April 10, 1992 (when 10-year bond yields moved by7.7 standard deviations).
Stress testing can be considered as a way of taking into account extreme events that do occurfrom time to time but that are virtually impossible according to the probability distributionsassumed for market variables. A five-standard-deviation daily move in a market variable is onesuch extreme event. Under the assumption of a normal distribution, it happens about once every7,000 years, but, in practice, it is not uncommon to see a five-standard-deviation daily move onceor twice every 10 years.
Whatever the method used for calculating VaR, an important reality check is back testing. Itinvolves testing how well the VaR estimates would have performed in the past. Suppose that we arecalculating a 1-day 99% VaR. Back testing would involve looking at how often the loss in a dayexceeded the 1-day 99% VaR calculated for that day. If this happened on about 1% of the days, wecan feel reasonably comfortable with the methodology for calculating VaR. If it happened on, say,7% of days, the methodology is suspect.
16.9 PRINCIPAL COMPONENTS ANALYSIS
One approach to handling the risk arising from groups of highly correlated market variables isprincipal components analysis. This takes historical data on movements in the market variablesand attempts to define a set of components or factors that explain the movements.
The approach is best illustrated with an example. The market variables we will consider are tenU.S. Treasury rates with maturities between three months and 30 years. Tables 16.3 and 16.4 showValue at Risk 361
 
Table 16.3 Factor loadings for U.S. Treasury data
PCL PC2 PC3 PC4 PCS  PC6 PC7 PC8 PCI PCIO
3m 0.21 -0.57 0.50 047 -0.39 -0.02 0.01 0.00 0.01 0.006m 0.26 -0.49 0.23 -0.37 0.70 0.01 -0.04 -0.02 -0.01 0.0012m 0.32 -0.32 -0.37 -0.58 -0.52 ~—0.23 -0.04 -0.05 0.00 0.012y 0.35 0.10 -0.38 0.17 0.04 0.59 0.56 0.12 0.12 -0.053y 0.36 0.02 -0.30 0.27 0.07 0.24 -0.79 0.00 ~0.09 -0.004y 0.36 0.14 -0.12 0.25 0.16 -0.63 0.15 0.55 —0.14 -0.08Sy 0.36- 0.17 -0.04 0.14 0.08 -0.10 009 -0.26 0.71 0.48Ty 034 0.27 O15 0.01 0.00 -0.12 0.13 -0.54 0.00 -0.68l0y 0.31 0.30 0.28 -0.10 -0.06 0.01 0.03 -0.23 -0.63 0.5230y 0.25 0.33 046 -0.34 -0.18 0.33 -0.09 0.52 0.26 —0.13
 
 
results produced by Frye for these market variables using 1,543 daily observations between 1989and 1995,"4 The first column in Table 16.3 shows the maturities of the rates that were considered.The remaining ten columns in the table show the ten factors (or principal components) describingthe rate moves. The first factor, shown in the column labeled PCI, corresponds to a roughlyparaliel shift in the yield curve. When we have one unit of that factor, the three-month rateincreases by 0.2i basis points, the six-month rate increases by 0.26 basis points, and so on. Thesecond factor is shown in the column labeled PC2. It corresponds to a “twist” or “steepening” ofthe yieland 30 years move in the other direction. The third factor corresponds to a “bowing” of the yieldcurve. Rates at the short end and long end of the yield curve move in one direction; rates in themiddle move in the other direction. The interest rate move fora particular factor is known as factorloading. In our example, the first factor’s loading for the three-month rate is 0.21.!°
Because there are ten rates and ten factors, the interest rate changes observed on any given daycan always be expressed as a linear sum of the factors by solving a set of ten simultaneousequations. The quantity of a particular factor in the interest rate changes on a particular day isknown as the factor score for that day.
The importance of a factor is measured by the standard deviation of its factor score. Thestandard deviations of the factor scores in our example are shown in Table 16.4 and the factors arelisted in order of their importance. The numbers in Table 16.4 are measured in basis points. Aquantity of the first factor equal to one standard deviation, therefore, corresponds to the three-month rate moving by 0.21 x 17.49 = 3.67 basis points, the six-month rate moving by0.26 x 17.49 = 4.55 basis points, and so on.
      
