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I'm wondering about ways to compute data and confidence ellipses around a bivariate median. For example, I can easily compute a data ellipse or a confidence ellipse for the bivariate mean of the following data (here just showing a data ellipse)

library("car")
set.seed(1)
df <- data.frame(x = rnorm(200, mean = 4, sd = 1.5),
                 y = rnorm(200, mean = 1.4, sd = 2.5))
plot(df)
with(df, dataEllipse(x, y, level = 0.68, add = TRUE))

enter image description here

But I'm struggling with how I'd do this for a bivariate median? In the univariate case I could just bootstrap resample to generate the required interval, but I not sure how to translate this into the bivariate case?

As pointed out by @Andy W, the median is not uniquely defined. In this instance we used the spatial median, by finding a point that minimises the L1 norm of the distances between observations at that point. An optimisation was used to compute the spatial median from the observed data points.

In addition, the x, y data pairs in the actual use case are two eigenvectors of a principal coordinates analysis of a dissimilarity matrix, hence x and y should be orthogonal, if that provides a particular avenue of attack.

In the actual use case, we want to compute the data/confidence ellipse for groups of points in the Euclidean space. For example:

enter image description here

The analysis is a multivariate analogue of a Levene's test of homogeneity of variances among groups. We use spatial medians or standard group centroids as the measure of multivariate central tendency, and wish to add the equivalent of the data ellipse in the figure above for the spatial median case.

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    $\begingroup$ The median in higher dimensions is not uniquely defined. You may be interested in boxplots generalized to higher dimensions though, e.g. The Bagplot: A Bivariate Boxplot (Rousseeuw et al., 1999). $\endgroup$ – Andy W Apr 18 '16 at 17:52
  • $\begingroup$ +1 Thanks @AndyW - I'd entirely forgotten about the bagplot (guess that's what you get for not teaching my EDA lectures for some years now - totally slipped my mind!) I should have indicated the type of median I had in mind --- I'll update the post, but we've calculated the spatial median, the point that minimises the L1 norm of the distances of the data points to that point. $\endgroup$ – Gavin Simpson Apr 18 '16 at 17:59
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    $\begingroup$ If you know that the $x$ and $y$ directions are orthogonal, why not estimate their medians independently? In other words, is there something special about the $L^1$ median for your application? $\endgroup$ – whuber Apr 18 '16 at 18:15
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    $\begingroup$ @whuber Ah, I might have misled there. I'll add a new fig that is a real eg of the use case. A dissimilarity matrix computed from the original data is embedded in a Euclidean space using PCoA. But what I neglected to mention is that we compute the spatial medians in this Euclidean space for groups of data points. Hence whilst the x and y are orthogonal over all groups, within an one group there may be correlation. See the updated figure in a minute for an illustration. Apologies for this; I didn't appreciate the importance of certain aspects of the real uses case when I posted the Q. $\endgroup$ – Gavin Simpson Apr 18 '16 at 18:30
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    $\begingroup$ I guess one approach can be based on bootstrapping: get the bootstrap distribution of your geometric median estimates and then mark a region that contains $1-\alpha$ fraction of the estimates. If you happy to assume that the estimates follow a normal distribution, then it's easy: fit a 2d Gaussian and draw a corresponding ellipse. If not, you can e.g. get the kernel density estimate of the 2d distribution and then find the region encompassing $1-\alpha$ of probability density. $\endgroup$ – amoeba Apr 18 '16 at 21:47
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This is a nice question.

I'll follow @amoeba's suggestion and bootstrap the spatial medians, using depth::med() with method="Spatial". However, there is a slight complication: med doesn't like it when there are duplicate data points, so we can't do a straightforward bootstrap. Instead, I'll draw a bootstrap sample and then jitter each point by a tiny amount - less than the minimum distances in each of the $x$ and $y$ dimensions in the original data sample - before calculating the spatial median.

Finally, I'll calculate the smallest ellipse covering a specified proportion (95%) of bootstrapped medians and plot.

library(depth)      # for med()
library(MASS)           # for cov.rob()
library(cluster)    # for ellipsoidhull()

# create data
set.seed(1)
df <- data.frame(x = rnorm(200, mean = 4, sd = 1.5),
                 y = rnorm(200, mean = 1.4, sd = 2.5))

# find minimum distances in each dimension for later jittering
foo <- outer(X=df$x,Y=df$x,FUN=function(xx,yy)abs(xx-yy))
delta.x <- min(foo[upper.tri(foo)])/2
foo <- outer(X=df$y,Y=df$y,FUN=function(xx,yy)abs(xx-yy))
delta.y <- min(foo[upper.tri(foo)])/2

# bootstrap spatial medians, using jittering
n.boot <- 1000
pb <- winProgressBar(max=n.boot)
boot.med <- matrix(NA,nrow=n.boot,ncol=2)
for ( ii in 1:n.boot ) {
    setWinProgressBar(pb,ii,paste(ii,"of",n.boot))
    index <- sample(1:nrow(df),nrow(df),replace=TRUE)
    bar <- df[index,] + 
      data.frame(x=runif(nrow(df),-delta.x,delta.x),
                 y=runif(nrow(df),-delta.y,delta.y))
    boot.med[ii,] <- med(bar,method="Spatial")$median
}
close(pb)

# specify confidence level
pp <- 0.95

# find smallest ellipse containing the specified proportion of bootstrapped medians
fit <- cov.rob(boot.med, quantile.used = ceiling(pp*n.boot), method = "mve")
best_ellipse <- ellipsoidhull( boot.med[fit$best,] )

plot(df)
points(boot.med,pch=19,col="grey",cex=0.5)
points(df)
lines(predict(best_ellipse), col="red")
legend("bottomright",bg="white",pch=c(21,19,NA),
    col=c("black","grey","red"),pt.bg=c("white",NA,NA),lwd=c(0,0,1),
    legend=c("Observations","Bootstrapped medians","Confidence ellipse"))

confidence ellipse

Finally, note that the bivariate spatial median is asymptotically normally distributed (Brown, 1983, JRSS, Series B), so we could also dispense with the "jittered bootstrap" above and directly calculate the ellipse, trusting that $n=200$ is "asymptotical enough". I may edit this post to include this parametric confidence ellipse if I find the time in the next days.

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