[R] contour for plotting confidence interval on scatter plot of bivariate normal distribution

[Greg Snow]

The key part of the ellipse function is:

    matrix(c(t * scale[1] * cos(a + d/2) + centre[1], t * scale[2] *
        cos(a - d/2) + centre[2]), npoints, 2, dimnames = list(NULL,
        names))

Where (if I did not miss anything) the variable 't' is derived from a
chisquare distribution and the confidence level, scale[1] and scale[2]
are the standard deviations of the 2 variables, d is the eccentricity
based on the correlation and a is just a sequence from 0 to 2*pi.  So
if you use 't' as 1 instead of derived based on confidence then you
would get a "1 SD" ellipse in the sense that any 1 dimensional slice
through the mean point would cut the ellipse at 1 SD from the mean.
You could then change t to 2 for the "2 SD" curve, etc.




On Sat, Mar 3, 2012 at 12:25 PM, drflxms <drflxms at googlemail.com> wrote:
> Thank you very much for your thoughts!
>
> Exactly what you mention is, what I am thinking about during the last
> hours: What is the relation between the den$z distribution and the z
> distribution.
> That's why I asked for ecdf(distribution)(value)->percentile earlier
> this day (thank you again for your quick and insightful answer on
> that!). I used it to compare certain values in both distributions by
> their percentile.
>
> I really think you are completely right: I urgently need some lessons in
> bivariate/multivariate normal distributions. (I am a neurologist and
> unfortunately did not learn too much about statistics in university :-()
> I'll take your statement as a starter:
>
> "Once you go into two dimensions, SD loses all meaning, and adding
> nonparametric density estimation into the mix doesn't help, so just stop
> thinking in those terms!"
>
> This makes me really think a lot! Is plotting the 0,68 confidence
> interval in 2D as equivalent to +-1 SD really nonsense!?
>
> By the way: all started very harmless. I was asked to draw an example of
> the well known target analogy for accuracy and precision based on "real"
> (=simulated) data. (see i.e.
> http://en.wikipedia.org/wiki/Accuracy_and_precision for a simple hand
> made 2d graphic).
>
> Well, I did by
>
> set.seed(138813)
> x<-rnorm(n); y<-rnorm(n)
> plot(x,y)
>
> I was asked whether it might be possible to add a histogram with
> superimposed normal curve to the drawing: no problem. "And where is the
> standard deviation", well abline(v=sd(... OK.
>
> Then I realized, that this is of course only true for one of the
> distributions (x) and only in one "slice" of the scatterplot of x and y.
> The real thing is is a 3d density map above the scatterplot. A very nice
> example of this is demo(bivar) in the rgl package (for a picture see i.e
> http://rgl.neoscientists.org/gallery.shtml right upper corner).
>
> Great! But how to correctly draw the standard deviation boundaries for
> the "shots on the target" (the scatterplot of x and y)...
>
> I'd be grateful for hints on what to read on that matter (book, website
> etc.)
>
> Greetings from Munich, Felix.
>
>
> Am 03.03.12 19:22, schrieb peter dalgaard:
>>
>> On Mar 3, 2012, at 17:01 , drflxms wrote:
>>
>>> # this is the critical block, which I still do not comprehend in detail
>>> z <- array()
>>> for (i in 1:n){
>>>        z.x <- max(which(den$x < x[i]))
>>>        z.y <- max(which(den$y < y[i]))
>>>        z[i] <- den$z[z.x, z.y]
>>> }
>>
>> As far as I can tell, the point is to get at density values corresponding to the values of (x,y) that you actually have in your sample, as opposed to den$z which is for an extended grid of all possible (x_i, y_j) combinations.
>>
>> It's unclear to me what happens if you look at quantiles for the entire den$z. I kind of suspect that it is some sort of approximate numerical integration, but maybe not of the right thing....
>>
>> Re SD: Once you go into two dimensions, SD loses all meaning, and adding nonparametric density estimation into the mix doesn't help, so just stop thinking in those terms!
>>
>
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-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow at imail.org
801.408.8111