    
> fates bety
id curve. Rates between 3 months and 2 years move in one direction; en 3 years
Table 16.4 Standard deviation of factor scores (basis points)
PCL PC2 PC3 PC4 PCS PC6 PC7 PC8& PCI PCIO1749 6.05 3.10 22.17 1.97 1.69 1.27 1.24 0.80 0.79
 
" See J. Frye, “Principals of Risk: Finding VAR through Factor-Based Interest Rate Scenarios,” in VAR:Understanding and Applying Value at Risk, Risk Publications, London, 1997, pp. 275-88.
'S The factor loadings have the property that the sum of their squares for each factor is 1.0.362 CHAPTER 16
Factor loading
   
0.6
0.2
Maturity (years)
  
Figure 16.6 Three most important factors driving yield curve movements
The technical details of how the factors are determined are not covered here. It is sufficient for usto note that the factors are chosen so that the factor scores are uncorrelated. For instance, in ourexample, the first factor score (amount of parallel shift) is uncorrelated with the second factor score(amount of twist) across the 1,543 days.
The variances of the factor scores (i.e., the squares of the standard deviations) have the propertythat they add up to the total variance of the data. From Table 16.3, the total variance of theoriginal data (i.e., the sum of the variance of the observations on the three-month rate, thevariance of the observations on the six-month rate, and so on) is
17.49? + 6.05° +3.10° +---+0.79° = 367.9
From this it can be seen that the first factor accounts for 17.497/367.9 = 83.1% of the variation inthe original data; the first two factors account for (17.497 + 6.05°)/367.9 = 93.1% of the variation inthe data; the third factor accounts for a further 2.8% of the variation. This shows most of the risk ininterest rate moves is accounted for by the first two or three factors. It suggests that we can relate therisks in a portfolio of interest rate dependent instruments to movements in these factors instead ofconsidering all ten interest rates. The three most important factors from Table 16.3 are plotted inFigure 16.6.'°
Using Principal Components Analysis to Calculate VaR
To illustrate how a principal components analysis can be used to calculate VaR, suppose we haveaportfolio with the exposures shown in Table 16.5 to interest rate moves. A one-basis-point change
'© Similar results to those described here, in respect of the nature of the factors and the amount of the total risk theyaccount for, are obtained when a principal components analysis is used to explain the movements in almost any yieldcurve in any country.Value at Risk 363
Table 16.5 Change in portfolio value for a one-basis-point ratemove ($ millions)
 
1-year 2-year 3-year > 4- year 5-yearrate . rate rate rate rate '+10 +4 —8 -7 +2
in the one-year rate causes the portfolio value to increase by $10 million; a one-basis-point changein the two-year rate causes it to increase by $4 million; and so on. We use the first two factors tomodel rate moves. (As mentioned in the preceding section, this captures over 90% of theuncertainty in rate moves.) Using the data in Table 16.3, our exposure to the first factor (measuredin millions of dollars per factor score basis point) is
10 x 0.32+4 x 0.35 —8 x 0.36 —7 x 0.36 + 2 x 0.36 = —0.08and our exposure to the second factor is10 x (—0.32) + 4 x (-0.10) — 8 x 0.02 —7 x 0.14 +2 x 0.17 = —4.40
Suppose that f| and /5 are the factor scores (measured in basis points). The change in the portfoliovalue is, to a good approximation, given by
8P = —0.08 f, — 4.40f,
The factor scores are uncorrelated and have the standard deviations given in Table 16.4. Thestandard deviation of 5P is therefore
¥0.082 x 17.49? + 4.40? x 6.05? = 26.66
The 1-day 99% VaR is therefore 26.66 x 2.33 = 62.12. Note that the data in Table 16.5 are suchthat we have very little exposure to the first factor and significant exposure to the second factor.Using only one factor would significantly understate VaR (see Problem 16.13). The duration-basedmethod for handling interest rates, mentioned in Section 16.4, would also significantly understateVaR as it considers only parallel shifts in the yield curve.
A principal components analysis can in theory be used for market variables other than interestrates. Suppose that a financial institution has exposures to a number of different stock indices. Aprincipal components analysis can be used to identify factors describing movements in the indicesand the most important of these can be used to replace the market indices in a VaR analysis. Howeffective a principal componenis analysis is fora group of market variables depends on how closelycorrelated they are.
As explained earlier in the chapter, VaR is usually calculated by relating the actual changes in aportfolio to percentage changes in market variables (the 4x;’s). A principal components analysis istherefore often carried out on percentage changes in market variables rather than on their actualchanges.364 CHAPTER 16
SUMMARY
 
