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English is not my native language and I'm neither a mathematician nor a statistican by trade so please excuse me if I fail to explain my problem in a concise way.

I'm trying to visualize a multivariate reference region of normally distributed measurements. A multivariate reference region can mathematically be described as an $n$-dimensional ellipsoid of the form: $$ (x-cx)'A(x-cx)=1 $$ where $cx$ is a $n \times 1$ vector representing the center of the ellipsoid and where the eigenvectors and eigenvalues of $A$ ($n \times n$ pos definite matrix) contains information on the orientation and semiaxis of the ellipsoid (wiki).

In order to present this I would like to plot the 2D intersection ellipse when $n-2$ positions of the vector $x$ is known. In the above equation $A$ and $cx$ will also be known.

Example: Let's say I have an multivariate reference region ($A$ and $cx$ known) for age, systolic blood pressure and diastolic blood pressure ($n=3$). Then I would like to plot the ellipse describing what systolic and diastolic blood pressure could be expected in subjects with the exact age of $60$ for instance. Then the vector $x=(60, s, d)$ where $s$ and $d$ are the unknowns describing the ellipse I want to visualize.

With the help of this article I've managed to find the orientation and dimensions of the 2D intersection ellipse of a 3-dimensional ellipsoid. It is, however, beyond me to generalize the solution into $n>3$ dimensions. One aspect that might simplify the problem is that the intersection plane will be perpendicular to the axes of the coordinate system.

Suggestions? Solutions? Anyone?

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Assuming the first two components of $x$ are free we can have the quadratic form evaluated to \begin{align} 1 =& \, ((x_1, x_2, rx) - (cx_1, cx_2, rcx))^t A ((x_1, x_2, rx) - (cx_1, cx_2, rcx)) \\ =& \, a_{11} (x_1-cx_1)^2 + a_{22} (x_2 - cx_2)^2 + (a_{12} + a_{21}) (x_1-cx_1)(x_2 - cx_2) + \\ & ((a_{13}+a_{31}) (x_3 - cx_3) + \ldots + (a_{1n}+a_{n1}) (x_n-cx_n)) (x_1 - cx_1) + \\ & ((a_{23}+a_{32}) (x_3 - cx_3) + \ldots + (a_{2n}+a_{n2}) (x_n-cx_n)) (x_2 - cx_2) + \\ & \sum_{i,j \ge 3} a_{ij} (x_i-cx_i)(x_j-cx_j) \\ =& \, a (x_1-cx_1)^2 + b (x_2-cx_2)^2 + c (x_1-cx_1)(x_2-cx_2) + d (x_1 - cx_1) + e (x_2 - cx_2) + (f + 1) \end{align} The coefficients $a, b, c, d, e, f$ contain only known values (the $x$ vector coordinates $x_3$ to $x_n$, the center vector coordinates $cx_i$ and matrix components $a_{ij}$).

Introducing $u = x_1 - cx_1$ and $v = x_2 - cx_2$ the above equation turns into $$ a u^2 + b v^2 + c u v + d u + e v + f = 0 $$ which is the general form of a conic section in the $u, v$-plane, see conic sections.

If $b = a_{22} \ne 0$ we can try to solve for $v$: \begin{align} 0 &= v^2 + \left(\frac{c}{b} u + \frac{e}{b} \right) v + \frac{a}{b} u^2 + \frac{d}{b} u + \frac{f}{b} \\ &= \left( v + \frac{cu+e}{2b} \right)^2 - \left( \frac{cu+e}{2b} \right)^2 + \frac{a}{b} u^2 + \frac{d}{b} u + \frac{f}{b} \end{align} which gives $$ v = -\frac{cu+e}{2b}\pm\sqrt{\left( \frac{cu+e}{2b} \right)^2 - \frac{au^2 + du + f}{b}} $$

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  • $\begingroup$ It is possible that I have made an error somewhere. The basic idea you should get is: You have two variables and if you start sorting all terms into terms with variables and those which are constant you should end up with a similar conic section in the two variables, like that formula that leads to the one for $u$ and $v$. $\endgroup$ – mvw Jan 3 '16 at 20:56
  • $\begingroup$ This is the solution I was looking for. Beautiful! Upon implementation in Stata I fail miserably however... I have interpreted the coefficients a-f like this for n=3: a=A[1,1], b=A[2,2], c=A[1,2]+A[2,1], d=(A[1,3]+A[3,1])(x3-cx3), e=(A[2,3]+A[3,2])(x3-cx3) and f=A[3,3](x3-cx3)(x3-cx3). Have I interpreted this correctly? I get a negative value within the square root when there should be a solution. Most likely I've got some typo in the code I can't find, but I would like to check that I've understood you correctly. $\endgroup$ – Jonas Selmeryd Jan 3 '16 at 21:05
  • $\begingroup$ A "1" disappered in your solution above when introducing u and v. When correcting for this it seems to work fine! Thanks! $\endgroup$ – Jonas Selmeryd Jan 3 '16 at 21:20
  • $\begingroup$ The constant part $f+1$ of the first equation was written that way, so that both sides of the equation can be substracted by $1$. $\endgroup$ – mvw Jan 3 '16 at 21:23
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    $\begingroup$ Thanks John. Happy to read this. $\endgroup$ – mvw Jul 17 '18 at 9:18
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Without loss of generality suppose $x_3,\ldots,x_n$ are fixed. The resulting quadratic equation in $x_1$ and $x_2$ has a quadratic, a linear and a constant part. You can obtain those by separating $A,$ $c$ and $x$ into blocks separating the first two coordinates from the other ones.

The quadratic part is still given by the upper left $2\times2$ submatrix of the original matrix $A.$ That gives us the orientation of the axes of the intersection ellipse. The linear part is the sum of the original linear part (first two elements of the matrix product $Ac$) and the upper right $2\times(n-2)$ part of $A$ applied to the separate blocks $(x_1\ x_2)$ and $(x_3 \ldots x_n)$. That gives us the center of the intersection. For the actual size of the two axes we need to complete the squares in the quadratic and linear terms.

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