Value at Risk By A.V. Vedpuriswar

Value at RiskBy A.V. Vedpuriswar

June 14, 2014

Introduction

¨ VAR tells us the maximum loss a portfolio may suffer at a given confidence interval for a specified time horizon.

¨ If we can be 95% sure that the portfolio will not suffer more than $ 10 million in a day, we say the 95% VAR is $ 10 million.

2

Computing Value at Risk

¨ VaR is the product of :– Value of portfolio– Z factor ( depends on confidence level)– Volatility– Time ( VaR scales in proportion to square root of

time)

3

Computing Value at Risk

¨ The usual practice is to calculate daily volatility by observing the daily opening and closing prices of the portfolio over a period of time, say 6 months.

¨ Then we obtain the volatility for the actual period under consideration by multiplying by the square root of the time.

¨ Thus the volatility for 5 trading days will be that for one day multiplied by √5.

4

How Banks disclose VaR

5

VaR at UBS

6Ref : Company Annual report

VaR at UBS


VaR at Credit Suisse


VaR at Goldman Sachs


14

¨ Average revenue = $5.1 million per day¨ Total no. of observations = 254. Std dev = $9.2 million¨ Confidence level = 95% ¨ No. of observations < - $10 million = 11¨ No. of observations < - $ 9 million = 15¨ Find 95% VAR.

Problem

Solution¨ The point such that the no. of observations to the left = (254) (.05) =

12.7¨ (12.7 – 11) /( 15 – 11 ) = 1.7 / 4 ≈ .4¨ So required point = - (10 - .4 x 1) = - $9.6 million¨ VAR = E (W) – (-9.6) = 5.1 – (-9.6) = $14.7 million¨ If we assume a normal distribution,¨ Z at 95% ( one tailed) confidence interval = 1.645¨ VAR = (1.645) (9.2) = $ 15.2 million

1515

Problem

The VAR on a portfolio using a one day time horizon is USD 100 million. What is the VAR using a 10 day horizon ? Solution¨ Variance scales in proportion to time.¨ So variance gets multiplied by 10¨ And std deviation by √10¨ VAR = 100 √10 = (100) (3.16) = 316¨ (σN

2 = σ12 + σ2

2 ….. = Nσ2)Note; This approach is valid only when the daily variances are independent.

1616

Problem

If the daily VAR is $12,500, calculate the weekly, monthly, semi annual and annual VAR. Assume 250 days and 50 weeks per year.SolutionWeekly VAR ＝ (12,500) (√5)

＝ 27,951 Monthly VAR ＝ ( 12,500) (√20)

＝ 55,902Semi annual VAR ＝ (12,500) (√125)

＝ 139,754 Annual VAR ＝ (12,500) (√250)

＝ 197,642

17

Variance Covariance Method

ProblemA fund has a portfolio consisting of 40% fixed income and 60% equity. The estimated 95% annual VAR assuming 250 trading days for the entire portfolio was $ 1,367,000 based on the portfolio’s market value of $ 12,500,000. The correlation between bond and stock returns is 0.The annual loss on the equity part of the portfolio is expected to exceed $ 1,153,000 5% of the time.What will be the daily expected loss that will be exceeded 5 % of the time for the bond portfolio?Solution¨ 1,367,000^2 = 1,153,000^2 + x^2¨ X = $ 734,357¨ Daily VAR = 734357/√250¨ = 46,445 18

ProblemConsider a bond position of $ 10 million, a modified duration of 3.6 years and an annualized bond volatility of 2%.Calculate the 10 day 99% VAR assuming 252 business days in a year.

Solution¨ 10 day volatility = .02X√(10/252)=.003984¨ So 99% VAR = 2.33 X 3.6 X 10,000,000 X .003984¨ = $ 334,186

19

ProblemConsider a position consisting of a $100,000 investment in asset A and a $100,000 investment in asset B. Assume that the daily volatilities of both assets are 1% and that the coefficient of correlation between their returns is 0.3. What is the 5-day 99% VaR for the portfolio?

