data visualizationellipsegeometrymathematical-statisticsr
Question: Best way to derive an ellipse formula when given bunch of points (say 60 points which when connected draws an ellipse)?
Background: Using R Momocs library function conf_ell which returns confidence ellipse coordinates. Would be nice to abstract those points into a formula.
You can see my ellipse points in the image below;
Any suggestions much appreciated,
Thanks
For the BIC and AIC, you can simply use AIC function as follow:
> model <- arima(x=sunspots, order=c(2,0,2), method="ML")
> AIC(model)
[1] 23563.39
> bic=AIC(model,k = log(length(sunspots)))
> bic
[1] 23599.05
The function AIC can provide both AIC and BIC. Look at ?AIC.
My answer is: you cannot see the condition of indeterminacy of factor $F$ on the above 3D plot because you will need 4D space to see it.
Let us, for a moment, reduce the whole picture by one dimension by dropping one of two X variables, while leaving there the prerequisites of factor analysis. (Please don't take the action for re-doing FA on a single variable - it is impossible. It is just imaginary deletion of one of the variables in order to spare one dimension.) So, we have some variable $X$ (centered), in the subject space of, say, N=3 individuals. The values are the coordinates onto the individuals:
ID X
1 2
2 -2
3 0
As the things go in FA, we must decompose $X$ into $F$ the common factor, and $U$, the unique factor, both orthogonal, but neither of them coinciding or orthogonal with $X$. $F$ and $U$ will define a plane, call it "plane U". The angle between $X$ and $F$ is determined from the analysis and it gives loading $a$ - the coordinate of $X$ on $F$.
We soon discover that the solution is not unique relative axes-individuals 1,2,3. Look at the left picture. Here "plane U" (grey) is defined to coincide with horizontal plane defined by axes 1-2 (beige). It may look a bit reclinate, but it's an illusion - it is actually a bit rotated about axis 3 because angle FX is somewhat less than angle UX. Now look at the right picture. Here, clearly, "plane U" is rotated about vector $X$ to become almost perpendicular to the horizontal "plane 12". In both cases we did not alter the coordinates of $X$ onto $U$ and $F$, including loading $a$ - we only spinned arbitrarily the same plane about the straight line $X$. Thereby we changed coordinates of endpoints of $F$ and $U$ onto axes 1,2,3. The coordinates, which are the values of factor $F$ and unique factor $U$.
Thus, we've just observed the indeterminacy of factor values. Factor can be determined in FA up to its loadings and its variance only; a infinite number of solutions exist in regards to factor values, - true factor values will always remain under question.
We've shown the indeterminacy by spinning "plane U" around axis of $X$, that is, a 2D space was revolving about 1D space in 3D space. A plane can revolve about some straight line in it, in a space; a line can revolve about some point in it, in a plane; in general: a q-dimensional space can freely spin about its q-1 dimensional subspace in a q+1 dimensional superspace.
Having grasped that, let's return to the initial picture posted with the question. Bringing back the temporarily removed second X variable, we now have 2D "plane X" and consequently 3D "space U" (it consists of orthogonally intersecting planes U1 and U2). That latter may freely spin about "plane X". As it spins - without changing any of the loadings ($a$'s) or vector lengths (variances) - the endpoint of $F$ rushes within the parent space, the N-dimensional space of subjects. But to be able to show it we need 3+1= 4D space (the "q+1 dimensional superspace"), which we cant'd draw in our world. So, we can't see the indeterminacy of factor values on that (geometrically correct) 3D picture, but it is there.
What about component values/scores and factor scores? Both are computed as linear combinations of variables, and so their vectors lie in "plane X". Component scores are true component values. Factor scores are approximations of unknown true factor values. Both component and factor scores can be fully determined in the analysis. If we apply once again to the pictures of this answer, showing the "reduced one-variable example", we'll find that the component or the factor scores variate should lie within 1D space X, the $X$ itself. So, no revolving can occur. The length of the component/variate vector is defined in the analysis, and its endpoint gets fixed in 3D space of individuals. No indeterminacy.
To conclude (staring again at the initial plot): What lies in the space X of the observed variables - is fixed, up to the case values. What transcends that space - namely the m-dimensional common factor space (m=1 in our situation) + the p-dimensional unique factors space, orthogonally intersecting with the former - is freely turnable, in a lump, about the space X in the grand space of N subjects. Therefore factor values are not fixed, while component values or estimated factor scores are.
Best Answer
A straightforward way, especially when you expect the points to fall exactly on an ellipse (yet which works even when they don't), is to observe that an ellipse is the set of zeros of a second order polynomial
$$0 = P(x,y) = -1 + \beta_x\,x + \beta_y\,y + \beta_{xy}\,xy + \beta_{x^2}\,x^2 + \beta_{y^2}\,y^2$$
You can therefore estimate the coefficients from five or more points $(x_i,\,y_i)$ using least squares (without a constant term). The response variable is a vector of ones while the explanatory variables are $(x_i,\,y_i, \,x_iy_i,\,x_i^2,\,y_i^2)$.
Here are 60 points drawn with considerable error.
The fit is shown as a black ellipse. The center and axes of the true underlying ellipse are plotted for reference.
When the points are known to fall on an ellipse, you may use any five of the points to estimate its parameters. (When you work out the normal equations you will obtain an explicit formula for the ellipse in terms of those ten coordinates.) It's best to choose the points situated widely around the ellipse rather than clustered in one place.
This is the
Rcode used to do the work. The three lines in the middle after "Estimate the parameters" illustrate the use of least squares to find the coefficients.