10 ms·

2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,-1000 MAD: 54.5 STDV: 229.4 edit: OK I get that you wanted an example of what "too much" weight is. If you're loo

• eanzenberg 10y ago
  In many cases outliers are extremely important. One that comes to mind is high spenders in mobile games. Trust me, if analysis was as simple as getting rid of outliers, treating everything as Gaussian, and retrieving simple summary statistics, then good data scientists wouldn't be paid $150k+ :)
  • cissou 10y ago
    same thing with venture capital ;) sometimes average are uninteresting, you just want one good outlier
• trhway 10y ago
  >Except in some specific fields (maths, physics), people are much more interested in the MAD than the STDV, but all they get to make decisions is the STDV. com'n guys, it all comes down to whether you like more romb or circle :) Interesting that MOND (modified Newtonian), if true, would suggest that a circle at very big distances looks like square (notice not like romb :), so physics may start to like it more.
• spikels 10y ago
  The MAD of that data is not 54.5. Here's how you calc MAD: (1) Find the median of your data which is -2 (2) Generate the absolute deviations of your data from this median which is {4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,4,0,998} (3) Find the median of the absolute deviations which is 4. It's ironic that Taleb prefers a statistic that ignores extreme examples (i.e. black swans) but he nevers seems to make sense to me. I've found MAD useful in dealing with noisy data.
  • vacri 10y ago
    It depends on whether the numbers provided are the actual data points themselves, or the deviation from median (the second is what the article provided).
  • cissou 10y ago
    I went with Taleb's proposed definitions: "Do you take every observation: square it, average the total, then take the square root? Or do you remove the sign and calculate the average?" edit: apparently this is consistent with https://en.wikipedia.org/wiki/Average_absolute_deviation https://en.wikipedia.org/wiki/Average_absolute_deviation I'm not sure what you referred to
    • spikels 10y ago
      In my experience MAD refers to either Median Absolute Deviation or the Mean Absolute Deviation. I was using the median version which is a pretty common "robust" statistic. Although I have occasionally seen the mean version it seems to be less common in practice. https://en.wikipedia.org/wiki/Median_absolute_deviation https://en.wikipedia.org/wiki/Median_absolute_deviation Take a look at the Wikipedia you linked. No version of Average Absolute Deviation is consistent with Taleb's definition. No squaring, no square root. Sounds more like a geometric mean. This is exactly what is so frustrating about Taleb. His ideas only partly makes sense. He often seems to see the problem but his solutions are poorly thought out. Of course, he thinks his solutions are perfect and everyone else is an idiot.
      • jamez1 10y ago
        In what field do you work that the median absolute deviation is used at all, let alone more than the mean absolute deviation? When he talked about mean absolute deviation being sqrt(pi/2) sigma did that not make it abundantly clear what he was discussing? >No squaring, no square root. Sounds more like a geometric mean Do you even know what the geometric mean is? (It has a root function so your statement just sounds stupid) Dispersion functions are built off the distance function under the metric you want to use. Standard deviation uses the L2 metric, which implies a euclidean distance function. (L2 corresponds to summing pow(u-x,2) and pow(sum,-2) as your functions) Mean absolute deviation takes the L1 metric, which implies pow(1) and pow(-1). This becomes summing pow(abs(u-x),1) and then pow(sum,-1), which, needless to say is the same thing as averaging the absolute differences. Hence the lack of any squaring or square rooting