Solution¨ The standard deviation of the daily change in the investment in each

asset is $1,000. The variance of the portfolio’s daily change is¨ 1,0002 + 1,0002 + 2 x 0.3 x 1,000 x 1,000 = 2,600,000¨ The standard deviation of the portfolio’s daily change is $1,612.45. ¨ The standard deviation of the 5-day change is¨ 1,612.45 x √5 = $3,605.55¨ From the tables of N(x) we see that Z = 2.33. ¨ The 5-day 99 percent value at risk is therefore 2.33 x 3,605.55 =

$8,401.

20Ref : John C Hull, Options, Futures and Other Derivatives

2121

We have a portfolio of $10 million in shares of Microsoft. We want to calculate VAR at 99% confidence interval over a 10 day horizon. The volatility of Microsoft is 2% per day.

Problem

Solution¨ σ = 2% = (.02) (10,000,000) =

$200,000¨ Z (p = .01) = Z (p =.99) = 2.33¨ Daily VAR =(2.33) (200,000) = $ 466,000¨ 10 day VAR = 466,000 √10 = $

1,473,621

Ref : John C Hull, Options, Futures and Other Derivatives

2222

Problem

Consider a portfolio of $5 million in AT&T shares with a daily volatility of 1%. Calculate the 99% VAR for 10 day horizon.Solution¨ σ = 1% = (.01) (5,000,000) = $ 50,000¨ Daily VAR = (2.33) (50,000) = $ 116,500 ¨ 10 day VAR = $ 116,500 √10 = $ 368,405


2323

ProblemNow consider a combined portfolio of AT&T and Microsoft shares. Assume the returns on the two shares have a bivariate normal distribution with the correlation of 0.3. What is the portfolio VAR.?Solution¨ σ2 = w1

2 σ12 + w2

2 σ22 + 2 ῤw1 w2 σ1 σ2

¨ = (200,000)2 + (50,000)2 + (2) (.3) (200,000) (50,000)

¨ σ = 220,277¨ Daily VAR = (2.33) (220,277) = 513,129¨ 10 day VAR = (513,129) √10 =

$1,622,657¨ Effect of diversification = (1,473,621 + 368,406) –

(1,622,657)= 219,369


24

Problem

Consider a portfolio made up of 5 year 5 % coupon government bonds. The bonds are trading at $ 100. The historical annual volatility is 1 %. Expected YTMs are normally distributed with zero mean and volatility of 1%. Calculate the 95% one year VAR.

SolutionWorst YTM = actual YTM + 1.65 x Volatility= 5 + 1.65 x 1 = 6.65%If YTM is 6.65%, bond price will be 93.1708So the VAR is 100-93.17 = $ 6.83

25

Problem Consider the following single bond of $10 million, a modified duration of 3.6 yrs and annualized yield of 2%. Calculate the 10 day holding period VaR of the position with a 99% confidence interval, assuming there are 252 days in a year.SolutionVAR = $10,000,000* 0.02*3.6* √10/252 * 2.33 = $334,186

26

Problem Assume that a risk manager wants to calculate VAR for an S&P 500 futures contract using the historical simulation approach. The current price of the contract is 935 and the multiplier is 250. Given the historical price data shown below for the previous 300 days, what is the VAR of the position at 99% using the historical simulation methodology?Returns: -6.1%,-6%,-5.9%,-5.7%, -5.5%, -5.1%..........4.9%, 5%, 5.3%, 5.6%, 5.9%, 6%Solution The 99% return among 300 observations would be the 3rd worst observation among the returns. Among the returns given above -5.9% is the 3rd worst return, the 99% VAR for this position is therefore (935)*250* (0.059) = $13,791.

27

Problem Consider the portfolio of an American trader with two foreign currencies, Canadian dollar and euro. These two currencies are uncorrelated and have a volatility against the dollar of5% and 12% respectively. The portfolio has $2 million invested in CAD and $1 million in Euro. What is the portfolio VAR at 95% confidence level?Solution¨ Variance of the portfolio return¨ = {(2 (.05)}2 + {(1) (.12)}2 =.01 + .0144 = .0244¨ Std devn = √.0244 = $ .156205 million¨ VAR = (1.65) (156,205) = $257,738¨ VAR for Canadian dollar part= $ (1.65) (.05) (2) million

= $165,000¨ VAR for Euro part = $ (1.65) (.12) (1) million

= $ 198,000¨ Undiversified VAR = $ 363,000¨ Thus the diversified VAR is significantly lower.

