RMBI4210 — Statistical Methods for Risk Management Solution to Homework Problems
Course instructor: Prof. Y.K. Kwok
1. Recall that y is the yield per period so that D=1∑n ci i
B i=1 (1 + y)i m
(cash flow of dollar amount ci occurs at time mi years), where
∑n i=1
The annualized yield λ = my. Note that c is the coupon rate per period. Say, m = 4 and annualized coupon rate is 8%, then c = 2%. On the coupon payment dates, ci = cBT , i = 1,2,…,n−1; while on the maturity date cn = (1+c)BT. Also, n = mT, where T is the number of years until maturity. Taking the derivative of B with respect to y, we have
B =
ci(1 + y)−i.
dB 1∑n mDB
ici(1+y)−i =−1+y
D = −1 + y 1 dB = −(1 + y) 1 dB, where λ = my.
dy =−1+y
m Bdy Bdλ
so that
The formula in the lecture note assumes annual coupon payments, where
i=1
D = − 1 + λ dB , λ = my. B dλ
The multiplication factor with m coupon payments per year becomes 1 + y instead of
1 + λ. We recall
B=Bc[1− 1 ]+ BT , T y (1 + y)n (1 + y)n
where c is the coupon rate per period and y is the yield per period. Therefore, dlnB = 1dB=−1+cn(1+y)−1+1+y(1+y)−1(−n).
dy BT B dy y Combining these relations, we have
D = −1 + y 1 dB = 1 + y − m B dy my
c[(1+y)n −1]+y
1 + y + n(c − y) .
mc[(1+y)n −1]+my
For fixed value of m, we take T → ∞, which is equivalent to take n → ∞. This gives
lim D = 1 + 1 − lim 1 + y + n(c − y) . T→∞ m λ n→∞ mc[(1+y)n −1]+my
To evaluate the limit as n → ∞, we observe that the numerator is linear in n while the denominator contains the power function (1 + y)n, y > 0. We then have
lim D = 1 + 1. T→∞ mλ
1
2. The term 1 represents the “deterministic” return received by an investor holding a B(t,T)
zero-coupon bond to maturity. The right-hand side is the expected return from time t to T generated by rolling over a $1 investment in one-period maturity bonds, each of which has a yield equal to the future spot rate rt, assuming that the investor cannot quit the annual rolling over strategy in the period [t, T ]. The relationship represents an equilibrium condition, in which the expected returns for equal holding periods are themselves equal. On one hand, one may argue that in an environment of economic equilibrium, the returns on zero-coupon bonds of similar maturity cannot be significantly different since investors would not hold the bonds with the lower return. On the other hand, the subjective expectation of an individual investor determines the expected return for the rolling over strategy. She would choose among the two strategies based on the one with higher expected return. For example, under the current low interest rate environment, suppose the investor expects future hikes in interest rates, she would prefer the rolling over strategy to the long-term bond investment strategy.
3. We write
c 1 ( c) Bt=i+(1+i)T BT−i .
With the passage of one calendar year, we have
c 1 ( c)
We then have
Bt+1=i+(1+i)T−1 BT−i . ( c) i
Bt+1−Bt= BT−i (1+i)T =iBt−c. Suppose the time interval becomes ∆t instead of one year, then we obtain
∆B B −B ( c) =t+∆t t=i− ∆t.
BBt Bt In the continuous time limit ∆t → 0, we deduce that
1 dB =i(t)− c(t). B(t) dt B(t)
Note that i(t) and c(t) in the differential equation are visualized as cash flow rates so that i(t)dt and c(t)dt are dollar amounts collected over (t, t + dt). Given that the governing equation for B(t) is
dB(t) = i(t)B(t) − c(t), t < T, dt
with B(T ) = BT . The closed form solution is seen to be ∫[∫T∫]
∫T i(s) ds
B(t)=e− tT i(s) ds BT +
For the coupon amount c(u)du received within the differential time interval (u, u+du), the
c(u)euT i(s) dsdu , t
1 1 d3B(di)3 < 0. 6B di3
Therefore, the actual bond value is below the quadratic approximation to the right of the tangency point. On the other hand, since convexity of the bond value is greater than zero, so the actual bond value lies above the linear approximation.
Similarly, when di < 0, both d2B (di)2 and d3B (di)3 are positive. Therefore, the actual di2 di3
bond value lies above the quadratic approximation and linear approximation curves to the left of the tangency point.