 
A value-at-risk (VaR) calculation is aimed at making a statement of the form “We are X percentcertain that we will not lose more than V dollars in the next N days.” The variable V is the VaR,X is the confidence level, and WN is the time horizon.
One approach to calculating VaR is historical simulation. This involves creating a database
consisting of the daily movements in all market variables over a period of time. The first simulationtrial assumes that the percentage changes in each market variable are the same as those on the firstday covered by the database; the second simulation trial assumes that the percentage changes arethe same as those on the second day; and so on. The change in the portfolio value, 8P, is calculatedfor each simulation trial, and the VaR is calculated as the appropriate percentile of the probabilitydistribution of 5P.
An alternative is the model-building approach. This is relatively straightf
tions can be made:
1. The change in the value of the portfolio (5?) is linearly dependent on percentage changes inmarket variables.
2. The percentage changes in market variables are multivariate normally distributed.
The probability distribution of 5P is then normal, and there are analytic formulas for relating thestandard deviation of 4P to the volatilities and correlations of the underlying market variables. TheVaR can be calculated from well-known properties of the normal distribution.
When a portfolio includes options, 5P is not linearly related to the percentage changes in marketvariables. From a knowledge of the gamma of the portfolio, we can derive an approximatequadratic relationship between 5P and the percentage changes in market variables. Either theCornish—Fisher expansion or Monte Carlo simulation can then be used to estimate VaR.
In the next chapter we discuss how volatilities and correlations can be estimated for the model-building approach.
SUGGESTIONS FOR FURTHER READING
 
Boudoukh, J., M. Richardson, and R. Whitelaw, “The Best of Both Worlds,” RISK, May 1998,pp. 64-67.
Dowd, K., Beyond Value at Risk: The New Science of Risk Management, Wiley, New York, 1998.
Duffie, D., and J. Pan. “An Overview of Value at Risk,” Journal of Derivatives, 4, no. 3 (Spring 1997),7-49,
Embrechts, P., C. Kluppelberg, and T. Mikosch, Modeling Extremal Events for Insurance and Finance,Springer, New York, 1997.
Frye, J., “Principals of Risk: Finding VAR through Factor-Based Interest Rate Scenarios,” in VAR:Understanding and Applying Value at Risk, Risk Publications, London, 1997, pp. 275-88.
Hendricks, D., “Evaluation of Value-at-Risk Models Using Historical Data.” Economic Policy Review,Federal Reserve Bank of New York, 2 (April 1996), 39-69.
Hopper, G., “Value at Risk: A New Methodology for Measuring Portfolio Risk,” Business Review,Federal Reserve Bank of Philadelphia, July/August 1996, 19-29.Value at Risk 365
Hua, P., and P. Wilmott, “Crash Courses,” RISK, June 1997, pp. 64-67.
Hull, J. C., and A. White, “Value at Risk When Daily Changes in Market Variables Are Not NormallyDistributed,” Journal of Derivatives, 5 (Spring 1998), 9-19.
Hull, J. C., and A. White, “Incorporating Volatility Updating into the Historical Simulation Methodfor Value at Risk,” Journal of Risk, 1, no. 1 (1998), 5-19.
Jackson, P., D. J. Maude, and W. Perraudin, “Bank Capital and Value at Risk,” Journal of Derivatives,4, no. 3 (Spring 1997), 73-90.
Jamshidian, F., and Y. Zhu, “Scenario Simulation Model: Theory and Methodology,” Finance andStochastics, | (1997), 43-67.
Jorion, P., Value at Risk: The New Benchmark for Controlling Market Risk, Trwin, Burr Ridge, IL, 1997.Risk Publications, “Value at Risk,” RISK supplement, June 1996.
QUESTIONS AND PROBLEMS(ANSWERS IN SOLUTIONS MANUAL)
 
