Transcript Market Risk Modelling

Value at Risk
By 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.
1
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)
2
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.
3
How Banks disclose VaR
4
VaR at UBS
Ref : Company Annual report
5
VaR at UBS
Ref : Company Annual report
6
VaR at UBS
Ref : Company Annual report
7
VaR at UBS
Ref : Company Annual report
8
VaR at UBS
Ref : Company Annual report
9
VaR at Credit Suisse
Ref : Company Annual report
10
VaR at Credit Suisse
Ref : Company Annual report
11
VaR at Goldman Sachs
Ref : Company Annual report
12
Problem
 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.
Solution
 The point such that the no. of observations to the left = (254) (.05) = 12.7
 (12.7 – 11) /( 15 – 11 )
=
 So required point = - (10 - .4 x 1)
 VAR = E (W) – (-9.6)
≈ .4
1.7 / 4
=
- $9.6 million
= 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
13
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
 (σN2 = σ12 + σ22 ….. = Nσ2)
Note; This approach is valid only when the daily variances are
independent.
14
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.
Solution
Weekly VAR
＝ (12,500) (√5)
＝ 27,951
Monthly VAR
＝ ( 12,500) (√20)
＝ 55,902
Semi annual VAR
＝ (12,500) (√125)
＝ 139,754
Annual VAR
＝ (12,500) (√250)
＝ 197,642
15
Variance Covariance Method
16
Problem
A 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
17
Problem
Consider 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
18
Problem
Consider 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.
Ref : John C Hull, Options, Futures and Other Derivatives
19
Problem
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.
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
20
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
Ref : John C Hull, Options, Futures and Other Derivatives
21
Problem
Now 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 = w12 σ12 + w22 σ22 + 2 ῤw1 w2 σ1 σ2

= (200,000)2 + (50,000)2 + (2) (.3) (200,000) (50,000)

σ

Daily VAR = (2.33) (220,277)
=
513,129

10 day VAR = (513,129) √10
=
$1,622,657
=
220,277
 Effect of diversification = (1,473,621 + 368,406) – (1,622,657)
= 219,369
22
Ref : John C Hull, Options, Futures and Other Derivatives
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.
Solution
Worst YTM = actual YTM + 1.65 x Volatility
= 5 + 1.65 x 1 = 6.65%
If YTM is 6.65%, bond price will be 93.1708
So the VAR is 100-93.17 = $ 6.83
23
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.
Solution
VAR = $10,000,000* 0.02*3.6* √10/252 * 2.33 = $334,186
24
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.
25
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
=

Undiversified VAR
= $ (1.65) (.12) (1) million
$ 198,000
=
$ 363,000
 Thus the diversified VAR is significantly lower.
26
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
=
=
$ 258,225
 Marginal VAR
(1.65) (156,500)
= 258,225 – 257,738
= $ 487
27
Problem
An 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.
Ref : John C Hull, Options, Futures and Other Derivatives
28
Problem
Some 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 zerocoupon 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
0.000000288
1.463 x 0.0005
=
 The standard deviation is therefore $0.000537 million.
 TheRef
10-day
99%
VaR Options,
is $0.000537
x √10
2.33 =Derivatives
$0.00396 million
: John
C Hull,
Futures
andx Other
29
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
30
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

31
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
Problem
Consider a portfolio with standard deviation of daily returns of 0.1
and autocorrelation of 0.3. Calculate the 5 day volatility.
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
33
Monte Carlo Simulation
34
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.
35
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
36
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.
37
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.
38
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.
39
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 tradeoff 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.
40
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.
41
Formula used typically in Monte Carlo for stock price
modelling
42
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.
43
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.
44
Historical Simulation
45
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.
46
Steps
Collect data
Generate scenarios
Calculate portfolio returns
Arrange in order.
47
What is VAR (90%) ?
% Returns
- 16
- 14
- 10
-7
-5
-4
-3
-1
0
1
2
4
6
7
8
9
11
12
14
18
21
23
Frequency
1
1
1
2
1
3
1
2
3
1
2
1
1
1
1
1
1
1
2
1
1
1
Cumulative Frequency
1
2
3
5
6
9
10
12
15
16
18
19
20
21
22
23
24
26
27
28
29
30
10% of the observations, i.e, (.10) (30)
= 3 lie below -7
48
So VAR = -7
Advantages and Disadvantages of Historical simulation
Advantages
Disadvantages
Simple
Reliance on the past
No normality
assumption
Length of estimation
period
Non parametric
Weighting of data
49
Comparison of different VAR modeling techniques
50
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
51
Model Validation
52
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.
53
Basel Committee Standards(2)
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.
54
Basel Committee Standards(3)
The Basel Committee prescribes an increase in capital
requirements if, based on a sample of 250 observations (a oneyear 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
55
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
56
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
Ref : John C Hull, Options, Futures and Other Derivatives
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% < 5%