28

Problem

Suppose we increase the Canadian dollar position by $10,000. What is the marginal VAR?

Solution¨ Variance = {(2.01) (.05)}2 + {(1) (.12)}2=.0101

+ .0144= .0245¨ σ = √.0245 = $.1565 million¨ VAR = (1.65) (156,500) = $ 258,225¨ Marginal VAR = 258,225 – 257,738

= $ 487

ProblemAn American trader owns a portfolio of options on the US dollar-sterling exchange rate. The delta of the portfolio is 56.0. The current exchange rate is $/£ 1.5000. Derive an approximate linear relationship between the change in the portfolio value and the percentage change in the exchange rate. If the daily volatility of the exchange rate is 0.7%, estimate the 10-day 99% VaR.

Solution¨ The approximate relationship between the daily change in the portfolio

value, ΔP, and the daily change in the exchange rate, ΔS, is ΔP = 56 ΔS¨ For a unit change in £, $ will change by 1.5. It follows that¨ ΔP = 56 x 1.5 Δx¨ Or ΔP = 84 Δx¨ The standard deviation of Δx equals the daily volatility of the exchange

rate, or 0.7 percent.The standard deviation of ΔP is therefore 84 x 0.007 = $ 0.588.

¨ So the 10-day 99 percent VaR for the portfolio is¨ 0.588 x 2.33 x √10 = $ 4.33 for an investment of £1.29


ProblemSome time ago a company entered into a forward contract to buy £1 million for $1.5 million. The contract now has 6 months to maturity. The daily volatility of a 6-month zero-coupon sterling bond (when its price is translated to dollars) is 0.06% and the daily volatility of a 6-month zero-coupon dollar bond is 0.05%. The correlation between returns from the two bonds is 0.8. The current exchange rate is $/ £ 1.53. Calculate the standard deviation of the change in the dollar value of the forward contract in 1 day. What is the 10-day 99% VaR? Assume that the 6-month interest rate in both sterling and dollar is 5% per annum with continuous compounding.

Solution¨ The contract is a long position in a sterling bond plus a short position in a dollar

bond.¨ The value of the sterling bond is 1.53e-0.05x0.5 or $1.492 million. ¨ The value of the dollar bond is 1.5e-0.05x0.5 or $1.463 million. ¨ The variance of the change in the value of the contract in one day is :¨ 1.4922 x 0.00062 + 1.4632 x 0.00052 – 2 x 0.8 x 1.492 x 0.0006 x 1.463 x

0.0005 = 0.000000288¨ The standard deviation is therefore $0.000537 million. ¨ The 10-day 99% VaR is $0.000537 x √10 x 2.33 = $0.00396 million

30


Problem

Consider the contract on the dollar/euro exchange rate (EC) traded on the CME. The notional amount is 125,000 Euros. Assume that the annual volatility is 12% and the current price is $1.05 per Euro. Find daily VAR at 99% confidence level.

Solution ¨ VAR = (2.33) [(.12) / √252] × (125,000 × 1.05)¨ = $ 2310

31

32

Problem

Consider a portfolio with a one day VAR of $1 million. Assume that the market is trending with an auto correlation of 0.1. Under this scenario, what would you expect the two day VAR to be? Solution¨ V2 = 2σ2 (1 + ῤ)¨ = 2 (1)2 (1 + .1) = 2.2 ¨ V = √2.2 = 1.4832¨

Auto correlation over longer periods¨ Consider a period of 5 days.¨ We assume the first day’s movement will have an

impact on the second day's movement through the correlation coefficient.

¨ The first day’s movement will affect the third day’s movement through the square of the correlation coefficient and so on.

