Say you fit a model using R's lm() function. What you get in return are coefficients that represent an estimate of the linear function that gave rise to the data. The assumption is the response of the model is normally distributed with a mean equal to the linear function and a standard deviation equal to the standard deviation of the residuals. Using notation we express this as follows for a simple linear model with intercept and slope coefficients:

\[ Y_{i} \sim N(\beta_{0} + \beta_{1}x_{i},\sigma) \]

The implication of this in the world of statistical computing is that we can simulate responses given values of x. R makes this wonderfully easy with its simulate() function. Simply pass it the model object and the number of simulations you want and it returns a matrix of simulated responses. A quick example:

set.seed(1)
# generate data from the linear function y = 3 + 4.2x with random noise
x <- seq(12, 23, length = 100)
y <- 3 + 4.2 * x + rnorm(100, 0, 5)
mod <- lm(y ~ x)
summary(mod)

## 
## Call:
## lm(formula = y ~ x)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -11.700  -3.029   0.078   2.926  11.487 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    3.900      2.504    1.56     0.12    
## x              4.180      0.141   29.70   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 4.51 on 98 degrees of freedom
## Multiple R-squared:   0.9,   Adjusted R-squared:  0.899 
## F-statistic:  882 on 1 and 98 DF,  p-value: <2e-16

Regressing y on x estimates the coefficients to be about 3.9 and 4.18, pretty close to the true values of 3 and 4.2. It also estimates the standard deviation of the error term to be 4.51, not too far from the 5 we specified in rnorm() when we generated the noise. Now let's use our model to simulate responses (i.e., y values).

head(simulate(mod, 5))

##   sim_1 sim_2 sim_3 sim_4 sim_5
## 1 51.26 55.90 58.09 58.91 54.40
## 2 54.71 62.14 49.79 63.08 53.18
## 3 50.87 62.15 63.88 52.26 49.64
## 4 56.16 53.96 53.72 53.69 55.50
## 5 52.96 45.60 63.38 54.04 60.39
## 6 64.35 67.65 63.20 54.68 63.57

The simulate() function returns a data frame. In this case the data frame has 100 rows and 5 columns. (I use the head() function to only show the first six rows.) Each column represents a simulated set of responses from our linear model given our x values. In the first row we have 5 values drawn from a Normal distribution with mean \( 3.89 + 4.17x_{1} \) and a standard deviation of 4.51. In the second row we have a value drawn from a Normal distribution with mean \( 3.89 + 4.17x_{2} \) and a standard deviation of 4.51. And so on. We could do this “manually” as follows:

simResp <- matrix(0, 100, 5)
for (i in 1:5) {
    simResp[, i] <- rnorm(5, c(coef(mod)[1] + coef(mod)[2] * x), rep(summary(mod)$sigma, 
        5))
}
simResp[1:6, ]

##       [,1]  [,2]  [,3]  [,4]  [,5]
## [1,] 52.52 48.93 54.80 52.14 52.90
## [2,] 61.30 47.77 56.02 53.17 46.46
## [3,] 57.37 53.98 53.25 46.90 63.04
## [4,] 57.90 64.48 49.14 54.33 63.41
## [5,] 55.30 56.91 67.99 54.80 59.03
## [6,] 52.52 48.93 54.80 52.14 52.90

But the simulate() function is much faster and easier to use.

Now 5 simulations is rather puny. A better number would be 1000. Then we can use our simulated y values to generate 1000 linear models, and thus have 1000 coefficient estimates. We can then use those estimates to create kernel density plots that allow us to visualize the amount of uncertainty in our coefficient estimates. Let's do that. Notice how I convert the output of the simulate() function to a matrix. That allows me to feed the 1000 responses to the lm() function in one fell swoop instead of coding a for loop to iterate through each column.

simResp <- data.matrix(simulate(mod, 1000))
modSim <- lm(simResp ~ x)

modSim is the usual linear model object, but with the results of 1000 models instead of 1. Here are the coefficients for the first 5 models:

coef(modSim)[1:2, 1:5]

##               sim_1 sim_2 sim_3 sim_4 sim_5
## (Intercept) -0.5932 3.949 3.861 5.683 5.038
## x            4.3836 4.166 4.199 4.062 4.113

Let's create a scatterplot of these coefficients.

plot(modSim$coefficients[1, ], modSim$coefficients[2, ], xlab = "intercept", 
    ylab = "slope", main = "Scatterplot of coefficients from simulated models")

We see that the variability of the slope estimates is pretty small (3.8 – 4.6), while the intercept estimates vary widely (from below 0 to 10). This jives with our original model (see output above) where the slope coefficient was signficant with a small standard error while the intercept was not significant and had a relatively large standard error.

