We have two linear fits, one for each data-set (unfortunately they include weights but I'm willing to ignore that if there's a nice analytic solution to this). Data-set d1 and d2 are different in the sense that x is the same variable but includes different discrete points (so x in d1 might be 100,120,500,... and x in d2 might be 110,110,335,...), and their y's are not the same. So in R jargon:

f1 <- lm(y~x,weights=w,data=d1)
f2 <- lm(y~x,weights=w,data=d2)

I then want to take the trend line of each of these fits and use it in the following equation:

z = 1 / exp( 0.02254*f1^2 / f2^2 )

I did that and plotted out this z, which is dandy. But! using the predict function in R, I managed to get the 95% confidence interval trend-lines of f1 and f2, which makes a nice graph. I'd love to be able to calculate/plot the same confidence interval trend-lines for the z line.

The two ways I thought would work are:

  1. to get the lower CI for z I take the lower CI for f1 and the upper CI for f2 and plug them into the z equation. To get the upper CI I do the opposite. I doubt this is correct.
  2. I generate random y data for specific x data from the f1 model (I'm not sure how to do that) and then generate random y data for the same x data from the f2 model. This way I have y's for d1 and d2 that share the same x data. I then can plug those discrete y's into the z equation and get my discrete z points. I then try to fit some model to that, though these discrete z points might not even follow a linear trend. this might be difficult or just silly.

So, this is my problem... Any brilliant ideas?


1 Answer 1


(All credit goes to Robert Shriver, I'm just posting this for him!) This is not an analytic solution, but I think that out of all the iterative solutions this ones' the nicest:

Take the mean regression parameters (intercept and slope) for both regressions (f1 and f2) along with the variance-covariance matrix of the intercept and slope parameters for each regression and plug them into a Multivariate normal distribution to simulate some new intercepts and slopes.

nn <- 100000
nx <- 100

bootf1 <- mvrnorm(nn, f1$coefficients, vcov(f1))
bootf2 <- mvrnorm(nn, f2$coefficients, vcov(f2))

This will be 10,000 random draws of the slope and intercept parameters for each regression. Use each one of these to determine the mean value of z across the range of x values you are interest in.

f1a <- bootf1[,2]
f1b <- bootf1[,1]
f2a <- bootf2[,2]
f2b <- bootf2[,1]

f1a <- rep(f1a, 1, nx)
f1b <- rep(f1b, 1, nx)
f2a <- rep(f2a, 1, nx)
f2b <- rep(f2b, 1, nx)

x  <- seq(-100, 1100, length=nx)
xs <- matrix(rep(x,nn), ncol=nx)

f1y <- f1b+f1a*xs
f2y <- f2b+f2a*xs

z = 1 / exp( 0.02254*f1y^2 / f2y^2 )

Then you can find the lower and upper CI of z as well as its mean by finding the .025 and .975 quantile of z at each x value.

setquantile <- function(x){
  r <- quantile(x, probs=c(0.025, 0.975))

ci <- apply(z, 2, setquantile)
mu <- apply(z, 2, mean)

plot(x, mu, type="l")
lines(x, ci[1,])
lines(x, ci[2,])

That's it. Please let me know if you find something better. Hope this helped someone!


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