| |
You are here: Experts > Science > Mathematics > Probability & Statistics > non gaussian distribution
Expert: ronny fisher - 10/22/2009
Question
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
---------- FOLLOW-UP ----------
QUESTION: Hi Ron:
Yes my question was terrible (wasn't really a question) as the hosptial filter system I am on is not very helpful on these chat boards.
Anyway what I didn't get to write is that for someone like me who's statisical calculus is two decades ago, I found it difficult to understand visually why a fat tailed distribution should make these rare events LESS common than a normal distribution (which is what Taleb said). He points out that the chance of being further out than 1sd is less likely in a fat tail which is counter intuative to what all the popular literature that I have read says.
Mark Elliott
Dept of Anesthesiology
Providence Healthcare
Vancouver, BC
Answer
mark -
the counter-intuitiveness you express about 'rare events' [which are called tail events in the trade] is not really so paradoxical. being at least 1 SD above the mean, which has a normal probability of about .16, is not, in many people's books, exactly a 'rare event'. that designation usually applies to events with a smaller probability, say .01 or less.
if one confines attention to such rarer tail events, the paradox disappears.
for example, consider a random quantity Y having a t-distribution with 20 df.
this distribution has fatter tails, compared with a normal distribution [as defined in the previous answer]. but
(*) P(Y > ySD(Y)) > P(N(0,1) > y) if y > 1.7631.
for y = 1.7631, the two expressions in (*) both equal .0389 - or just about .04.
so it is certainly true that any really rare tail event (Y > ySD) is more likely
to occur than the corresponding normal tail event (N(0,1) > y).
this is in accord with the more common notion of what it means to have fatter tails
than the normal distribution. the [standard] normal curve decreases very rapidly
in the tails - faster than exponential decay - because there is a quadratic term
-(x^2)/2 in the exponent.
by contrast, the t-density curves decrease only algebraically fast - like 1/x^k, [where k
is a positive power that depends on the df]. so eventually [i.e, for large enough y], the probability of a tail event (Y > ySD) is orders of magnitude GREATER than the corresponding normal tail event (N(0,1) > y).
by contrast, the less rare tail events have greater probability for the normal curve than
for one with fatter tails. is this so paradoxical? if normal curves have smaller probabilities [further out] in the tails, they must have greater mass nearer the center [the mean] - since all probability curves have a total mass of 1.
ronny
Add to this Answer Ask a Question
|
|
