# What is the difference between the MCD and the MVE estimators?

As far as I understand,

• the Minimum Covariance Determinant (MCD) estimator looks for the subset of h data points whose covariance matrix has the smallest determinant.
• the Minimum Volume Ellipsoid (MVE) searches for the ellipsoid with the smallest volume that covers h data points.

Once either subset is found, the robust estimate of the covariance matrix is given by that of those h points.

Can you explain me the conceptual difference between the MCD and MVE? Isn't the determinant related to the volume?

EDIT

User603 have argued that both estimators are solutions to different optimisation problems. Even if I do not understand everything in its answer, I do trust him. But then, what about this formula for the volume of the ellipsoid $(x-v)'A^{-1}(x-v)=1$ $$V = \frac{4}{3}\pi\sqrt{\det(A)}$$ according to which minimising V is equivalent to minimising $\det(A)$...

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 yes, for an ellipsoid! (see the last paragraph in my comment, about the strong relation between both problems in the case of symmetrically distributed $x$'s). Also, you really don't need to trust user603 -- it's quiet well spelled out in the linked paper. – user603 Jan 29 '12 at 16:04 do you mean that in the multinormal case, both estimators are equivalent? – user7064 Jan 29 '12 at 16:06 no:> even in multi-dimension, they are solution to different (but related) problems. Both belong to the class of minimum-volume estimators (together with the S and $\tau$ estimators). – user603 Jan 29 '12 at 17:41

First off, it is easier to answer your question in the uni variate case because then both estimators have an explicit solution as a function of the series of sorted observations $x_{(1)}\leq x_{(2)}\leq,...,x_{(n-1)}\leq x_{(n)}$.

The uni-variate version of the MVE is also known as the short estimator and is the solution to

$\underset{1\leq i\leq(n-h+1)}{\arg.\min.}\;\;\;x_{(i+h-1)}-x_{(i)}\;\;\;[1]$

The uni-variate version of the MCD is also known as the truncated likelihood estimator and is the solution to:

$\underset{1\leq i\leq(n-h+1)}{\arg.\min.}\;\;\;\displaystyle\frac{1}{h}\displaystyle\sum_{j=1}^{i+h-1}x_{(j)}^2-\left(\frac{1}{h}\displaystyle\sum_{j=1}^{i+h-1}x_{(j)}\right)^2\;\;\;[2]$

Now, for symmetrical distribution, the objective function in $[2]$ is proportional to the truncated likelihood (the likelihood computed over $h$ observations). This is also the reasons why it has much higher efficiency than $[1]$ (while enjoying a similar breakdown point).

You will find more details on this in this very clear paper:

• Maxbias Curves of Robust Location Estimators based on Subranges, Croux and Haesbroeck, Metrika, 2001

So these estimators are solutions to different problems, with the definitions above it is easy to show that the solution to equation [1] does in general not equal the solution to equation [2] (for the case of symmetrically distributed $x$'s both problems are strongly related, but [2] is statistically much more efficient).

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 Thanks for your answer. I do not get everything yet... mainly because I do not manage to make the link between my definitions and yours. Moreover, I thought that minimising the volume of the ellipsoid was equivalent to minimising the determiant... – user7064 Jan 29 '12 at 15:20 for the mcd: for a scalar $a$, $\det(a)=a$, so you could add a $\det$. These are not 'my' definitions btw. Where did you get 'your' definition from? Check out the linked the paper also (it avail. online ungated), it compares both estimators in greater lengths. – user603 Jan 29 '12 at 15:41 it would make things easier if you could explain what part(s) you didn't get. – user603 Jan 30 '12 at 1:06