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Simple problems on Value at Risk

Simple problems on Value at Risk

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  • 1. 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
  • 3.
    • 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
  • 4. Problem What is VAR (90%) ? % 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 12 0 3 15 1 1 16 2 2 18 4 1 19 6 1 20 7 1 21 8 1 22 9 1 23 11 1 24 12 1 26 14 2 27 18 1 28 21 1 29 23 1 30
  • 5.
    • 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 ?
  • 7. Solution
    • Variance gets multiplied by 10, std deviation by √10
    • VAR = 100 √10 = (100) (3.16) = 316
    • ( σ N 2 = σ 1 2 + σ 2 2 ….. = N σ 2 )
  • 8. 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.
  • 9. 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
  • 10. 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?
  • 11. Solution
    • V 2 = 2 σ 2 (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.
  • 15. 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
  • 16. 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.
  • 17. 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
  • 18. 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.?
  • 19. Solution
    • σ 2 = w 1 2 σ 1 2 + w 2 2 σ 2 2 + 2 ῤ Pw 1 W 2 σ 1 σ 2
    • = (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?
  • 21. 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
  • 22. Problem
    • Suppose we increase the Canadian dollar position by $10,000. What is the marginal VAR?
  • 23. 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
  • 24. 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.
  • 25. Solution
    • = 68.7
    • CFAR old = (1.65) (60) = $99 million (95% level)
    • CFAR new = (1.65) (68.7) = $113.4 million (95% level)
  • 26. 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?
  • 27. Solution
    • Trader's portion of firm VAR
    • = (.08) (.9) (120)
    • = $ 8.6 million
  • 28. Problem on VAR cash flow mapping
    • Consider a long position in a $1 million Treasury bond.
    • Maturity : 0.8 years
    • Coupon : 10% payable semiannually
    • Explain how mapping can be done while calculating VaR,
    Annualized yield & volatility 3 Month 6 Month 1 Year Annualised yield 5.50 6.00 7.00 Volatility 0.06 0.10 0.20 Correlations between daily returns 3 Month 6 Month 1 Year 3 month 1.0 0.9 0.6 6 month 0.9 1.0 0.7 1 year 0.6 0.7 1.0
  • 29. 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.
  • 30. 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 =  1 2 +  2 2 + 2  1  2
    • Here  = .068  1 = .06  2 = .10  = .90
    • or .068 2 =  2 (.06) 2 + (1-  ) 2 (.10) 2 + 2 (.9)(  ) (1-  )(.06)(.10)
  • 31. Solution Cont…
    • or .068 2 =  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
  • 32. 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
  • 33.
    • Let  be the position in the 6 month bond and (1-  ) in the 12 month bond. Then we can write:
    •  2 =  2  1 2 + (1-  ) 2  2 2 + 2  (1-  )  1  2
    • 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
  • 34. 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
  • 35. 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 =  1 2 +  2 2 +  3 2 + 2  12  1  2 + 2  23  2  3 + 2  13  1  3
    • = [(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
  • 36. Solution Cont…
    • ≈ 2,628,536
    •  =
    • = $1621.3
    • 10 day 99% VAR
    • = 1621.3 x  10 x 2.33
    • = $11,946
  • 37. Problem
    • Some time back, 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 volatilities are:
    • 6 month zero coupon sterling bond : .06%
    • 6 month zero coupon dollar bond : .05%
    • The correlation between the returns of the two bonds is 0.8. The current exchange rate is 1.53. What is the standard deviation of the daily change in value of the forward contract? What is the 10 day 99% VAR? Assume the 6 month interest rates in both sterlings and dollars is 5% per annum with continuous compounding.
  • 38. Solution
    • The contract is to sell $ and buy £
    •  Short position in dollar bond
    • Long position in sterling bond.
    • Value of the sterling bond = 1.53 e -.05x.5 = $1.492 million
    • Value of the dollar bond = 1.5 e -.05x.5 = $1.463 million
    • Variance of daily change in value of contract
    • = 1.492 2 x .0006 2 + 1.463 2 x .0005 2 – (2) (.8) (1.492) (.0006) (1.463) (.0005)
  • 39. Solution
    • = 8.01383 x 10 -7 + 5.35092 x 10 -7 – 10.47742 x 10 -6
    • = 2.8873 x 10 -7 = .28873 x 10 -6 ≈ .000000288
    • Std devn = = .0005373
    • 10 day 99% VAR = 2.33  10 (.0005373)
    • = $ .00396 million