 
16.1. Consider a position consisting of a $100,000 investment in asset A and a $100,000 investment inasset B. Assume that the daily volatilities of both assets are 1% and that the coefficient ofcorrelation between their returns is 0.3. What is the 5-day 99% value at risk for the portfolio?
16.2. Describe three alternative ways of handling interest-rate-dependent instruments when the model-building approach is used to calculate VaR. How would you handle interest-rate-dependentinstruments when historical simulation is used to calculate VaR?
16.3. A financial institution owns a portfolio of options on the U.S. dollar-sterling exchange rate. Thedelta of the portfolio is 56.0. The current exchange rate is 1.5000. Derive an approximate linearrelationship between the change in the portfolio valuc and the percentage change in the exchangerate. If the daily volatility of the exchange rate is 0.7%, estimate the 10-day 99% VaR.
16.4. Suppose that you know that the gamma of the portfolio in the previous question is 16.2. Howdoes this change your estimate of the relationship between the change in the portfolio value andthe percentage change in the exchange rate? Calculate a new 10-day 99% VaR based onestimates of the first two moments of the change in the portfolio value.
16.5. Suppose that the daily change in the value of a portfolio is, to a good approximation, linearlydependent on two factors, calculated from a principal components analysis. The delta of aportfolio with respect to the first factor is 6 and the delta with respect to the second factoris —4. The standard deviations of the factor are 20 and 8, respectively. What is the 5-day90% VaR?
16.6. Suppose that a company has a portfolio consisting of positions in stocks, bonds, foreignexchange, and commodities. Assume there are no derivatives. Explain the assumptions under-lying (a) the linear model and (b) the historical simulation model for calculating VaR.
16.7. Explain how an interest rate swap is mapped into a portfolio of zero-coupon bonds withstandard maturities for the purposes of a VaR calculation.
16.8. Explain the difference between value at risk and conditional value at risk.
16.9, Explain why the linear model can provide only approximate estimates of VaR for a portfoliocontaining options.366
CHAPTER 16
 
16.10.
16.11.
16.12.
16.13.
16.14.
16.15.
Verify that the 0.3-year zero-coupon bond in the cash-flow mapping example in Appendix 16A ismapped in a $37,397 position in a three-month bond and a $11,793 position in a six-monthbond.
Suppose that the 5-year rate is 6%, the seven year rate is 7% (both expressed with annualcompounding), the daily volatility of a 5-year zero-coupon bond is 0.5%, and the daily volatilityof a 7-year zero-coupon bond is 0.58%. The correlation between daily returns on the two bondsis 0.6. Map a cash flow of $1,000 received at time 6.5 years into a position in a five-year bondand a position in a seven-year bond. What cash flows in five and seven years are equivalent tothe 6.5-year cash flow?
Some time ago a company has entered into a six-month forward contract to buy £1 million for$1.5 million. The daily volatility of a six-month zero-coupon sterling bond (when its price istranslated to dollars) is 0.06% and the daily volatility of a six-month zero-coupon dollar bond is0.05%. The correlation between returns from the two bonds is 0.8. The current exchange rate is1.53. Calculate the standard deviation of the change in the dollar value of the forward contractin one day. What is the 10-day 99% VaR? Assume that the six-month interest rate in bothsterling and dollars is 5% per annum with continuous compounding.
The text calculates a VaR estimate for the example in Table 16.5 assuming two factors. Howdoes the estimate change if you assume (a) one factor and (b) three factors.
A bank has a portfolio of options on an asset. The delta of the options is —30 and the gammais —5. Explain how these numbers can be interpreted. The asset price is 20 and its volatility is1% per day. Using the quadratic model, calculate the first three moments of the change in theportfolio value. Calculate a 1-day 99% VaR using (a) the first two moments and (b) the firstthree moments.
Suppose that in Problem 16.14 the vega of the portfolio is —2 per 1% change in the annualyolatility. Derive a model relating the change in the portfolio value in one day to delta, gamma,and vega. Explain, without doing detailed calculations, how you would use the model tocalculate a VaR estimate.
ASSIGNMENT QUESTIONS
 
