1. Advanced Term Structure Practice Using HJM Models Practical Issues Copyright David Heath, 2004

2. HJM motion of term structure: u Equivalent martingale measure: (Harrison-Kreps / Harrison-Pliska) u If no arbitrage then there is an equivalent probability measure (same null sets as given probability) for which discounted prices of basic securities are martingales. u If it’s unique then prices for all claims must be expected discounted values of payoffs.

3. Picking the “right” model u Theory says: u Martingale measure Q must be “equivalent” to the physical measure P u That means sets of probability 1 must be the same u Some sets of probability 1: u The value of the squared variation is in principle computable for sure (in the limit) u The value of covariances is too u Initial forward curve is “known” with prob 1

4. u This means: u The initial forward curve must be correct u The variances and covariances in the model must match the “true” values. u Where to get the “true” values? u Estimate from the past u Is this enough? u Under some circumstances it is u In general there are more things to match u Example: B t 2 – t has same “squared variation process” as its negative, but these processes are very different

5. Summary Need to choose Initial forward curve Volatility functions These are the same for the physical measure as for the pricing measure

6. Too much to choose!  Actually, choosing f(0,  ) is often easy: u From Treasury strips, or u From Treasury bonds, or u From swaps curve u It’s tougher in some “smaller” markets u Not enough bonds u Markets not liquid u Bonds may have special features

7. Choosing  u For each T we must choose an adapted stochastic process for u These must vary smoothly with respect to T  Narrowing down a useful   Limit dependence on  :

8. Still too much freedom?  Limit  ’s to be of the form u (Still must choose d = number of factors.)  Then: “estimate” the using historical data. “Calibrate” s to match option prices at different maturities, or “estimate”  using some model of information inflow (announcements, …).

9. The resulting model form:   is determined by “model choice” u s is to permit adjustment for “implied volatility” u d (the number of factors) and the are estimated from historical data u The drift under the martingale measure is determined by the above choices

10. An unfortunate fact u I’d like to choose u Unfortunately, the resulting equation has the property that solutions explode with positive probability in any time interval u (Think of simpler equation y’=ky 2.) u We therefore use

11. How we estimate the model u We assume we have values of historical forward curves u We choose M to be larger than any forward rate in our data set. u We discretize our evolution equation to get

12. Written in terms of our data u Now define the column vector X (j) so that its k th component,, is given by the above expression. These X’s are, of course, random u Now compute the natural estimator for the variance-covariance matrix of the X’s: u It’s the matrix V whose (k 1,k 2 ) entry is:

13. u The expected value is given by: u We can, without loss of generality, pretend that the first term is equal to 1 (since that can change our results only by a multiplicative constant, and we can “adjust” the s() we’ll use to compensate for that).

14. Method of moments estimation Define column vectors (These are what we want to estimate.) u Then we have u Using V as an estimate for E(V) (method of moments) we need to find our estimates such that But how?

15. Facts about matrices  Definition: A square matrix B is positive definite provided x’Bx>0 whenever x is a column vector not equal to 0. It’s positive semidefinite if x’Bx  0 for all x. u Fact: Our V will always be positive semidefinite, and will almost always be positive definite.

16. Principal components If V is a symmetric real positive definite matrix then there are matrices U and  such that:   is a diagonal matrix with diagonal entries positive and non-increasing UU’=I (the identity matrix) V=U  U’ Set T=U  0.5. IF Z is a column vector of independent standard normal random variables, then the variance-covariance matrix of TZ is V

17. More facts u The columns of U are called the “principal components” of V. u The “best” lower-dimensional approximation to the distribution of X results from replacing the “right-most” columns of U by 0. u “Best” means, in a particular sense, “minimal mean-squared error.”

18. No error analysis so far! u Recall: In theory one can calculate the variance-covariance structure exactly from any short evolution of the forward curve (using finer and finer subdivisions). u With a fixed sample, we can’t do that. u Let’s look at a simpler problem: estimating the variance of a random variable known to be normally distributed, mean 0, unknown variance

19.  Suppose y 1, …, y n are independent observations from an N(0,  2 ) distribution  The natural estimator for  2 is u What is the standard error? It’s  Thus with n=50 observations, we typically get about 20% error in  2, which means about 10% in  and in at-the-money option prices. Conclusion: we need lots of observations!

20. Trying this with real data u We make a further discretization. We fit the forward curve in a “piecewise flat” way:  We think of discrete (relative) maturities of 0, 0.25, 0.5, 1.0, 2.0, 3.0, 5.0, 7.0 and 9.8 years. We use a forward curve which is constant between these maturities and select one  k from each resulting interval. u For the “forward curve at the end” we keep the “endpoints” at the same “absolute dates”. u We do this for lots of values of t j. u We set “dt” to be one week.

21. Our data u We were given data about “forward Libor” rates. The next slide has an example of this data for observation date March 8, 1991. u In each row, the entries are: u The date of a start of a “quarter” u The number of days in this quarter u The “actual daycount over 360” rate rate for this quarter

