2.04, 2.52), quite shorter than the ones produced by the inversion and Wald methods in the previous section. Did the bootstrap really do that well? The answer here is no, but it is not the bootstrap’s fault. Recall that \(\bar{y}_0\) is an observed value of the random variable \(\overline{\mathcal{Y}}_0\) which has variance \(\sigma^2/m\). This source of variability is ignored in the bootstrap Monte Carlo simulation above, which treats \(\bar{y}_0\) as a fixed constant. Compare the interval just obtained with the output from
invest(mod, y0 = mean(c(309, 296, 419)), mean.response = TRUE)We can get a hold of this extra source of variability by again
sampling with replacement from the centered residuals. In particular, we
resample the responses \(y_{0j}\) \((j = 1, 2, \dotsc, m)\), by randomly
selecting \(m\) additional residuals
\(e_j^\star\) from the centered
residuals and calculating the bootstrap responses \[y_{0j}^\star = \bar{y}_0 + e_j^\star, \quad j =
1, 2, \dotsc, m.\] Let \(\bar{y}_0^\star = \sum_{j=1}^m
y_{0j}^\star/m\). The correct bootstrap replicate of \(x_0\) is given by \(\widehat{x}_0^\star = f^{-1}\left(\bar{y}_0^\star;
\widehat{\boldsymbol{\theta}}^\star\right)\). To implement this,
all we need to do is make the following changes to boot.fun
(the changes are colored magenta):
boot.fun <- function(data, i) {
boot.mod <- nls(fit + res[i] ~ theta1 / (1 + exp(theta2 + theta3 * log(conc))),
start = list(theta1 = 1000, theta2 = -1, theta3 = 1), data = data)
## Simulate the correct variance
Y0 <- y0 + sample(data$res, size = length(y0), replace = TRUE)
## Make sure the original estimate also gets returned
if (all(i == 1:n)) x0.fun(mod, y0) else x0.fun(boot.mod, Y0)
}Rerunning the previous example with the modified
boot.fun function produces a more realistic 95% confidence
interval for \(x_0\) of (1.818, 2.950)
and an estimated standard error for \(\widehat{x}_0\) of 0.2861 (similar to that
obtained by the delta method earlier). Figure 5 shows a histogram and normal Q-Q plot of
the 9,999 bootstrap replicates of \(\widehat{x}_0\). The bootstrap distribution
of \(\widehat{x}_0\) is comparable to
the approximate posterior density of \(x_0\) given in Racine-Poon (1988, fig. 9). The bootstrap
distribution is skewed to the right and clearly not normal, suggesting
that the Wald interval may not be realistic for this problem.
Summary
We introduced the investr package for computing point
estimates along with inversion and Wald-based confidence intervals for
linear and nonlinear calibration problems with constant variance. We
also showed how the boot package can be used for constructing
approximate \(100(1 - \alpha)\%\)
calibration intervals, which for convenience, may be incorporated into a
future release of investr. The authors are currently working on
extending the package to handle the case of heteroscedastic errors,
random coefficient models [e.g., objects of class "lme"
from the recommended nlme package; Pinheiro et al. (2014)]), and even multivariate
calibration problems (e.g., objects of class "mlm").
Acknowledgements
The authors would like to thank two anonymous reviewers and the Editor for their helpful comments and suggestions.
The views expressed in this paper are those of the authors, and do not reflect the official policy or position of the United States Air Force, Navy, Department of Defense, or the U.S. Government.↩︎