 
 
16.16.
16.17.
16.18.
16.19,
A company has a position in bonds worth $6 million. The modified duration of the portfolio is5.2 years. Assume that only parallel shifts in the yield curve can take place and that the standarddeviation of the daily yield change (when yield is measured in percent) is 0.09. Use the durationmodel to estimate the 20-day 90% VaR for the portfolio. Explain carefully the weaknesses of thisapproach to calculating VaR. Explain two alternatives that give more accuracy.
Consider a position consisting of a $300,000 investment in gold and a $500,000 investment insilver. Suppose that the daily volatilities of these two assets are 1.8% and 1.2%, respectively, andthat the coefficient of correlation between their returns is 0.6. What is the 10-day 97.5% value atrisk for the portfolio? By how much does diversification reduce the VaR?
Consider a portfolio of options on a single asset. Suppose that the delta of the portfolio is 12, thevalue of the asset is $10, and the daily volatility of the asset is 2%. Estimate the I-day 95% VaRfor the portfolio. :
Suppose that the gamma of the portfolio in Problem 16.18 is —2.6. Derive a quadraticrelationship between the change in the portfolio value and the percentage change in theunderlying asset price in one day.Value at Risk 367
 
16.20.
16.21.
16.22.
a. Calculate the first three moments of the change in the portfolio value.
b. Using the first two moments and assuming that the change in the portfolio is normallydistributed, calculate the 1-day 95% VaR Tor the portfolio.
c. Use the third moment and the Cornish—Fisher expansion to revise your answer to (b).
A company has a long position in a two-year bond and a three-year bond as well as a shortposition in a five-year bond. Each bond has a principal of $100 and pays a 5% coupon annually.Calculate the company’s exposure to the l-year, 2-year, 3-year, 4-year, and 5-ycar rates. Use thedata in Tables 16.3 and 16.4 to calculate a 20-day 95% VaR on the assumption that rate changesare explained by (a) one factor, (b) two factors, and (c) three factors. Assume that the zero-coupon yield curve is flat at 5%.
A bank has written a call option on one stock and a put option on another stock. For the firstoption the stock price is 50, the strike price is 51, the volatility is 28% per annum, and the time tomaturity is nine months. For the second option the stock price is 20, the strike price is 19, thevolatility is 25% per annum, and the time to maturity is one year. Neither stock pays a dividend,the risk-free rate is 6% per annum, and the correlation between stock price returns is 0.4.Calculate a 10-day 99% VaR:
a. Using only delta
b. Using delta, gamma, and the first two moments of the change in the portfolio value
c. Using delta, gamma, and the first three moments of the change in the portfolio value
d, Using the partial simulation approach
e. Using the full simulation approachA common complaint of risk managers is that the model-building approach (either linear orquadratic) does not work well when delta is close to zero, Test what happens when delta is closeto zero in using Sample Application E in the DerivaGem Application Builder software. (You cando this by experimenting with different option positions and adjusting the position in theunderlying to give a delta of zero.) Explain the results you get.368 CHAPTER 16
APPENDIX 16ACash-Flow Mapping
In this appendix we explain one procedure for mapping cash flows to standard maturity dates. Wewill illustrate the procedure by considering a simple example of a portfolio consisting of a longposition in a single Treasury bond with a principal of $1 million maturing in 0.8 years. We supposethat the bond provides a coupon of 10% per annum payable semiannually. This means that thebond provides coupon payments of $50,000 in 0.3 years and 0.8 years. It also provides a principalpayment of $1 million in 0.8 years. The Treasury bond can therefore be regarded as a position ina 0.3-year zero-coupon bond with a principal of $50,000 and a position in a 0.8-year zero-couponbond with a principal of $1,050,000.
The position in the 0.3-year zero-coupon bond is mapped into an equivalent position in3-month and 6-month Zero-coupon bonds. The position in the 0.8-year zero-coupon bond ismapped into an equivalent position in 6-month and |-year zero-coupon bonds. The result is thatthe position in the 0.8-year coupon-bearing bond is, for VaR purposes, regarded as a position inzero-coupon bonds having maturities of three months, six months, and one year.
The Mapping Procedure
Consider the $1,050,000 that will be received in 0.8 years. We suppose that zero rates, daily bondprice volatilities, and correlations between bond returns are as shown in Table 16.6.
The first stage is to interpolate between the 6-month rate of 6.0% and the l-year rate of 7.0% toobtain a 0.8-ycar rate of 6.6%. (Annual compounding is assumcd for all rates.) The present valueof the $1,050,000 cash flow to be received in 0.8 years is
1,050,000
magne = 997,662
We also interpolate between the 0.1% volatility for the 6-month bond and the 0.2% volatility forthe l-year bond to get a 0.16% volatility for the 0.8-year bond.
Table 16.6 Data to illustrate mapping procedure
 
