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bio website quantdec.com
location Northeastern US
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Consultant (environmental and spatial stats a specialty), expert witness, and teacher. I can be reached through (outdated but still valid) links posted on my web site.

Twitter: @WilliamAHuber // ASA-P website: http://amstatphilly.org/


Why waste time learning, when ignorance is instantaneous?

--T(iger) Hobbes.

For any complex problem there is a simple solution. And it's always wrong.

--[Mis?]attributed to H.L. Mencken by Dava Sobel, Longitude.


Mar
1
comment How do you draw credible regions/intervals on a 2D PDF?
It would be good to know what these are "confidence regions" of. They are not confidence regions of the mean, mode, or median, for instance: such regions would be tremendously smaller in extent than shown here. To be clear, I'm not questioning your beautiful work, but I am asking for clarification of what it is intended to mean and how it is supposed to be interpreted.
Mar
1
answered How do you draw credible regions/intervals on a 2D PDF?
Mar
1
comment How do you draw credible regions/intervals on a 2D PDF?
The usual terminology for this is a 90% credible region, Jay. There typically exist infinitely many such regions in multivariate situations, so it helps to provide some constraints, such as asking for a credible region of the smallest diameter that encloses the mode (if the parameters are comparable to one another), for example, or to insist that the credible region be bounded by an isocontour of the density.
Mar
1
comment How do you draw credible regions/intervals on a 2D PDF?
NB: This is a confidence region for the mean, not for the PDF. I have asked the OP for clarification. (A CI for the PDF would be a pair of surfaces enclosing the PDF; it could not be depicted in any clear fashion with a contour plot or array plot.)
Mar
1
comment How do you draw credible regions/intervals on a 2D PDF?
Jay, what sort of "confidence intervals" do you want? The most natural ones would be confidence intervals for the density estimates themselves: that is, confidence intervals for values of the PDF. However, Platomaniac's answer seems to think you intended confidence intervals for the mean. The two are completely different things, so please clarify your intentions.
Feb
28
comment Symmetry-finding packages
@belisarius Thanks. That seems like a version of the reference supplied by the OP.
Feb
28
comment Symmetry-finding packages
Because we are an English language site, it would be appropriate (and kind) of you to point out that the entire text of that thesis--including the comments in the code--is in Greek.
Feb
28
reviewed Leave Open How can I change the default output number form to accounting form?
Feb
28
reviewed Leave Open Permanent minors
Feb
28
reviewed Leave Open How can I control the number format of exported data?
Feb
28
comment Showing a rectangular plot on an almost-closed sphere
This is on the right track, but the example does not seem mathematically correct, because it illustrates a series of maps from the plane to the sphere rather than a single, fixed map to the sphere. In fact, it appears to be a series of maps from the square to the sphere which, in the limit, will map only the square to the entire sphere. A mathematically correct answer would look like the texture on the sphere is fixed (immobile) and that the sphere is merely being more or less chopped off near its top.
Feb
27
comment Showing a rectangular plot on an almost-closed sphere
Thanks, Steve. I wasn't sure, because under a stereographic projection those axes won't look like the one in your image: there has to be a huge amount of distortion in the northern hemisphere to accommodate all the numbers between $2$ and $\infty$ in size :-).
Feb
27
comment Showing a rectangular plot on an almost-closed sphere
I answered before you added that image, Steve. But what precisely do you mean by "actual axes" on the surface of the sphere? What values do they purport to show? Explain that and perhaps we can find a solution that meets your needs :-).
Feb
27
comment Eigenvalues and Determinant of a large matrix
It makes sense only to evaluate this determinant relative to the determinant of $m$. When you divide DET by Det[m] you will get values around $10^{-13}$ down to $10^{-16}$ in size, indicating that only one to four decimals of accuracy have actually been lost out of the original $17$. In other words, DET is just the opposite of a "huge number": it is extremely small, relative to a natural measure of numerical size for this problem.
Feb
27
comment Showing a rectangular plot on an almost-closed sphere
@Daniel That's right. The importance of stereographic projection is that it is a rational transformation between the affine and projective planes, so that changing the pole of the projection--which I think is what you might have in mind with "affine slices"--just gives an alternative affine picture of the same cubic (or the same rational function, if you began with a rational function).
Feb
27
comment Mathematica vs Sigmaplot (Non LinearModelFit)
@rc Constant error variance is assumed by (unweighted) least-squares fitting. When that variance changes, intuitively it makes sense to put less weight on the points with larger variance--and in fact using appropriately weighted least squares is one way to fix up the procedure. But there are great advantages to re-expressing the data to produce constant error variances: interpretation is simpler, prediction is simpler and more reliable, and in physical applications often the re-expressions have meaning. These issues are extensively discussed on stats.stackexchange.com, Mary.
Feb
27
comment Mathematica vs Sigmaplot (Non LinearModelFit)
Unless you are confident the error variance is constant across different values of $x$, you should instead be fitting the model $\log(y) = \log(a_0) + a_1 t + a_2 t^2$. Use LinearModelFit; it's fast, easy, and can output a huge amount of additional information.
Feb
27
answered Showing a rectangular plot on an almost-closed sphere
Feb
26
comment Confirming the existence of a function related to a matrix
The requirements on $f$ imply $m_{ij} + m_{ji}$ = $f(i,j)+f(j,i)$ = $2f(i,j)$, giving a unique solution--as well as a formula for it--if any solution exists at all.
Feb
26
revised How can I get the solution of complicated implicit function?
added 654 characters in body