¨ Then the combined variance will be:¨ σ2 + σ2 + σ2 + σ2 + σ2 + (2) () σ2 + (2) ()2 σ2 + (2)

()3 σ2 + (2) ()4 σ2 + (2) () σ2 + (2) ()2 σ2 + (2) ()3 σ2 + (2) () σ2 + (2) ()2 σ2 + (2) () σ2

¨ = 5 σ2 + (8) () σ2 + (6) ()2 σ2 + (4) ()3 σ2 + (2) ()4 σ2

Solution¨ σ = 0 .1; = .3¨ Variance = (5) (.1)2 + (4) (2) (.3) (.1)2

¨ + (3) (2) (.3)2 (.1)2 + (2) (2) (.3)3 (.1)2

¨ + (2) (.3)4 (.1)2 ¨ = .05 + .024 + .0054 + .00108 + .000162¨ = .080642¨ Volatility = .284

34

ProblemConsider a portfolio with standard deviation of daily returns of 0.1 and autocorrelation of 0.3. Calculate the 5 day volatility.

35

Monte Carlo Simulation

36

What is Monte Carlo VAR?

¨ The Monte Carlo approach involves generating many price scenarios (usually thousands) to value the assets in a portfolio over a range of possible market conditions.

¨ The portfolio is then revalued using all of these price scenarios.

¨ Finally, the portfolio revaluations are ranked to select the required level of confidence for the VAR calculation.

37

Generate Scenarios

¨ The first step is to generate all the price and rate scenarios necessary for valuing the assets in the relevant portfolio, as well as the required correlations between these assets.

¨ There are a number of factors that need to be considered when generating the expected prices/rates of the assets:– Opportunity cost of capital – Stochastic element – Probability distribution

38

Opportunity Cost of Capital¨ A rational investor will seek a return at least equivalent

to the risk-free rate of interest.¨ Therefore, asset prices generated by a Monte Carlo

simulation must incorporate the opportunity cost of capital.

39

Probability Distribution¨ Monte Carlo simulations are based on random draws

from a variable with the required probability distribution, usually the normal distribution.

¨ The normal distribution is useful when modeling market risk in many cases.

¨ But it is the returns on asset prices that are normally distributed, not the asset prices themselves.

¨ So we must be careful while specifying the distribution.

40

Calculate the Value of the Portfolio¨ Once we have all the relevant market price/rate

scenarios, the next step is to calculate the portfolio value for each scenario.

¨ For an options portfolio, depending on the size of the portfolio, it may be more efficient to use the delta approximation rather than a full option pricing model (such as Black-Scholes) for ease of calculation.

¨ Δ (Option) = Δ(ΔS)¨ Thus the change in the value of an option is the product

of the delta of the option and the change in the price of the underlying.

41

Other approximations¨ There are also other approximations that use delta,

gamma (Γ) and theta (Θ) in valuing the portfolio.¨ By using summary statistics, such as delta and gamma,

the computational difficulties associated with a full valuation can be reduced.

¨ Approximations should be periodically tested against a full revaluation for the purpose of validation.

¨ When deciding between full or partial valuation, there is a trade-off between the computational time and cost versus the accuracy of the result.

¨ The Black-Scholes valuation is the most precise, but tends to be slower and more costly than the approximating methods.

42

Reorder the Results ¨ After generating a large enough number of scenarios and

calculating the portfolio value for each scenario: – the results are reordered by the magnitude of the

change in the value of the portfolio (Δportfolio) for each scenario

– the relevant VAR is then selected from the reordered list according to the required confidence level

¨ If 10,000 iterations are run and the VAR at the 95% confidence level is needed, then we would expect the actual loss to exceed the VAR in 5% of cases (500).

¨ So the 501st worst value on the reordered list is the required VAR.

¨ Similarly, if 1,000 iterations are run, then the VAR at the 95% confidence level is the 51st highest loss on the reordered list.

43

Formula used typically in Monte Carlo for stock price modelling

44

Advantages of Monte Carlo

¨ We can cope with the risks associated with non-linear positions.

¨ We can choose data sets individually for each variable. ¨ This method is flexible enough to allow for missing

data periods to be excluded from the VAR calculation. ¨ We can incorporate factors for which there is no actual

historical experience. ¨ We can estimate volatilities and correlations using

different statistical techniques.

45

Problems with Monte Carlo

¨ Cost of computing resources can be quite high.¨ Speed can be slow.¨ Random Numbers may not be all that random.¨ Pseudo random numbers are only a substitute for

true random numbers and tend to show clustering effects.

¨ Monte Carlo often assumes normal distribution. ¨ But it can be performed with alternative distributions. ¨ Results (value at risk estimate) depend critically on

the models used to value (often complex) financial instruments.