 
Maturity 3-month 6-month d-yearZero rate (% with annual compounding) 5.50 6.00 7.00Bond price volatility (% per day) 0.06 0.10 0.20Correlation between 3-month 6-month d-yeardaily returns bond bond bond3-month bond 1.0 0.9 0.66-month bond 0.9 1.0 0.7
1-year bond 0.6 0.7 1.0Value at Risk 369
Table 16.7 The cash-flow mapping
$50,000 received $1,050,000 received Total
 
- in 0.3 years in 0.8 yearsPosition in 3-month bond ($) 37,397 37,397Position in 6-month bond (3) 11,793 319,589 331,382
Position in i-year bond ($) 678.074 678.074
Suppose we allocate @ of the present value to the 6-month bond and 1 — a@ of the present valueto the {-year bond. Using equation (16.2) and matching variances, we obtain
0.0016? = 0.0017a? + 0,0027(1 — a)? +2 x 0.7 x 0.001 x 0.0020(1 — a)
This is a quadratic equation that can be solved in the usual way to give a = 0.320337. This meansthat 32.0337% of the value should be allocated to a 6-month zero-coupon bond and 67.9663% ofthe value should be allocated to a 1-year zero coupon bond. The 0.8-year bond worth $997,662 istherefore replaced by a 6-month bond worth
997,662 x 0.320337 = $319,589and a l-year bond worth997,662 x 0.679663 = $678,074
This cash-flow mapping scheme has the advantage that it preserves both the value and the varianceof the cash flow. Also, it can be shown that the weights assigned to the two adjacent zero-couponbonds are always positive.
For the $50,000 cash flow received at time 0.3 years, we can carry out similar calculations (secProblem 16.10). It turns out that the present value of the cash flow is $49,189. It can be mappedinto a position worth $37,397 in a three-month bond and a position worth $11,793 in a six-monthbond
The results of the calculations are summarized in Table 16.7. The 0.8-year coupon-bearing bondis mapped into a position worth $37,397 in a three-month bond, a position worth $331,382 inasix-month bond, and a position worth $678,074 in a one-year bond. Using the volatilities andcorrelations in Table 16.6, equation (16.2) gives the variance of the change in the price of the0.8-year bond with n= 3, a, = 37,397, a = 331,382, a3 = 678,074, o, = 0.0006, 0, = 0.001,0; = 0.002, and p)7 = 0.9, p,;3 = 0.6, p23 = 0.7. This variance is 2,628,518. The standard deviation-of the change in the price of the band is therefore \/2,628,518 = 1,621.3. Because we are assuming
that the bond is the only instrument in the portfolio, the 10-day 99% VaR is
1621.3 x V10 x 2.33 = 11,946or about $11,950.37