Value at Risk : Problems

By A V Vedpuriswar

September 12, 2009

2

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

Illustration

Find 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) = - $9.6 million

VAR = E (W) – (-9.6) = 5.1 – (-9.6) = $14.7 million

If we assume a normal distribution,

Z at 95% confidence interval, 1 tailed = 1.645

VAR = (1.645) (9.2) = $ 15.2 million

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Problem% 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%) ?

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

= 3 lie below -7

So VAR = -7

Solution

6

Problem

The VAR on a portfolio using a one day horizon is USD 100 million. What is the VAR using a 10 day horizon ?

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Solution

Variance gets multiplied by 10, std deviation by √10

VAR = 100 √10 = (100) (3.16) = 316

(σN2 = σ1

2 + σ22 ….. = Nσ2)

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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.

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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

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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?

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Solution

V2 = 2σ2 (1 + ῤ)

= 2 (1)2 (1 + .1) = 2.2

V = √2.2 = 1.4832

12

Problem

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

13

Solution

10% of the time loss may exceed VAR

So no. of observations = (.10) (250)

= 25

14

Problem

Suppose 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. Calculate VAR.

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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

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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.

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Solution

σ= 1% = (.01) (5,000,000) = $ 50,000

Daily VAR = (2.33) (50,000) = $ 116,500

10 day VAR = $ 111,6500 √10 = $ 368,405

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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.?

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Solution

σ2 = w12 σ1

2 + w22 σ2

2 + 2 ῤPw1 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

20

Problem

Consider a portfolio with two foreign currencies, Canadian dollar and euro. These two currencies are uncorrelated and have a volatility against the dollar of 5% 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?

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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) = .165 = $165,000

VAR for Euro part = (1.65) (.12) (1) = .198 = $ 198,000

Undiversified VAR = $ 363,000

Thus the diversified VAR is significantly lower

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Problem

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

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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

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Problem

A firm’s current cash flows have a volatility of $60 million. A new project will have cash flows with a volatility of $20 million. The correlation coefficient between the two sets of cash flows is 0.3. Calculate the combined volatility and cash flow at risk at 95% confidence level.

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Solution

= 68.7

CFAR old = (1.65) (60) = $99 million (95% level)

CFARnew = (1.65) (68.7) = $113.4 million (95% level)

)20)(60)(3)(.2(2060 22

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Problem

A trader has an allocation equal to 8% of the firm’s capital. The returns of this unit have a beta with respect to the overall returns of 0.9. The firm’s daily VAR is $120 million. What should be the VAR allocated to the trader’s unit?

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Solution

Trader's portion of firm VAR

= (.08) (.9) (120)

= $ 8.6 million

Problem on VAR cash flow mapping

Consider 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,

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 month interest rate = 5.50%

6 month interest rate = 6.00%

.3 years = (.3) (12) = 3.6 months.

Solution Cont…

Effective interest rate for 3.6 months zero coupon bond = 5.50 + .6/3(.5) = 5.51%

Present value = = 49,201

Volatility = = .068%.

Let us allocate to a 3 month bond and 1 - of the present value to a 6 month bond.

Then we can write: 2 = 12 + 2

2 + 2 12

Here = .068 1 = .06 2 = .10 = .90

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

3.)0551.1(

000,50

)04(.3

6.06.

Solution Cont…

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) (49,201) = $37,406 in 3 month bond

$ (.2397) (49,201) = $11,795 in 6 month bond

Solution Cont…

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

Volatility = [.1 + (3.6/6)(0.1) ] = 0.16

Present value of cash flows = =$997,662

)01(.6

6.306.

8.)066.1(

000,050,1

Let be the position in the 6 month bond and (1-) in the 12 month bond. Then we can write:

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

We get RHS =.001026 + .01848 + .006096

≈ .0256

Solution Cont…

So the position is equivalent to

(.320337) (997,662) = $319,589 in 6 month bond

(.679663) (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 Total

t = .3 t = .8

3 month bond 37,406 -- 37,406

6 month bond 11,795 319,589 331,384

12 month bond -- 678,074 678,074

Solution Cont…

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

Then 2 = 12 + 2

2 + 32 + 21212 + 22323 +

21313

= [(37,406)2 (.06)2 + (331,384)2 (.10)2 + (678,074)2 (.20)2

+ (2) (37,406) (331,384) (.06) (.10) (.90)

+ (2) (331,384) (678,074) (.10) (.20) (.70)

+ (2) (37,406) (678,074) (.06) (.20) (.60)] x 10-

4

= [5,037,152 + 1,098,153,555 + 18,391,373,980 + 133,874,099+6,291,680,484+365,242,119]x10-4

Solution Cont…

≈ 2,628,536

=

= $1621.3

10 day 99% VAR

= 1621.3 x 10 x 2.33

= $11,946

536,628,2