46

Historical Simulation

47

Introduction¨ Unlike the Monte Carlo approach, it uses the actual

historical distribution of returns to simulate the VAR of a portfolio.

¨ Real data plus ease of implementation, have made historical simulation a very popular approach to estimating VAR.

¨ Historical simulation avoids the assumption that returns on the assets in a portfolio are normally distributed.

¨ Instead, it uses actual historical returns on the portfolio assets to construct a distribution of potential future portfolio losses.

¨ This approach requires minimal analytics. ¨ All we need is a sample of the historic returns on the

portfolio whose VAR we wish to calculate.

48

Steps

¨ Collect data¨ Generate scenarios¨ Calculate portfolio returns¨ Arrange in order.

49

% Returns Frequency Cumulative Frequency

- 16 1 1- 14 1 2- 10 1 3- 7 2 5- 5 1 6- 4 3 9- 3 1 10- 1 2 120 3 151 1 162 2 184 1 196 1 207 1 218 1 229 1 23

11 1 2412 1 2614 2 2718 1 2821 1 2923 1 30

What is VAR (90%) ?

10% of the observations, i.e, (.10) (30)

= 3 lie below -7So VAR = -7

50

Advantages and Disadvantages of Historical simulation

¨Advantages¨ Simple¨ No normality

assumption¨ Non parametric

¨Disadvantages¨ Reliance on the past¨ Length of estimation

period¨ Weighting of data

51

Comparison of different VAR modeling techniques

52

Simulation vs Variance Covariance methods¨ Simulation approaches are preferred by global banks

due to: – flexibility in dealing with the ever-increasing range of complex

instruments in financial markets – the advent of more efficient computational techniques in recent

years – the falling costs in information technology

¨ However, the variance-covariance approach might be the most appropriate method for many smaller firms, particularly when: – they do not have significant options positions – they prefer to outsource the data requirement component of

their risk calculations to a company such as RiskMetrics – significant savings can often be made by using outsourced

volatility and correlation data, compared to internally storing the daily price histories required for simulation techniques

53

Model Validation

54

Basel Committee Standards (1)

¨ Banks that prefer to use internal models must meet, on a daily basis, a capital requirement that is the higher of either: – the previous day's value at risk – the average of the daily value at risk of the preceding 60

business days multiplied by a minimum factor of three¨ VAR must be computed on a daily basis. ¨ A one-tailed confidence interval of 99% must be used. ¨ The minimum holding period should be 10 trading

days . ¨ The minimum historical observation period should be

one year.

55

¨ Banks should update their data sets at least once every three months.

¨ Banks can recognize correlations within broad risk categories.

¨ Provided the relevant supervisory authority is satisfied with the bank's system for measuring correlations , they may also recognize correlations across broad risk factor categories.

¨ Banks' internal models are required to accurately capture the unique risks associated with options and option-like instruments.

¨ The Basel Committee has also specified qualitative factors that banks must meet before they are permitted to use internal models.

Basel Committee Standards(2)

56

¨ The Basel Committee prescribes an increase in capital requirements if, based on a sample of 250 observations (a one-year observation period), the VAR model underpredicts the number of exceptions (losses exceeding the 99% confidence level).

¨ For such purposes, three 'zones' have been distinguished by the Committee.

¨ Green Zone : 0-4 exceptions¨ Yellow zone : 5-9 exceptions¨ Red zone : 10 or more exceptions

Basel Committee Standards(3)

57

Problem

Based on a 90% confidence level, how many exceptions in back testing a VAR model should be expected over a 250 day trading year?

Solution¨ 10% of the time loss may exceed VAR ¨ So no. of observations = (.10) (250) ¨ = 25

Problem

We are currently feeding a model with 600 days of data. The VAR confidence level is 99%. Nine exceptions are observed. Should we reject the model? Suppose it had been 12. Would we reject the model?Solution¨ 1 – Binomdist (8, 600, .01, True) = .152¨ So we cannot reject the model at 5% significance

level. ¨ 1 – Binomdist (11, 600, .01, True) = .019¨ So we would reject the model at 5% significance

level


Problem

We back test a VAR model with 1000 days of data. The VAR confidence level is 99%. 17 exceptions are observed. Should the model be rejected at 5% significance level?

