Pairwise correlation for credit portfolio

[Liming]

Dear David,

There’s something from your note (on page http://www.bionicturtle.com/learn/article/unexpected_loss_ul_for_a_portfolio_of_credit_assets_9_min_briefing/) that I would like to clarify with you, when you mentioned that “My example here (from Ong Table 6.2) only requires a single default correlation; but a portfolio of N credit assets contains [N(N+1)]/2 pairwise correlations”.
1) this statement seems to contradicts with a statement made in slide 6 in the video from page http://www.bionicturtle.com/products/screencast/2009_6.c._credit_risk_portfolio_sample/ , where you said that 'for N asset portfolio, the pairwise correlation is N(N-1) /2.
2) does it mean that for 3 credit assets, there will be 3*4 /2 = 6 pairwise correlations? For example, if we have three assets each with their own UL: A, B, C. It follows that the portfolio UL would be A^2+B^2+C^2+AB+BA+AC+CA+BC+CB, which can be reduced to A^2+B^2+C^2+2AB+2AC+2BC. Is my understanding correct?
However, when you proceed on that page to say that “Actually, since the diagonals in a correlation matrix are 1.0s, the number of pairwise correlations = COMBIN(N,2) where COMBIN(N,2) = [N(N+1)]/2 - N”, does it mean that the actual pairwise correlation will be, in this case, 6-3=3? I don’t really understand this part. Can you kindly differentiate them and clarify? What’s more, I thought it is the first formula ([N(N+1)]/2) that will be relevant to FRM exam, not the second one?

Thanks!

Cheers!
Liming
27/09/09

 

[David Harper CFA FRM]

Hi Liming,

That is a very interesting observation! (I will make a question out of it this week, because i think it's very instructive)...I am pretty sure the difference arises because sometimes the source author refers to a covariance matrix and sometimes a correlation matrix.

BTW, the numbers are all triangle numbers:
http://en.wikipedia.org/wiki/Triangular_number
i.e., the number of cells in a triangle that is "extracted" from the square matrix

if we think about an C * C covariance matrix (ie., covariance in each cell), then the number of unique entries required is:
C(C+1)/2
e.g., 4 assets implies 4*5/2 = 10 cell entries
however, 4 of those are on the diagonal:
covariance(1,1), covariance(2,2), covariance(3,3), covariance(4,4); i.e., variance(1), variance(2), variance(3), variance(4)
the other six are "pairwise" covariances; e.g., covariance(1,3)

what if instead a correlation matrix R*R. In this case, the diagonal contains 1.0s as correlation(something,itself) = 1.0
and the number of unique entries = R(R-1)/2
e.g., 4 assets implies 3*4/2 = 6

...but the 6 is just the next smallest triangle number (down from 10), they are necessarily related
C(C-1)/2 + C = C(C-1)/2 + 2C/2 = (C(C-1)+2C)/2 = (C^2 - C + 2C) / 2 = C(C+1)/2

so hopefully this shows, that the difference is whether we are counting the diagonal or not
and, in regard to unique cell entries, the key point (IMO) is:
covariance matrix has variances in the diagonal, and
correlation matrix has 1.0s in the diagonal

...which formula is relevant to the FRM exam? The FRM does not have a standard on this (e.g., i think some authors have used either formula)...i don't recall the formula per se being tested, rather it is employed to give color to the "problem" with a parametric approach that is called the "curse of dimensionality"

hope this helps, David

append: I realize you asked about the default *correlation* matrix and i broadened the topic, but hopefully you can see that the difference is "merely" whether your definition includes the diagonal of 1.0s.