Next we create kernel density plots of our coefficient estimates. In addition I add a vertical line to represent the mean of the 1000 coefficient estimates and add the mean itself to the plot using the text() function. I use the difference in the range of the axes to allow me to automatically determine suitable coordinates.

d1 <- density(modSim$coefficients[2, ])
d2 <- density(modSim$coefficients[1, ])
# slope
plot(d1, main = "distribution of slope coefficients")
abline(v = mean(modSim$coefficients[2, ]))
text(x = mean(modSim$coefficients[2, ]) + diff(range(d1$x)) * 0.05, y = diff(range(d1$y)) * 
    0.05, labels = round(mean(modSim$coefficients[2, ]), 2))

# intercept
plot(d2, main = "distribution of intercept coefficients")
abline(v = mean(modSim$coefficients[1, ]))
text(x = mean(modSim$coefficients[1, ]) + diff(range(d2$x)) * 0.05, y = diff(range(d2$y)) * 
    0.05, labels = round(mean(modSim$coefficients[1, ]), 2))

If you don't compare the scale of the x-axis in the two plots the coefficients appear to have similar variability. Perhaps a better way to do it is to put both plots in one window and set the limits to the x-axis for both plots to be the range of the estimated intercept coefficients.

par(mfrow = c(2, 1))
# slope
plot(d1, main = "distribution of slope coefficients", xlim = range(d2$x))
abline(v = mean(modSim$coefficients[2, ]))
text(x = mean(modSim$coefficients[2, ]) + diff(range(d2$x)) * 0.05, y = diff(range(d1$y)) * 
    0.05, labels = round(mean(modSim$coefficients[2, ]), 2))
# intercept
plot(d2, main = "distribution of intercept coefficients", xlim = range(d2$x))
abline(v = mean(modSim$coefficients[1, ]))
text(x = mean(modSim$coefficients[1, ]) + diff(range(d2$x)) * 0.05, y = diff(range(d2$y)) * 
    0.05, labels = round(mean(modSim$coefficients[1, ]), 2))

par(mfrow = c(1, 1))

Now we see just how uncertain our intercept estimate is compared to the slope estimate.

Finally, why don't we go ahead and plot all 1000 fitted lines over the original data along with our original fitted line and the traditional 95% confidence band we get using our original model.

plot(x, y, col = "red", pch = 19)
abline(mod)  # original fitted line
apply(coef(modSim), 2, abline, col = "gray", lty = 3)

# predicted responses from original model
modOrig <- predict(mod, interval = "confidence")
# add 95% confidence band
lines(x, modOrig[, "lwr"], col = "blue")
lines(x, modOrig[, "upr"], col = "blue")

The plotting of the 1000 lines essentially provides us a confidence band for the original fitted line. We see the blue lines are slightly closer to the fitted line than the band produced by our simulated models. That's because the blue lines represent the 95% confidence band. In other words, we would expect 95% of our 1000 estimated lines to fit within those blue lines. The gray lines we see outside of the blue lines are those 5% that don't fall within the 95% confidence band.

Working through Gelman and Hill’s book can be maddening. The exposition is wonderful but recreating their examples leads to new and exciting levels of frustration. Take the height and earnings example in chapter 4. It’s a simple linear regression of earnings on height. You’d think you could download the data from the book web site, use the code in the book, and reproduce the example. They sure lead you to think that. But it doesn’t work that way. For starters, when you load the data it has 2029 records. However the output of the regression in the book shows n = 1192. So subsetting needs to be done. As far as I can tell, the subsetting is not discussed in the book.

Now the author has an “earnings” folder on his site under an “examples” directory, which contains a R file called “earnings_setup.R“. (Never mind the book itself doesn’t appear to mention this directory on the web site.) So this seems to be the place where we find out how to subset the data. The key line of code is

ok <- !is.na (earn+height+sex+age) & earn>0 & yearbn>25

which creates a logical vector to subset the data. But when you run it and request the dimensions of the data frame you have 1059 records, not 1192! After trial and error I believe the subset to reproduce the results in the book should be

ok <- !is.na (earn+height+sex) & earn>0

That gave me 1192. For the record, here’s my full code:

heights <- read.dta ("http://www.stat.columbia.edu/~gelman/arm/examples/earnings/heights.dta")
attach(heights)
male <- 2 - sex # make male 1 (and therefore female = 0)
ok <- !is.na (earn+height+sex) & earn>0
heights.clean <- as.data.frame (cbind (earn, height, sex, male)[ok,])
heights.clean$log.earn <- log(heights.clean$earn)