Solution¨ The probability of the VAR being exceeded on a given day = 1 - .99 = .01¨ The probability of the VAR being exceeded on 17 days or

more¨ =1 – Binomdist (16, 1000, .01, True) = 2.64%¨ 2.64% < 5%¨ So the model should be rejected.


ProblemA $10 million one year loan has a 1.25% probability of default. If there is no default, profit of $ 0.2 million will be made. If there is a default, the recovery can be anything from 0 to the full loan value. Calculate the 99% VAR and conditional VAR. Solution ¨ Default zone = 1.25% starting from 98.75% going up to

100%.¨ Loss at 99% point =[ .25/1.25] X 10 = $ 2 million.¨ So the 99% VAR is $ 2million.¨ Expected shortfall = {2 + 10]/2 = $ 6 million

60

61

Stress Testing

62

Introduction

¨ Stress testing involves analysing the effects of exceptional events in the market on a portfolio's value.

¨ These events may be exceptional, but they are also plausible.

¨ And their impact can be severe.¨ Historical scenarios or hypothetical scenarios can be

used.

63

Two approaches to Stress testing¨Single-factor stress testing (sensitivity testing)

involves applying a shift in a specific risk factor to a portfolio in order to assess the sensitivity of the portfolio to changes in that risk factor.

¨Multiple-factor stress testing (scenario analysis) involves applying simultaneous moves in multiple risk factors to a portfolio to reflect a risk scenario or event that looks plausible in the near future.

64

Extreme Value Theory¨ EVT is a branch of statistics dealing with the extreme deviations

from the mean of statistical distributions. ¨ The key aspect of EVT is the extreme value theorem. ¨ According to EVT, given certain conditions, the distribution of

extreme returns in large samples converges to a particular known form, regardless of the initial or parent distribution of the returns.

¨ This distribution is characterized by three parameters – location, scale and shape (tail).

¨ The tail parameter is the most important as it gives an indication of the heaviness (or fatness) of the tails of the distribution.

¨ The EVT approach is very useful because the distributions from which return observations are drawn are very often unknown.

¨ EVT does not make strong assumptions about the shape of this unknown distribution.

Gaussian Copulas

¨ Consider variables, V1 and V2 that are not normally distributed.

¨ Map the two variables on to a normal distribution.¨ Apply correlation¨ Create bivariate normal distribution.

65

Illustration¨ Consider 2 variables that have a uniform distribution.

Using Gaussian copula, and assuming a copula correlation of 0.3, define a correlation structure.

66

0.25 0.50 0.750.250.500.75

67

0.25 0.50 0.750.25 -.675, -.675 -.675, 0 -.675, +.6750.50 0, -.675 0, 0 0,

+ .6750.75 +.675,

-.675+.675, 0 +.675, .+

675

0.25 0.50 0.750.25 .095 .1633 .21570.50 .1633 .2985 .41340.75 .2157 .4134 .5953

Illustrationx percentil

eZ perce

ntileZ

0.1

5.00 -1.65 2.00 -2.06

0.2

20.00 -.84 8.00 -1.41

0.3

38.75 -.29 18.00 -.92

0.4

55.00 .13 32.00 -.47

0.5

68.75 .49 50.00 0

0.6

80.00 .84 68.00 .47

0.7

88.75 1.21 82.00 .92

0.8

95.00 1.65 92.00 1.41

0.9

98.75 2.24 98.00 2.06

68

Correlation coefficient = 0.5

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.90.1 .006 .017 .028 .037 .044 .048 .049 .050 .0500.2 .013 .043 .081 .120 .156 .181 .193 .198 .2000.3 .017 .061 .124 .197 .273 .331 .364 .381 .3870.4 .019 .071 .149 .248 .358 .449 .505 .535 .5480.5 .019 .076 .164 .281 .417 .537 .616 .663 .6830.6 .020 .078 .173 .301 .456 .600 .701 .763 .7930.7 .020 .079 .177 .312 .481 .642 .760 .837 .8770.8 .020 .080 .179 .318 .494 .667 .798 .887 .9360.9 .020 .080 .180 .320 .499 .678 .816 .913 .970

