4 ms·

A finite set of numbers has a mean, yes. A probability distribution with heavy enough tails does not have a mean (e.g. the Cauchy distribution).

mumblemumble 7y ago
Sure, but let's not scope creep the conversation. This is in the context of talking about summary stats calculated against sets of discrete observations, not properties of abstract probability distributions. So we're talking about doing some basic arithmetic on a list of numbers, not taking integrals.
- SiempreViernes 7y ago
  No, the point is that for certain sets of numbers, calculating the mean does not give you a summary statistic. It just gives the average of a bunch of numbers, not a number that is representative of your population.
  - throwaway41968 7y ago
    This is why using Gaussian models to predict something that's fundamentally fat-tailed (e.g. IQ vs. many forms of societal achievemnt) is doomed to failure.
- CrazyStat 7y ago
  If your data comes from a probability distribution that doesn't have a mean then calculating the mean of your data is basically meaningless (no pun intended). In a Cauchy distribution, for example, the mean of a million datapoints is equally likely to be a thousand units off from the true center of the distribution as any individual datapoint is. This is not scope creep. This is not an abstract academic concern. I've actually seen people run into this in practice--"I quadrupled the size my dataset, why is my sample mean still wildly inaccurate?" If you're not aware of the properties of abstract probability distributions then your basic arithmetic on a list of numbers may well be completely useless.
  - jerf 7y ago
    "If your data comes from a probability distribution that doesn't have a mean then calculating the mean of your data is basically meaningless (no pun intended)." I think this is actually the major takeaway for this sort of discussion. All statistical measures carry with them assumptions about the underlying distribution. When you blindly use one without verifying the underlying distribution, you are asking to be lied to by your statistics. It took me a while, but I've trained myself to actively ask the question "is this appropriate for the underlying distribution" whenever I see a "mean" or a "standard deviation". Spoiler: The answer is usually "no"! Central limit theorem notwithstanding, we encounter a lot of non-Gaussian and outright pathological distributions in the real world. Average is useful for more than just Gaussian, but it means different things in different distributions, most of them quite unlike what our Gaussian-trained intuition suggests. Standard deviation is really Gaussian-only, or at least, if you insist on using it, you ought to pair it with the skew, kurtosis, or other measures of how non-Gaussian your distribution is. Remembering that there's only one way to be Gaussian but an infinity of ways to be non-gaussian is helpful too. This can be a helpful video to visualize that: https://www.youtube.com/watch?v=iwzzv1biHv8 https://www.youtube.com/watch?v=iwzzv1biHv8
- contravariant 7y ago
  To add to the arguments already presented, it's not like you never encounter practical scenarios where a mean does not exist. A good example would be earthquakes. The commonly used Richter law [1] suggests that the number of earthquakes decreases exponentially with the magnitude, while the intensity increases exponentially with the magnitude. In seismically active regions the frequency and intensity even appear to be inversely proportional. This results in a model where there is no average intensity. Sure this model is bound to break down eventually, but until the first earthquake that quite literally breaks the mould we likely won't know what the upper bound may be. [1]:https://en.wikipedia.org/wiki/Gutenberg%E2%80%93Richter_law https://en.wikipedia.org/wiki/Gutenberg%E2%80%93Richter_law

mumblemumble 7y ago

Sure, but let's not scope creep the conversation. This is in the context of talking about summary stats calculated against sets of discrete observations, not properties of abstract probability distributions. So we're talking about doing some basic arithmetic on a list of numbers, not taking integrals.

SiempreViernes 7y ago
No, the point is that for certain sets of numbers, calculating the mean does not give you a summary statistic. It just gives the average of a bunch of numbers, not a number that is representative of your population.
- throwaway41968 7y ago
  This is why using Gaussian models to predict something that's fundamentally fat-tailed (e.g. IQ vs. many forms of societal achievemnt) is doomed to failure.
CrazyStat 7y ago
If your data comes from a probability distribution that doesn't have a mean then calculating the mean of your data is basically meaningless (no pun intended). In a Cauchy distribution, for example, the mean of a million datapoints is equally likely to be a thousand units off from the true center of the distribution as any individual datapoint is. This is not scope creep. This is not an abstract academic concern. I've actually seen people run into this in practice--"I quadrupled the size my dataset, why is my sample mean still wildly inaccurate?" If you're not aware of the properties of abstract probability distributions then your basic arithmetic on a list of numbers may well be completely useless.
- jerf 7y ago
  "If your data comes from a probability distribution that doesn't have a mean then calculating the mean of your data is basically meaningless (no pun intended)." I think this is actually the major takeaway for this sort of discussion. All statistical measures carry with them assumptions about the underlying distribution. When you blindly use one without verifying the underlying distribution, you are asking to be lied to by your statistics. It took me a while, but I've trained myself to actively ask the question "is this appropriate for the underlying distribution" whenever I see a "mean" or a "standard deviation". Spoiler: The answer is usually "no"! Central limit theorem notwithstanding, we encounter a lot of non-Gaussian and outright pathological distributions in the real world. Average is useful for more than just Gaussian, but it means different things in different distributions, most of them quite unlike what our Gaussian-trained intuition suggests. Standard deviation is really Gaussian-only, or at least, if you insist on using it, you ought to pair it with the skew, kurtosis, or other measures of how non-Gaussian your distribution is. Remembering that there's only one way to be Gaussian but an infinity of ways to be non-gaussian is helpful too. This can be a helpful video to visualize that: https://www.youtube.com/watch?v=iwzzv1biHv8 https://www.youtube.com/watch?v=iwzzv1biHv8
contravariant 7y ago
To add to the arguments already presented, it's not like you never encounter practical scenarios where a mean does not exist. A good example would be earthquakes. The commonly used Richter law [1] suggests that the number of earthquakes decreases exponentially with the magnitude, while the intensity increases exponentially with the magnitude. In seismically active regions the frequency and intensity even appear to be inversely proportional. This results in a model where there is no average intensity. Sure this model is bound to break down eventually, but until the first earthquake that quite literally breaks the mould we likely won't know what the upper bound may be. [1]:https://en.wikipedia.org/wiki/Gutenberg%E2%80%93Richter_law https://en.wikipedia.org/wiki/Gutenberg%E2%80%93Richter_law