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Expert: ronny fisher - 10/21/2009
Question
Just read another article by Nicholas Nasser Taleb about fat tail distributions. I'll just copy and paste?
The point is mathematically simple, but does not register easily. I've enjoyed giving math students the following quiz (to be answered intuitively, on the spot). In a Gaussian world, the probability of exceeding one standard deviation is around 16 percent. What are the odds of exceeding it under a distribution of fatter tails (with same mean and variance)? The right answer: lower, not higher—the number of deviations drops, but the few that take place matter more. It was entertaining to see that most of the graduate students got it wrong. Those who are untrained in the calculus of probability have a far better intuition of these matters.
Answer
mark -
sorry for the rejection email that went out to you. it was sent inadvertently.
the allexperts system is a bit quirky.
i haven't seen the article by taleb you mention - and i didn't see that there was
a copy of it attached to your email.
i'm not sure what question you are asking, but here is a thought regarding what
you wrote: you do not define precisely what you mean by having "fatter tails" than
a normal distribution. here is what i think that means:
suppose Z is a standard normal random variable and suppose Y is another random
variable having a symmetric distribution.
Y can be said to have [a distribution with] fatter tails than Z if
(1) P(Y > y) >= P(Z > y) for all y > 0,
or equivalently [since Y is symmetric], if
(2) P(|Y| > y) >= P(|Z| > y) for all y > 0.
another way to phrase this is that |Y| is stochastically larger than |Z|.
this then implies that |Y| has the same distribution as |Y|' = |Z| + W, where W
is some non-negative random variable. then
var(Y) = EY^2 = E|Y|^2 = E(|Y|')^2 = E{Z^2 + 2W|Z| + W^2} >= EZ^2 = var(Z) = 1.
moreover, unless W is identically zero, so that |Y| and |Z| have the same distribution
[and then Y and Z also have the same (normal) distribution], var(Y) > var(Z) = 1.
in other words, you cannot have a symmetric Y with the same mean and SD as Z, which
also has fatter tails than Z [i.e. for which (1) holds for all y > 0, with strict
inequality for some y > 0].
you can think of y in (1) as the number of SDs above the mean. your claim may indeed hold for y = 1 [1 SD above the mean] - but it cannot hold for all y > 0.
so for example, the t-distribution with 3 df has mean = 0 and var = 3. if one normalizes it to have var = 1 and gives Y this distribution, (1) holds for y < 2.098224 and does not
hold if y > 2.098224.
is there something special about being 1 SD above the mean - as opposed to being 2 SDs
above - or 2.1 SDs above?
ronny
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