Looking for some intuition regarding the MCD estimator

The Minimum Covariance Determinant (MCD) estimator may be used to achieve robustness when estimating a covariance matrix. It looks for the subset of $h$ data points (out of $n > h$) whose covariance matrix has the smallest determinant and it computes the covariance of those $h$ points.

Why does minimising the determinant do the job?

• i think i answered this question in the paragraph under formula [2] in stats.stackexchange.com/questions/21907/… ... – user603 Jan 30 '12 at 6:33
• Sorry @user603 but I am not totally satisfied with your comment. Let us say that the subset of h points contains an outlier. I still do not understand why the determinant will be large... thank you – user7064 Jan 30 '12 at 8:59
• but I never claimed such a thing...if you looked at assyptotic bias curves (say in the paper I linked to in my previous answer) you would see that this is not a particularly original point either. – user603 Jan 30 '12 at 10:34
• Sorry... definitely, I do not get your point. If you still have some time to explain me the rationale behind the MCD estimator, can you do it independently from the other question please. I really don't understand why we try to minimise the determinant to achieve robustness... – user7064 Jan 30 '12 at 11:02
• Technically, 'we' don't try to minimize the determinant to achieve robustness but to achieve high break down point (which is a provable property of an estimator). It turns out that volume minimizing estimators (and hence also truncated likelihood ones like MCD) achieve this property. Are you asking for a proof that MCD has high break down point? – user603 Jan 30 '12 at 12:45

Okay, i tried an intuitive overview.

Suppose that $X\sim F_p(m,C)$, a known continuous distribution on $\mathbb{R}^p$ (often assumed elliptical but this can be generalized). So $X$ is a collection of $n$ draws from $F_p$.

I show it for the estimator of scatter, the proof for the estimator of location follows easily.

Denote $S(X)$ an estimator of $C$ computed from the uncontaminated data $X$, $S(X_{\epsilon})$ another estimator of $C$ computed on $X_{\epsilon}$, the contaminated data. $X_{\epsilon}$ is $X$ but for $\lfloor n\epsilon\rfloor$ of the observations which have been replaced by draws from $G_q$.

• max-bias of an estimator $S$, for a rate of contamination $\epsilon$ of a dataset $X$:

$\verb+bias+(\epsilon,S,X)=\displaystyle\underset{X_{\epsilon}}{\sup}||S(X)-S(X_{\epsilon})||_{\kappa}$

where $||.||_{\kappa}$ is the condition number metric.

• breakdown point of an estimator $S$:

$\epsilon^*(S,X)=\min\{\epsilon:\verb+bias+(\epsilon,S,X)=\infty\}$

The max bias is often hard to manipulate theoretically because it depends on the distribution of $X_{\epsilon}$, whereas the definition of the breakdown point contains no probability distribution.

• maximum breakdown point of an affine equivariant estimator $S$

For an AE estimator S and a dataset $X$ in general linear position in $\mathbb{R}^p$:

$\epsilon^*(S,X)< \left\lfloor \frac{n-p-1}{2n}\right\rfloor$

• breakdown point of an affine equivariant estimator $S$

For an AE estimator S and a dataset $X$ in general linear position in the expression of the breakdown point simplifies to

$\epsilon^*(S,X)=\min\{\epsilon:\frac{\lambda_1(S(X_{\epsilon}))}{\lambda_p(S(X_{\epsilon}))}=\infty\}$

where $\lambda_p(C)$ ($\lambda_1(C)$) is the smallest (largest) eigen-value of a covariance matrix $C$.

There is then two types of breakdown. Implosion breakdown are caused by cases where $\lambda_p(S(X_{\epsilon}))=0$ (cases where $h$ observations lie on an affine hyper-plane) and explosion breakdown by cases where $\lambda_1(S(X_{\epsilon}))=\infty$.

The exact fit property of the MCD protects it from implosion breakdowns. For explosion breakdowns, it is enough to show that the chosen MCD H-subset $H_*$

• $H_*=\underset{H:|H|=h}{\Arg.\min.}\det(\underset{i\in H}{\text{cov}}(x_i))$

cannot belong to the set of $H$-subsets $H^+$:

• $H^+=\cup \{H:|H|=h, \lambda_1(\underset{i\in H}{\text{cov}}(x_i))=\infty\}$