Breakdown point of weighted median

I posted the same question on math.stackexchange, but haven't got any reply. So I will try my luck here:

In the context of robust statistics, there is the concept of breakdown point, which basically measures the proportion of outliers an estimator can handle before giving an arbitrary result: http://en.wikipedia.org/wiki/Robust_statistics#Breakdown_point

For example, arithmetic mean has breakdown point $0$, while median has it as $0.5$, the highest possible.

My question is that how should we estimate the breakdown point for weighted median, which is the data point with the smallest $m$ that satisfies the following, when the set of $n$ data points are sorted in ascending order:

$$\sum_{i=1}^{m} \Omega_{i} \ge \sum_{j=m+1}^{n}\Omega_{j}$$

where $\Omega_{i}$ is the weight for point $i$.

More specifically, what relationship can we derive between the weights and breakdown point?

• so the $\Omega_i$'s are the weights? Something seems to be missing...i can make its bdp go to 0 by giving weight 0 to all except one outlier – user603 Feb 23 '12 at 17:52
• I think I see the answer lurking right there in your formula :-). What happens to the weighted median when you greatly modify the first $m$ values? The last $n-m$ values? – whuber Feb 23 '12 at 17:53
• @user603, the weights are fixed beforehand, just the way they would be with most weighted data (e.g., frequency-weighted or probability-weighted datasets): you don't get to modify them. – whuber Feb 23 '12 at 18:12
• @whuber: but the bdp is computed under worst possible weighting scheme... – user603 Feb 23 '12 at 19:00
• @whuber:> in other words: just as the regular median will go the way of the majority of the observations, the weighted median will go the way of the observations with the majority of the weights....what makes it impossible for a few outliers to have a lot of weight? That's what i meant when i wrote 'something is missing' – user603 Feb 23 '12 at 19:06