from two to three…

A short post to give more details about the final remark in the course of Financial Econometrics, and more precisely the formula that can be found in the book of Philip Jorion,

Note that this formula can be found (perhaps written with slight changes) in several papers, e.g. the following sentence (on the,

or the following formula, on documents from the Bank of England website,

I recently pulished (in French, here) a paper on the Value-at-Risk, including the following graph,

Usually, three times the average over 60 trading days is the larger component, but during the financial crisis, it turned out that the daily component was almost three times higher than the average value over the past the months (this fact was mention by Paul Embrechts in some conference in Paris on risk measures).
The interpreation of the multiplicative k coefficient (which is from 2 to 3 in some publications, or which exceeds 3 in others) has been proposed in a paper of Gerhard Stahl, entitled three cheers. The idea is to use the Bienaymé-Tchebychev inequality. For random variables with finite variance, then

Recall that this inequality is simply a corrolary of Markov’s inequality

or for any increasing function

(taking function, applied to This upper bound can be far away from the true probability, see e.g. the gaussian case below, i.e. if,


> z = seq(0,3,by=.01)
> P = 2*dnorm(k)
> U = 1/z^2
> plot(z,P,type="l",lwd=2,col="red",xlab="",ylab="")

The ratio between the two is given below,

> plot(z,U/P,type="l",lwd=2,col="purple",xlab="",ylab="",ylim=c(0,10))

Note that it is possible to interprete the axis values as probabilities values, taking quantiles of the gaussian distribution

> plot(pnorm(z),U/P,type="l",lwd=2,col="purple",xlab="",
+ ylab="",ylim=c(0,10),xlim=c(.9,1))
> abline(h=3,lty=2)

The interpretation is that the upper bound is 3 times higher than the true probability in the Gaussian case when z is the quantile of the distribution associated with probability level 99%.
Note that

  • if z is the 95% quantile of the mathcal{N}(0,1) distribution, the ratio is 2 (1.92)
  • if z is the 99% quantile of the mathcal{N}(0,1) distribution, the ratio is 3 (3.04)
  • if z is the 99.55% quantile of the mathcal{N}(0,1) distribution, the ratio is almost 4 (3.88)
  • if z is the 99.75% quantile of the mathcal{N}(0,1) distribution, the ratio is 5 (5.04)

A more formal explaination is to assume that X is symmetric, and then

Thus, if, i.e., we have an upper bound for the,

where the upper bound is the upper bound for the Value-at-Risk for any distribution with finite variance and centred.
If, then, i.e.  But since, for a distribution, then

and further

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.