OK, so now(!) we can reproduce their example:

earn.logmodel.3 <- lm (log.earn ~ height + male + height:male, data=heights.clean)

The reason I was interested in this example was for another example in chapter 7 on "using simulation to represent uncertainty in regression coefficients" (p. 142). In other words, instead of using the standard errors and intervals obtained from the predict() function in R, we compute uncertainties by simulation. It seems easy enough using the sim() function from the book's arm package. You do something like the following:

library(arm)
sim.1 <- sim(earn.logmodel.3, 1000)

The sim function takes two arguments: your model and the number of simulations. It returns a vector of simulated residual standard deviations and a matrix of simulated regression coefficients. We can use these to calculate confidence intervals for coefficients that do not have standard errors in the regression output. Their example asks "what can be said about the coefficient of height among men?" If you look up at the model, you can see it contains an interaction term of height and male. To answer the question we can't just look at the standard error of the height coefficient or just the interaction coefficient. There is no simple way to do what they ask using regression output. So the book instructs us to use the output of the sim function to answer this as follows:

height.for.men.coef <- sim.1$beta[,2] + sim.1$beta[,4]
quantile(height.for.men.coef, c(0.025,0.975))

Except that doesn't work. More frustration. It produces an error that says "Error in sim.1$beta : $ operator not defined for this S4 class" (instead of giving us a 95% interval). With some googling and persistence I was able to determine that the following is how it should be done:

height.for.men.coef <- sim.1@coef[,2] + sim.1@coef[,4]
quantile(height.for.men.coef, c(0.025,0.975))
         2.5%         97.5% 
-0.0004378039  0.0507464098 

Notice that "@coef" replaces "$beta". And with that I was able to finally reproduce the example I was interested in!

Now about this simulation function. While I appreciate functions that save time and make life easy, I do like to know how they work. Fortunately Gelman and Hill provide pseudo-code in the book. It goes like this:

1. Run your regression to compute the vector of estimated parameters, the unscaled estimation covariance matrix , and the residual variance
2. Create n random simulations for the coefficient vector and residual standard deviation. For each simulation draw:
   1. Simulate where X is a random draw from the distribution with n - k degrees of freedom.
   2. Given the random draw of , simulate from a multivariate normal distribution with mean and variance matrix

Not too bad. Let's use this to manually run our own simulations, so we have an idea of how the sim() function works. (Plus you may not want to use the arm package as it requires loading 9 more packages.)

Step 1 is easy enough. That's just running your regression as you normally would. Next we need to extract our estimated parameters, the unscaled covariance matrix and the residual standard deviation. We also need to snag degrees of freedom for our chi-square random draw. Here's how we can get them:

# extract coefficents
earn.logmodel.3$coef

# extract residual standard error from model
summary(earn.logmodel.3)$sigma

# extract unscaled covariance matrix
summary(earn.logmodel.3)$cov.unscaled

# extract k and n - k; first two elements in vector
summary(earn.logmodel.3)$df

Let's use this information to do a single simulation.

s.hat <- summary(earn.logmodel.3)$sigma
n.minus.k <- summary(earn.logmodel.3)$df[2]
library(MASS) # need for mvrnorm function to simulate draws from multivariate normal dist'n
# simulate residual standard deviation
(s.hat.sim <- s.hat*sqrt(n.minus.k/rchisq(1, n.minus.k)))
[1] 0.8591814
# use simulated residual standard deviation to simulate regression coefficients
mvrnorm(1,earn.logmodel.3$coef, s.hat.sim^2*summary(earn.logmodel.3)$cov.unscaled)
(Intercept)       height         male  height:male 
7.906605029  0.025163124 -0.160828921  0.007904422 

That seems to work. How about we try doing 1000 simulations. Here's one way:

n <- 1000
sim.2.sigma <- rep(NA,n)
sim.2.coef <- matrix(NA,n,4)
for (i in 1:n){
 sim.2.sigma[i] <- s.hat*sqrt(n.minus.k/rchisq(1, n.minus.k))
 sim.2.coef[i,] <- mvrnorm(1,earn.logmodel.3$coef,sim.2.sigma[i]^2*summary(earn.logmodel.3)$cov.unscaled)
}

Now let's see how our simulation results compare to what we got using the sim() function:

height.for.men.coef.2 <- sim.2.coef[,2] + sim.2.coef[,4]
quantile(height.for.men.coef.2, c(0.025,0.975))
        2.5%        97.5% 
-0.001828216  0.049499381 

Looks similar to what we got above. Nice. It's probably better to just use the sim() function for this sort of thing, but at least now we know a little more about what it's doing.