So the model should be rejected.
Ref : John C Hull, Options, Futures and Other Derivatives
2.64%
Problem
A $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
59
Stress Testing
60
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.
61
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.
62
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.
63
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.
64
Illustration
Consider 2 variables that have a uniform distribution. Using
Gaussian copula, and assuming a copula correlation of 0.3,
define a correlation structure.
0.25
0.50
0.75
0.25
0.50
0.75
65
0.25
0.50
0.75
0.25
-.675, -.675
-.675, 0
-.675, +.675
0.50
0,
0.75
+.675, -.675
+.675, 0
+.675, .+ 675
0.25
0.50
0.75
0.25
.095
.1633
.2157
0.50
.1633
.2985
.4134
0.75
.2157
.4134
.5953
-.675
0, 0
0,
+ .675
66
Illustration
x
percentile
Z
percen
tile
Z
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
Correlation coefficient = 0.5
67
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.1
.006
.017
.028
.037
.044
.048
.049
.050
.050
0.2
.013
.043
.081
.120
.156
.181
.193
.198
.200
0.3
.017
.061
.124
.197
.273
.331
.364
.381
.387
0.4
.019
.071
.149
.248
.358
.449
.505
.535
.548
0.5
.019
.076
.164
.281
.417
.537
.616
.663
.683
0.6
.020
.078
.173
.301
.456
.600
.701
.763
.793
0.7
.020
.079
.177
.312
.481
.642
.760
.837
.877
0.8
.020
.080
.179
.318
.494
.667
.798
.887
.936
0.9
.020
.080
.180
.320
.499
.678
.816
.913
.970
68
VAR – Cash flow mapping
Problem
Consider a long position in a $1 million Treasury bond.
Maturity
:
0.8 years
Coupon
:
10% payable semiannually
Annualised yield
Volatility
Annualized yield & volatility
3 Month
6 Month
1 Year
5.50
6.00
7.00
0.06
0.10
0.20
3 month
6 month
1 year
Correlations between daily returns
3 Month
6 Month
1 Year
1.0
0.9
0.6
0.9
1.0
0.7
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 years
Cash flow of $1,050,000 in .8 years
So the position can be considered a combination of two
zero coupon bonds, maturity 0.3, 0.8 years.
Let us write the position as equivalent to a combination of
standard 3 month, 6 month and 1 year bonds.
3 years = (.3) (12)
=
3.6 months.
3 month interest rate
=
5.50%
6 month interest rate
=
6.00%
Effective interest rate for 3.6 months zero coupon bond
= 5.50 + (.6/3)(.5) = 5.6% = .056
Present value =
Volatility
=
50,000
(1.056).3
.6
.06  (. 04 )
3
=
49,189
=
.068%.
Let us allocate  to a 3 month bond and 1 -  to a 6 month bond.
Then we can write:
2
Here  = .068
1
=
12 + 22 + 2 12
= .06
2 = .10
 = .90
or .0682 = 2 (.06)2+ (1-)2(.10)2 + 2 (.9)() (1-)(.06)(.10)
or .0682 = 2 (.06)2 + (1-)2 (.10)2 + 2(.9) ()(1-)(.06)(.10)
Putting  = .7603,
LHS = .00462, RHS = .00208 + .00057 + .001968 = .00462
So we can write the position as equivalent to
$ (.7603) (49,189)
=
$37,397 in 3 month bond
$ (.2397) (49,189)
=
$11,791 in 6 month bond
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: .06 
Present value of cash flows =
1,050,000
(1.066).8
Volatility = [.1 + (3.6/6)(0.1) ]
3.6
(. 01)
6
=.066
= $997,662
= 0.16
If  is the position in the 6 month bond and (1-) in the 12
month bond, 2
= 2 12 + (1-)2 22 + 2  (1-) 12
Or (.16)2
= 2 (.1)2 + (1-)2 (.2)2 + 2 (.7) () (1-) (.1)(.2)
LHS =.0256
.006096 ≈.0256
Put =.320337; RHS =.001026 + .01848 +
So the position is equivalent to
(.3203) (997,662)
=
$319,589 in 6 month bond
(.6797) (997,662)
=
$678,074 in 12 month bond
 We can now write the portfolio in terms of 3 month, 6 month, 12 month
zero coupon bonds.

$50,000
$1,050,000

t = .3
t = .8
Total
 3 month bond 37,397
--
 6 month bond 11,791
319,589
331,380
 12 month bond
678,074
678,074
--
37, 397
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 + 22 + 32 + 21212 + 22323 + 21313

= [(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