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By definition outliers are just points which differ significantly from the others. Obviously if your distribution has heavy tails they are likely to be there.

By definition outliers are just points which differ significantly from the others. Obviously if your distribution has heavy tails they are likely to be there. They could also be noise or simply errors. The only things you know before analysing is that they are different which is why they are outliers.

whatshisface 4y ago
If you carefully read the article, what they're doing is classifying points by how unlikely they'd be according to their model of their system, and they are using that to find inaccuracies in their model. Without a model there is no difference between outliers and "regular" data points, because a data point will always match a distribution made from itself.
- WastingMyTime89 4y ago
  You don’t need a model for a point to be significantly different. It just needs to have characteristics making it stand out. Outliers are an important concept for the theory of estimators for exemple and that doesn’t involve modelling at all. This discussion is why I regret that people doing CS have so little exposure to frequentist statistics.
  - whatshisface 4y ago
    If you measure 5, 5, 5, 9999999999, then your measured distribution is a peak at 5 and another at 9999999999. You can't say that 9999999999 is a surprising measurement unless you have a theoretical distribution - usually an implicit one - to evaluate at it and find a low probability. In other words, if being surprised means having your expectations violated, an expression of surprise implies that you must have had expectations to violate.
    - strangedejavu 4y ago
      This argument doesn't make much sense to me. You can definitely associate a probability based just on the measured data. What if you measured 5, 5, 5, 5, 5, 5,5 (1000 more 5s) , 5, 999999999? I think it's pretty intuitive to say that 999999999 is unexpected relative to the 5s. And you certainly can have expectations that are generated purely based off of the measured distribution and not off of some theoretical model.
      - whatshisface 4y ago
        It is unexpected relative to the 5s, but it is also rare in your sample, meaning that its occurrence in the sample is in exact proportion to how often you'd expect it to occur, given the distribution you inferred from your sample. That means you have no grounds for singling it out as "abnormal."
        nl 4y ago
        And here we see the "Bayesian vs Frequentist" argument. If this gets resolved in a HN thread you should publish it.
    - WastingMyTime89 4y ago
      Obviously you can say that 9999999999 is very different from the rest of the sample consisting of 5 without an implicit distribution. It’s incredibly far from the median compared to all the other value.
      - whatshisface 4y ago
        But is it surprising to find it far from the median?
        WastingMyTime89 4y ago
        It doesn’t matter if it’s surprising. It might be a measurement mistake, a mixed population, a fat tail or a poor sample. It’s different therefore an outlier.
        nl 4y ago
        There are many statistical tests that will tell you that based on the observations alone (with no model). Basically this depends on how many 5s there are.
        whatshisface 4y ago
        Every single one of those tests makes an implicit assumption about the distribution, explicit in its derivation. Usually, it is that the distribution is Gaussian.

whatshisface 4y ago

If you carefully read the article, what they're doing is classifying points by how unlikely they'd be according to their model of their system, and they are using that to find inaccuracies in their model. Without a model there is no difference between outliers and "regular" data points, because a data point will always match a distribution made from itself.

WastingMyTime89 4y ago
You don’t need a model for a point to be significantly different. It just needs to have characteristics making it stand out. Outliers are an important concept for the theory of estimators for exemple and that doesn’t involve modelling at all. This discussion is why I regret that people doing CS have so little exposure to frequentist statistics.
- whatshisface 4y ago
  If you measure 5, 5, 5, 9999999999, then your measured distribution is a peak at 5 and another at 9999999999. You can't say that 9999999999 is a surprising measurement unless you have a theoretical distribution - usually an implicit one - to evaluate at it and find a low probability. In other words, if being surprised means having your expectations violated, an expression of surprise implies that you must have had expectations to violate.
  - strangedejavu 4y ago
    This argument doesn't make much sense to me. You can definitely associate a probability based just on the measured data. What if you measured 5, 5, 5, 5, 5, 5,5 (1000 more 5s) , 5, 999999999? I think it's pretty intuitive to say that 999999999 is unexpected relative to the 5s. And you certainly can have expectations that are generated purely based off of the measured distribution and not off of some theoretical model.
    - whatshisface 4y ago
      It is unexpected relative to the 5s, but it is also rare in your sample, meaning that its occurrence in the sample is in exact proportion to how often you'd expect it to occur, given the distribution you inferred from your sample. That means you have no grounds for singling it out as "abnormal."
      - nl 4y ago
        And here we see the "Bayesian vs Frequentist" argument. If this gets resolved in a HN thread you should publish it.
  - WastingMyTime89 4y ago
    Obviously you can say that 9999999999 is very different from the rest of the sample consisting of 5 without an implicit distribution. It’s incredibly far from the median compared to all the other value.
    - whatshisface 4y ago
      But is it surprising to find it far from the median?
      - WastingMyTime89 4y ago
        It doesn’t matter if it’s surprising. It might be a measurement mistake, a mixed population, a fat tail or a poor sample. It’s different therefore an outlier.
      - nl 4y ago
        There are many statistical tests that will tell you that based on the observations alone (with no model). Basically this depends on how many 5s there are.
        whatshisface 4y ago
        Every single one of those tests makes an implicit assumption about the distribution, explicit in its derivation. Usually, it is that the distribution is Gaussian.