Dear allstat,

 On 25/06/2008, Allan White <[log in to unmask]> wrote:
 > If the total risk is
 >  required for a number of risky assets then, assuming independence
 >  between assets, the correct procedure is of course to add the variances.


I'm not sure that that's the correct procedure, though, because the
 variance of the sum is computed as follows:
 var(a+b) = var(a) + var(b) + 2cov(a,b)

 Or, more generally: the variance of the sum of the variables = the sum
 of the covariance matrix entries. (Each covariance entry appears
 twice.)
 (See Note 1 for a derivation for three variables.)

 Question: why does the industry use the sum of the std. deviations?
 Preliminary answer: I couldn't tell you; perhaps it has a desirable property.

 More specific question: how does the industry formula for overall
 risk, i.e. overall variance, differ from the proper one?
 We have the following equations to estimate risk:
 [1] var(a + b) = var(a) + var(b) + 2*cov(a,b)
 [2] ( sd(a) + sd(b) )^2

 [1] is the correct formula, I believe. [2] is the industry formula
 (adapted to express risk). Let's see whether [2] tends to over- or
 underestimate risk.

 Now,
 (x+y)^2 - (x^2 + y^2) = xy,
 so if we subtract [1] from [2] we get
 [2] - [1] = sd(a) *sd(b) - 2*cov(a,b)

 Expressing both in terms of sd:
 [2] - [1] = sd(a)*sd(b) - cor(a,b)*sd(a)*sd(b)

 If all variables are non-negatively correlated, then
 [2] - [1] =< 0.
 In other words, if profits are correlated, the industry formula tends
 to underestimate risk. But if one profit tends to increase as the
 other one drops, the formula tends to overestimate risk.

 (Calculating the difference between [1] and [2] for N>2 variables is
 left as an exercise to the reader.)

 Answering the first question: I still don't know why the industry uses
 the sum of the standard deviations to estimate overall risk. I'd say
 tradition, but for the regulation: why would a regulation insist that
 one be lulled when risk is high, and alert when risk is low?

 It could still be due to tradition; and the regulation could be
 because the company wants a standard way of assessing risk, to avoid
 plausible-but-wrong assessments. Still, I expect that now that people
 tend to learn the basics of their trade at university, rather than
 learning everything at the company, this little mistake will die out.

 Yours,

 Sietse Brouwer

 (Note 1)
 Derivation for i = 3:

 Variance and covariance:
 var(A) = sum[ ( A - mean[A] )^2 ] / n
 cov(A,B)  = sum[ ( A - mean[A] )*( B - mean[B] ) ] / n

 Variance is mean-independent, so we can assume the mean to be zero,
 and work with deviation scores. Easier calculations, that. Write the
 deviation score of 'A' as 'a', for clarity's sake.
 var(a) = sum( a^2 ) / n
 cov(a,b) = sum( ab ) / n

 Then
 var(a + b + c) = sum[ (a+b+c)^2 ] / n
        = sum(a^2)/n + sum(b^2)/n + sum(c^2)/n +
                2*sum(ab)/n + 2*sum(ac)/n + 2*sum(bc)/n
        = var(a) + var(b) + 2*cov(a,b) + 2*cov(a,c) + 2*cov(b,c)
        = var(A) + var(B) + 2*cov(A,B) + 2*cov(A,C) + 2*cov(B,C)
 (This last because of mean-independence.)

 The two-variable case is easier to follow. There's a good explanation at
 http://visualstatistics.net/Visual%20Statistics%20Multimedia/covariance.htm ,
 subsection "Variance of a Sum and a Difference".


 --
 Sietse Brouwer -- [log in to unmask] -- +31 6 13456848
 Wildekamp 32 -- 6721 JD Bennekom -- the Netherlands
 MSN: [log in to unmask] -- ICQ: 341232104