69

VAR – Cash flow mappingProblemConsider a long position in a $1 million Treasury bond. ¨ Maturity : 0.8 years¨ Coupon : 10% payable semiannually

Annualized yield & volatility3 Month 6 Month 1 Year

Annualised yield 5.50 6.00 7.00Volatility 0.06 0.10 0.20

Correlations between daily returns3 Month 6 Month 1 Year

3 month 1.0 0.9 0.66 month 0.9 1.0 0.71 year 0.6 0.7 1.0

Explain how mapping can be done while calculating VaR.Ref : John C Hull, Options, Futures and Other Derivatives

Solution ¨ The current position involves the following:Cash flow of $50,000 in .3 yearsCash flow of $1,050,000 in .8 years¨ So the position can be considered a combination of

two zero coupon bonds, maturity 0.3, 0.8 years.¨ Let us write the position as equivalent to a

combination of standard 3 month, 6 month and 1 year bonds.

¨ 3 years = (.3) (12) = 3.6 months.¨ 3 month interest rate = 5.50%¨ 6 month interest rate = 6.00%

¨ Effective interest rate for 3.6 months zero coupon bond = 5.50 + (.6/3)(.5) = 5.6%

= .056¨ Present value = = 49,189¨ Volatility = = .068%.¨ Let us allocate to a 3 month bond and 1 - to a 6

month bond.¨ Then we can write: 2 = 1

2 + 22 + 2 12

¨ Here = .068 1 = .06 2 = .10 = .90¨ or .0682 = 2 (.06)2+ (1-)2(.10)2 + 2 (.9)() (1-)(.06)

(.10)¨ or .0682 = 2 (.06)2 + (1-)2 (.10)2 + 2(.9) ()(1-)(.06)

(.10)¨ Putting = .7603, ¨ LHS = .00462, RHS = .00208 + .00057 + .001968

= .00462¨ So we can write the position as equivalent to ¨ $ (.7603) (49,189) = $37,397 in 3 month bond¨ $ (.2397) (49,189) = $11,791 in 6 month bond

3.)056.1(000,50

)04(.36.06.

¨ Now consider $1,050,000 received after 0.8 years.¨ It can be considered a combination of 6 month and 12

month positions.¨ Interpolating the interest rate we get: =

.066¨ Present value of cash flows = =

$997,662¨ Volatility = [.1 + (3.6/6)(0.1) ] = 0.16¨ If is the position in the 6 month bond and (1-) in

the 12 month bond, 2 = 2 12 + (1-)2 2

2 + 2 (1-) 12

¨ Or (.16)2 = 2 (.1)2 + (1-)2 (.2)2 + 2 (.7) () (1-) (.1)(.2)

¨ LHS =.0256 Put =.320337; RHS =.001026 + .01848 + .006096 ≈.0256

¨ So the position is equivalent to ¨ (.3203) (997,662)= $319,589 in 6 month bond¨ (.6797) (997,662)= $678,074 in 12 month bond

)01(.66.306.

8.)066.1(000,050,1

¨ We can now write the portfolio in terms of 3 month, 6 month, 12 month zero coupon bonds.

¨ $50,000 $1,050,000 Total¨ t = .3 t = .8¨ 3 month bond 37,397 -- 37, 397¨ 6 month bond 11,791 319,589 331,380¨ 12 month bond -- 678,074 678,074

¨ Let 1, 2, 3 be the volatilities of the 3 month, 6 months, 12 months bonds and 12, 13, 23 be the respective correlations.

¨2 = 12 + 2

2 + 32 + 21212 + 22323 + 21313

¨ = [(37,397)2 (.06)2 + (331,380)2 (.10)2 + (678,074)2 (.20)2

¨ + (2) (37,397) (331,380) (.06) (.10) (.90)¨ + (2) (331,380) (678,074) (.10) (.20) (.70)¨ + (2) (37,397) (678,074) (.06) (.20) (.60)] x 10-4

¨ = ≈ 2,628,536¨ = = $1621.3¨ 10 day 99% VAR¨ = 1621.3 x 10 x 2.33¨ = $11,946

Value at Risk By A.V. Vedpuriswar

Documents

Transcript of Value at Risk By A.V. Vedpuriswar