On Mar 3, 2012, at 12:34 PM, drflxms wrote:

Thanx a lot Greg for the hint and letting me not alone with this!

Tried ellipse and it works well. But I am looking for something more
precise. The ellipse fits the real border to the nearest possible
ellipse. I want the "real" contour, if possible.

Meanwhile I found an interesting function named draw.contour
http://quantitative-ecology.blogspot.com/2008/03/plotting- contours.html
provided by "Forester", an "Assistant Professor in the Department of
Fisheries". - I love it how statistics brings people from completely
different specialities together. - I am a neurologist.

This function results not exactly in the same curve, I got with the code I posted last, but is in deed very close by (parallel). So I think both
solutions are probably true and differ only because of rounding and
method (..!?).

Still I am not completely happy with

1. the numeric solution of setting to 68% to get 1SD. I'd prefer a
symbolic, formula driven solution instead.
2. me not comprehending the code completely.

While 2. will be hopefully solved soon by me delving into the code, 1
remains...


You could take a look at answers to similar questions on StackOverflow:

http://stackoverflow.com/questions/6655268/ellipse-containing-percentage-of-given-points-in-r

http://stackoverflow.com/questions/7055848/plot-ellipse-bounding-a-percentage-of-points

http://stackoverflow.com/questions/7718569/how-to-plot-90-confidence-bands-with-locfit-in-r

http://stackoverflow.com/questions/7511889/ellipse-representing-horizontal-and-vertical-error-bars-with-r

Searching on "plot confidence ellipse" in the rhelp archives should be similarly productive.

There's also a non-parametric 2d boundary plotting method called a bagplot and I think it's implemented in a package by the same name.

--
David.

Best, Felix


Am 03.03.12 17:28, schrieb Greg Snow:
Look at the ellipse package (and the ellipse function in the package)
for a simple way of showing a confidence region for bivariate data on
a plot (a 68% confidence interval is about 1 SD if you just want to
show 1 SD).

On Sat, Mar 3, 2012 at 7:54 AM, drflxms <drfl...@googlemail.com> wrote:
Dear all,

I created a bivariate normal distribution:

set.seed(138813)
n<-100
x<-rnorm(n); y<-rnorm(n)

and plotted a scatterplot of it:

plot(x,y)

Now I'd like to add the 2D-standard deviation.

I found a thread regarding plotting arbitrary confidence boundaries from
Pascal Hänggi
http://www.mail-archive.com/r-help@r-project.org/msg24013.html
which cites the even older thread
http://tolstoy.newcastle.edu.au/R/help/03b/5384.html

As I am unfortunately only a very poor R programmer, the code of Pascal Hänggi is a myth to me and I am not sure whether I was able to translate the recommendation of Brain Ripley in the later thread (which provides
no code) into the the correct R code. Brain wrote:

You need a 2D density estimate (e.g. kde2d in MASS) then compute the
density values at the points and draw the contour of the density which includes 95% of the points (at a level computed from the sorted values via quantile()). [95% confidence interval was desired in thread instead
of standard deviation...]

So I tried this...

den<-kde2d(x, y, n=n) #as I chose n to be the same as during creating the distributions x and y (see above), a z-value is assigned to every
combination of x and y.

# create a sorted vector of z-values (instead of the matrix stored
inside the den object
den.z <-sort(den$z)

# set desired confidence border to draw and store it in variable
confidence.border <- quantile(den.z, probs=0.6827, na.rm = TRUE)

# draw a line representing confidence.border on the existing scatterplot
par(new=TRUE)
contour(den, levels=confidence.border, col = "red", add = TRUE)

Unfortunately I doubt very much this is correct :( In fact I am sure
this is wrong, because the border for probs=0.05 is drawn outside the
values.... So please help and check.
Pascal Hänggis code seems to work, but I don't understand the magic he
does with

pp <- array()
for (i in 1:1000){
      z.x <- max(which(den$x < x[i]))
      z.y <- max(which(den$y < y[i]))
      pp[i] <- den$z[z.x, z.y]
}

before doing the very same as I did above:

confidencebound <- quantile(pp, 0.05, na.rm = TRUE)

plot(x, y)
contour(den, levels = confidencebound, col = "red", add = TRUE)


My problems:

1.) setting probs=0.6827 is somehow a dirty trick which I can only use
by simply knowing that this is the percentage of values inside +-1sd
when a distribution is normal. Is there a way doing this with "native"
sd function?
sd(den.z) is not correct, as den.z is in contrast to x and y not normal any more. So ecdf(den.z)(sd(den.z)) results in a percentile of 0.5644 in
this example instead of the desired 0.6827.

2.) I would like to have code that works with any desired confidence. Unfortunately setting probs to the desired confidence would probably be
wrong (?!) as it relates to den.z instead of x and y, which are the
underlying distributions I am interested in. To put it short I want the
confidence of x/y and not of den.z.


I am really completely stuck. Please help me out of this! Felix


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David Winsemius, MD
West Hartford, CT

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