22. 19910308 92 0.06790763318 19910608 92 0.06895597405 19910908 91 0.0702446393 19911208 91 0.07449457699 19920308 92 0.07641909209 19920608 92 0.07905264791 19920908 91 0.08121356523 19921208 90 0.08345680601 19930308 92 0.08388518428 19930608 92 0.0849136007 19930908 91 0.08580050549 19931208 90 0.08749666101 19940308 92 0.08771156563 19940608 92 0.08840474026 19940908 91 0.09039488269 19941208 90 0.09092852189 19950308 92 0.08962092803 19950608 92 0.0893030357 19950908 91 0.09033209647 19951208 91 0.09079276802 19960308 92 0.09046775515 19960608 92 0.09101440106 19960908 91 0.09250467811 19961208 90 0.09269835718 19970308 92 0.09146662972 19970608 92 0.09138410867 19970908 91 0.09270728505 19971208 90 0.09267263896 19980308 92 0.09108505137 19980608 92 0.09071476543 19980908 91 0.09185052797 19981208 90 0.09174675845 19990308 92 0.09018690963 19990608 92 0.08966723663 19990908 91 0.09049240154 19991208 91 0.09047274176 20000308 92 0.08941867094 20000608 92 0.0893215848 20000908 91 0.09044857633 20001208 90 0.09132722271 20010308 Forward curve seen on March 8, 1991

23. Computing forward rates u For the week from t k to t k+1: u At t k compute the piecewise-flat continuously- compounded interest rate “explaining” the prices of pure discount bonds maturing at t k +0.25, t k +0.5, t k +1, t k +2, t k +3, t k +5, t k +7, and t k +9.8 years. u Similarly, at t k+1, compute rates for bonds maturing at t k +0.25, t k +0.5, t k +1, t k +2, t k +3, t k +5, t k +7, and t k +9.8 years. (The same absolute dates!)

24. Results for t k =June 5, 1991 At the start of the week.06147114853932487.06418599716067086.07028591418277555.07894618023869922.08562248112095414.09197819566016921.09200485752193654.09005201517594776 At the end of the week.06282817092590599.06547295594397448.07270227193025801.08144837689395647.08688266008163854.09130940580147720.09348600859670100.09060552897329775

25. Now we compute  f/f, getting 0.02207576105 0.02005046023 0.03437897587 0.03169496799 0.01471785148 -0.007271178282 0.01609861821 0.006146600898

26. . Finally we divide by. 0.1591905768 [0, 0.25] 0.1445859249 [0.25, 0.5] 0.2479103206 [0.5, 1.0] 0.2285556645 [1.0, 2.0] 0.1061319364 [2.0, 3.0] -0.05243321226 [3.0, 5.0] 0.1160887868 [5.0, 7.0] 0.04432376942 [7.0, 9.8] These do seem a bit “noisy”.

27. That was just one “X-vector”. u We need to do this for each week. (I had about 5 years’ worth of data.) u With 5*52, or about 250 data points we could estimate variances with standard errors of about 9% (and sd’s with 4.5%) u We then need to estimate the variance- covariance matrix from these observations

28. Estimated variance- covariance matrix 0.02899 0.02016 0.02336 0.01810 0.01051 0.00756 0.00386 0.00289 0.02016 0.03267 0.03625 0.02769 0.01613 0.01353 0.00739 0.00551 0.02336 0.03625 0.05308 0.04384 0.02720 0.02354 0.01520 0.01114 0.01810 0.02769 0.04384 0.03972 0.02584 0.02163 0.01441 0.01086 0.01051 0.01613 0.02720 0.02584 0.02023 0.01737 0.01239 0.00972 0.00756 0.01353 0.02354 0.02163 0.01737 0.02447 0.00895 0.00766 0.00386 0.00739 0.01520 0.01441 0.01239 0.00895 0.03180 0.01066 0.00289 0.00551 0.01114 0.01086 0.00972 0.00766 0.01066 0.02508

29. Principal components u My software gave them in rows, biggest last -0.000203 -0.004158 0.014010 -0.019113 0.012983 -0.002911 -0.001068 -0.000720 0.002154 -0.018453 0.026459 -0.001490 -0.034284 0.008992 0.002189 0.002544 0.010108 -0.051067 0.008814 0.039079 0.022068 -0.038964 -0.009173 -0.004753 0.063842 -0.040597 -0.031512 -0.012123 0.017485 0.061027 0.002508 -0.010106 -0.029053 -0.004099 0.008321 0.009764 0.005078 0.011601 0.060757 -0.110809 -0.078997 -0.022686 0.010380 0.025695 0.027516 0.071875 -0.076100 -0.008346 -0.077378 -0.059910 -0.025879 0.000313 0.029882 0.029506 0.123272 0.092637 0.108330 0.154711 0.224201 0.192153 0.128131 0.113873 0.083815 0.063391

30. 1 st Principal Component

31. 2 nd Principal Component

32. How important are multifactor models? u To check this we analyze two sets of problems with single- and multi-factor models u Problem 1: Caplets u Problem 2: Swaps with optionality

33. The caplets u For each year y from 1 to 9, consider an interest rate cap with only one period: It is effective precisely from (now plus y years) until (now plus y plus 0.25 years). The cap level is 6% (using actual/360), and the initial forward curve is flat at 6% (continuously compounded).

34. The swaps u Consider a swap with two floating rate payers. Every six months from now until ten years are up (except the first period) u Payer 1 pays interest at the 6 month Tbill rate on a notional of $1,000,000. u Payer 2 pays similarly, but at the smaller of: u the Tbill rate u the yield of a 10 year CMT + b basis points. u We consider contracts with b varying from -80 to +20. u The initial forward curve is upward sloping, starting at 6% and rising at 0.08% per year.

35. The models  Ho and Lee:  = constant  Vasicek:  (u)=c exp(- u) u HJM “Old 1-factor” u HJM “Old 2-factor” u HJM “Adjusted 1 factor”

36. HJM Old 1 factor (BARRA)

37. HJM Old second factor

38. For Ho-Lee use  =.01.

39. “HJM Adjusted” u Effort is to get volatility at of each forward rate correct. (Correlations will be wrong, of course.) u HJM Adjusted = ((HJM 1)^2 + (HJM 2)^2) ^ 0.5