Fat tails equal higher probability of non-extreme values according to Nassim Taleb? I just came across the following passage written by Nassim Taleb Link:
The fattest tail distribution has just one very large extreme deviation, rather than many departures form the norm. [...] if we take a distribution like the Gaussian and start fattening it, then the number of departures away from one standard deviation drops. The probability of an event staying within one standard deviation of the mean is 68 per cent. As we the tails fatten, to mimic what happens in financial markets for example, the probability of an event staying within one standard deviation of the mean rises tobetween 75 and 95 per cent.
So as far as I understand he states that the fatter the tails, the higher the probability of an observation falling into the window of 1 SD away from the mean. I have read definitions elsewhere that basically state that fat tails imply more extreme observations (i.e. a higher probability of extreme observations). Intuitively I feel that this contradicts what Mr. Taleb is saying and hence, I am having issues to relate what Mr. Taleb is saying with the aforementioned definition.
Could someone shed light on the point Mr. Taleb is trying to make? Maybe, my confusion stems from a misconception of what a fat tail is in the first place. Thank you very much in advance.
 A: The statement
"As ... the tails fatten ... the probability of an event staying within one 
standard deviation of the mean rises to between 75 and 95 per cent" 

may be true within narrow families of probability distributions, but it is false in general.  Here is a counterexample, edited from here, https://math.stackexchange.com/a/2510884/472987 , where the probability within a standard deviation stays constant (0.5) as the tails fatten.
Let $X = \mu + \sigma Z$, where $Z$ has the discrete distribution
$Z = -0.5$,   with probability (wp) $.25$
$= +0.5$,   wp $.25$
$= -1.2$,   wp $.25 - \theta/2$
$= +1.2$,   wp $.25 - \theta/2$
$= -\sqrt{0.155/\theta + 1.44}$,  wp $\theta/2$
$= +\sqrt{0.155/\theta + 1.44}$,  wp $\theta/2$.
The family of distributions of $X$ is indexed by three parameters: $\mu$, $\sigma$, and $\theta$, with ranges $(-\infty, +\infty)$, $(0, +\infty)$ and $(0,.5)$.
In this family, $E(X) = \mu$, $Var(X) = \sigma^2$, and the kurtosis of $X$ is as follows:
kurtosis $= E(Z^4) = .5^4 * .5 + 1.2^4 * (.5 - \theta) + (0.155/\theta + 1.44)^2 * \theta$.
Within this family,
(i) kurtosis tends to $\infty$ as $\theta \rightarrow 0$.
(ii) the distribution within the "shoulders" (i.e., within the $\mu \pm \sigma$ range) is constant for all values of kurtosis; it is simply the two points $\mu \pm \sigma/2$, wp $0.25$ each. This provides a counterexample to one interpretation of kurtosis, which states that larger kurtosis implies movement of mass away from the shoulders, simultaneously into the range between the shoulders and into the tails.
(iii) the "peak" of the distribution is also constant for all value of kurtosis; again, it is simply the two points $\mu \pm \sigma/2$, wp $0.25$ each. This provides a counterexample to the often given but obviously incorrect interpretation that larger kurtosis implies a more "peaked" distribution.
(iv) In this family, the central portion of the distribution actually becomes flatter as kurtosis increases, since the probabilities on  $\mu \pm 1.2\sigma$ and $\mu \pm 0.5\sigma$ converge to the same value, $0.25$, as the kurtosis increases. This provides a counterexample to the often-stated interpretation that higher kurtosis corresponds to "peakedness" and lower kurtosis corresponds to "flatness." Within this given family of distributions, higher kurtosis actually corresponds to a flatter peak.
