# Location of “x standard deviation(s)” for multivariate normals

In 1D with scalar parameters $$(\mu,\sigma^2)$$, it is common to represent normally distributed data with error bars spanning $$[\mu-\sigma, \mu+\sigma]$$.

In n-D with parameters $$(\mu, \Sigma)$$, where $$\mu\in\mathbb{R}^n$$ and $$\Sigma$$ is $$n\times n$$ SPD, one infers the "shape" of multivariate-normal distributions by looking at equiprobable locations, given by $$(x-\mu)^t\ \Omega\ (x-\mu) = \mathrm{C}$$, where $$C$$ is some constant, and $$\Omega = \Sigma^{-1}$$ is the precision matrix. These are, of course, ellipsoids.

Q1 But what is the location of the points corresponding to "one standard-deviation" away from $$\mu$$? Is it the ellipsoid obtained by taking all points $$x$$ such that $$\| x - \mu \| = 1$$? I.e. the mapping of the unit sphere centred at $$\mu$$ by the precision matrix?

Q2 If so, what are the coordinates of those points? The formula $$(x-\mu)^t\ \Omega\ (x-\mu)$$ is scalar, how does one obtain the n-D coordinates of those points on the "one st-dev" ellipsoid?

(Note: apologies if that sounds like homework, I can assure you that it is not).

• Please see stats.stackexchange.com/search?q=confidence+ellipse It's difficult to determine what you are asking, because you have already provided an equation for the ellipsoid. In what form are you hoping to have the "coordinates of those points"? – whuber Jan 10 at 17:19
• @whuber Sorry if my question was unclear to you, but the coordinates of those points were nowhere in my post. Please see my attempt at answering my own question below; I am happy to reformulate if it makes it clearer what I was after. – Sheljohn Jan 10 at 17:45

From the geometric interpretation of multivariate normals written on Wikipedia, the location of the points "one standard deviation away" from $$\mu$$ would be: $$\big\{\ \mu + U\Lambda^{1/2}x\; \big|\ \|x\|=1 \big\}$$ where $$\Sigma = U\Lambda U^t$$ such that:
• $$U$$ is a rotation matrix ($$UU^t = I$$ and $$\det(U) = 1$$) containing the eigen-vectors of $$\Sigma$$ (which are also the eigenvectors of $$\Omega$$) in column;
• $$\Lambda$$ is a diagonal matrix with the corresponding eigen-values, which are also the squared-lengths of the axes of the ellipsoid, and the reciprocals of $$\Omega$$'s eigenvalues.
In effect, this means that the ellipsoid corresponding to "one std from $$\mu$$" can be obtained by:
• scaling the unit sphere along the canonical axes with $$\Lambda^{1/2}$$;
• rotating the scaled sphere with $$U$$;
• and translating it by $$\mu$$.
• @whuber Thank you :) The "coordinates" are given by the formula $\mu + U\Lambda^{1/2} x$ for points $x$ on the unit sphere. So I can obtain points on that particular ellipsoid by sampling $\mathcal{N}(0,I)$, normalising those sampled points to have norm 1, and apply the transformation given. – Sheljohn Jan 10 at 17:58