# Variance of Multi-Dimensional OU process

I'm trying to implement this model shown here:

http://www.sciencedirect.com/science/article/pii/S0304407611000388

As part of the modelling process I have to calculate the unconditional variance of X see page 10).

$\sigma_R^2=\int_u^t \exp^{-A s}\Sigma \Sigma^T \exp^{-A^T s}ds$

They say they use a closed form result from here

http://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CDEQFjAA&url=http%3A%2F%2Fwww.markfisher.net%2F~mefisher%2Fpapers%2Fterm_prem.pdf&ei=-6yAU6K_M4ziO4m9gPgK&usg=AFQjCNEeJcmAiEzZbfcrWfTGP2uCP5GMFg&bvm=bv.67720277,d.ZWU

see the eqn directly below eqn (3.10) on p5 sadly I can not understand this result much less transform it in to the DNS/AFDNS framework

However the solution is presented here in the DNS/AFNS fraemwork

http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1974033

as

$\int_u^t \exp^{-A s}\Sigma \Sigma^T \exp^{-A^T s}ds=\Lambda \Gamma \Lambda^{-1}$

where the (i,j)th element of $\Gamma$ is $=\frac{\sigma_{ij}}{\lambda_i+ \lambda_j}(1-\exp^{-(\lambda_i + \lambda_j)\delta T})$ where $\sigma_{i,j}$ is the element (i,j) of the covariance matrix ($\Sigma \Sigma^T$) assumed constant, and $\Lambda$ is the eigenvector of $\kappa(s_t):=A$.

Unfortunately the solution I found does not say what the $\lambda$ are. I assume they must be the eigenvalues of A?

Q1.) Can someone please confirm my hunch about the $\lambda$ being the eigenvalues

Q2.) Can someone either show me or point me to a reference where I can see how this derivation is done.

• I cant get to the Nelson Siegel paper. Can you summarize or post it? – user12348 May 24 '14 at 16:10
• Here is an earlier version of the paper: google.co.uk/… – Bazman May 24 '14 at 18:29
• In this version they simply calculate the variance numerically using the equation in the footnote on p13. Hoever generally their notation is better because they use \Gamma for the market price of diffusing risk rather than \lambda. In the final paper they say that the use the closed form solution in Fisher and Gilles – Bazman May 24 '14 at 18:31
• to calculate the variance. – Bazman May 24 '14 at 18:35

## 2 Answers

This interesting question provides excellent links to Dynamic Nelson-Siegel Term Structure Models for interest rates for No Arbitrage and exposes key formulation in an interesting way.

Appendix in p37 of ssrn link says $\lambda$ is market price of diffusion risk. However, in the DNS model the $\lambda$ is eigenvalues of $\kappa$, which then part of covariance matrix elements as you described. See section 5.2 of the ssrn reference for an example. Actually you can the data from Bloomberg for the period and be able to validate their results.

• Hey there thanks for the response. Unfortunately I'm not sure this is correct. The paper is a bit sloppy in it's notation lambda_t corresponds to the market price of diffusion risk, but \lambda corresponds to the curvature factor in the DNS model see p12 eqn 29. Even more confusingly the i,j on p15 is used to differentiate between the different states on HMM but that is not what I believe they are being used for here either. The calculation is just for the unconditional variance. As the mp of diffusion risk affects the drift I don't think that lambda_t is the lambda they are referring to here – Bazman May 24 '14 at 18:11
• Here the i,j is being used to describe the i,jth element of the covariance matrix. Sorry if the error is mine but I think what is being calculated here is simply the variance over one time step in one state. – Bazman May 24 '14 at 18:20
• commenting here too, in case you do not get the edit. -- From page 18, the LHS is covariance matrix. $\Gamma$ is eigenvalues $\Lambda$ is eigenvector matrix, such that $\lambda_i$ and $\lambda_j$ are parts of $\Lambda$ that define the transition matrix which is the speed $\kappa$ which shares the $\lambda_{i \space or\space j}$ for different market regimes, and still $\lambda$ is mpor while i,j are the regime states. P16 says "we can solve (39) explicitly;" this explicit solution is resulting in eigenvectors as part of eigenvalues causing the confusion. – user12348 May 24 '14 at 23:04
• OK I think I understand what you are getting at but still not 100% sure. Can you write your solution out please. – Bazman May 25 '14 at 11:36
• Thanks for sticking with this I'm starting to come round to your way of thinking but not 100% sure.I think you are saying that each $Lambda$ is the eigenvector relating to $\kappa(s_t)$ as stated in the paper. So assuming we can take out the variance covariance matrix we have $\Lambda e^{-lambda_i}\Lambda^{-1}*\Lambda e^{-lambda_j}\Lambda^{-1}$ This gives $\Lambda e^{-\lambda_i- \lambda_j} \Lambda_{-1}$ which is of the right form but here lambda is simply the eigenvalues of $\kappa(s_t)$ not the market prices of risk. – Bazman May 25 '14 at 11:59
1. I verified numerically, $$\lambda$$ is the eigenvalue of $$A$$.

2. Also, if you looked at this paper: http://liu.diva-portal.org/smash/get/diva2:1140151/FULLTEXT02.pdf, equation 11 is a special case of the above general formula in your question. It is clearly showing that $$\lambda$$ is the eigenvalue.