di3 = −
t=1
6. (a) Bond A: maturity is 15 years and coupon rate is 10%
Time of payment t (year)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
t(t+1) Cash flows in Discount Cash flows nominal rate present
value value
2 100 0.8929 89.2857 6 100 0.7972 79.7194 12 100 0.7118 71.1780 20 100 0.6355 63.5518 30 100 0.5674 56.7427 42 100 0.5066 50.6631 56 100 0.4523 45.2349 72 100 0.4039 40.3883 90 100 0.3606 36.0610 110 100 0.3220 32.1973 132 100 0.2875 28.7476 156 100 0.2567 25.6675 182 100 0.2292 22.9174 210 100 0.2046 20.4620 240 1100 0.1827 200.9659
863.7827
in
Share of cash flows in present value in bond price
0.1034 0.0923 0.0824 0.0736 0.0657 0.0587 0.0524 0.0468 0.0417 0.0373 0.0333 0.0297 0.0265 0.0237 0.2327
Weighted time of payment
0.1034 0.1846 0.2472 0.2943 0.3285 0.3519 0.3666 0.3741 0.3757 0.3727 0.3661 0.3566 0.3449 0.3316 3.4899 7.8880
t(t+1) times share of discounted cash flows
0.2067 0.5537 0.9888 1.4715 1.9707 2.4634 2.9326 3.3665 3.7573 4.1002 4.3931 4.6356 4.8287 4.9746 55.8379 76.9145
Calculation results for Bond A: duration is 7.8880 and convexity is 76.9145. Bond B: maturity is 11 years and coupon rate is 5%
3
Time of payment t (year)
1 2 3 4 5 6 7 8 9 10 11
t(t+1) Cash flows in Discount Cash flows in Share of cash
Weighted time of payment
0.0764 0.1364 0.1827 0.2175 0.2428 0.2601 0.2709 0.2765 0.2777 0.2755 5.6820 7.8985
t(t+1) times share of discounted cash flows
0.1528 0.4093 0.7308 1.0875 1.4565 1.8207 2.1675 2.4882 2.7770 3.0304 68.1842 67.2073
nominal rate present value value
2 50 0.8929 44.6429 6 50 0.7972 39.8597 12 50 0.7118 35.5890 20 50 0.6355 31.7759 30 50 0.5674 28.3713 42 50 0.5066 25.3316 56 50 0.4523 22.6175 72 50 0.4039 20.1942 90 50 0.3606 18.0305 110 50 0.3220 16.0987 132 1050 0.2875 301.8499
584.3611
flows in present value
in bond price
0.0764 0.0682 0.0609 0.0544 0.0486 0.0433 0.0387 0.0346 0.0309 0.0275 0.5165
Calculation results for Bond B: duration is 7.8985 and convexity is 67.2073.
(b) Bond A has the rate of return of 12.06% at horizon H = D = 7.8880 if the in- terest rate jumps to 10% or 14%. Bond B has the rate of return of 12.03% at horizon H = D = 7.8985 if the interest rate jumps to 10% or 14%. These sample calculations show that the rates of return almost stay at the same level of 12% at horizon that equals duration.
Rates of return of Bond A when YTM changes
Horizon
Scenario (YTM)
Current 10% 14%
1
2
3
4
5
6
7 7.888 8
9 10
12.00% 27.35% 12.00% 18.36% 12.00% 15.50% 12.00% 14.10% 12.00% 13.27% 12.00% 12.72% 12.00% 12.33% 12.00% 12.06% 12.00% 12.03% 12.00% 11.80% 12.00% 11.62%
-0.45% 6.53% 8.97% 10.20% 10.95% 11.45% 11.81% 12.06% 12.09% 12.30% 12.47%
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Rates of return of Bond B when YTM changes
Horizon
Scenario (YTM)
Current 10% 14%
1
2
3
4
5
6
7 7.8985 8
9 10
12.00% 27.11% 12.00% 18.25% 12.00% 15.43% 12.00% 14.05% 12.00% 13.23% 12.00% 12.68% 12.00% 12.30% 12.00% 12.03% 12.00% 12.01% 12.00% 11.78% 12.00% 11.60%
-0.65% 6.42% 8.89% 10.15% 10.91% 11.42% 11.78% 12.03% 12.06% 12.27% 12.44%
(c) We would choose bond A. These two bonds have pretty much the same duration, but bond A has a higher value of convexity. As a result, bond A has a higher rate of return compared to that of bond B, no matter interest rates increase or decrease, and for any choice of horizon. As bond investors, they should embrace convexity.
7. Let t be the tax rate, xi be the number of units of bond i bought, ci be the coupon of bondi,pi bethepriceofbondi,i=1,2. Theparvalueofeachbondis100. Letp0 be the price of the zero-coupon bond.
To create a zero coupon bond, we require that the after-tax coupons match. This gives 100[x1(1 − t)c1 + x2(1 − t)c2] = 0,
which reduces to an equation independent of t:
x1c1 + x2c2 = 0.
Next, we require that the after-tax final cash flows match (mutually consistent). This gives another equation that relates the bond prices p1, p2 and p0.
x1[100 − (100 − p1)t] + x2[100 − (100 − p2)t] = [100 − (100 − p0)t]. The price of the zero-coupon bond will be
p0 = x1p1 + x2p2.
The last relation for matching the par payments at maturity is given by
x1 + x2 = 1.
All these relations are independent of the tax rate t. Combining x1 +x2 = 1, c1x1 +c2x2 =
0, and p0 = x1p1 + x2p2, we obtain
p0 = c2p1 −c1p2 = 0.07×92.21−0.1×75.84 = 37.64.
c2 −c1
0.07−0.1
5
8. Let P denote the principal left in the pool and r denote the annualized rate of return.
Year 1
Year 2
Year 3
Year 4
Year 5
Year 6
P = (20)(1.085) + −10(20)(−0.005) − 12.5(0.07) = 21.75 1.085
r = 21.75 − 20 = 23.33%; 20 − 12.5
P = (21.75)(1.08) + −10(21.75)(−0.005) − 12.5(0.065) = 23.68 1.08
r = 23.69 − 21.75 = 20.86%; 21.75 − 12.5
P = (23.68)(1.075) + −10(23.68)(−0.005) − 12.5(0.06) = 25.81 1.075
r = 25.81 − 23.68 = 19.02%; 23.68 − 12.5
P = (25.81)(1.07) + −10(25.81)(−0.005) − 12.5(0.055) = 28.14 1.07
r = 28.14 − 25.81 = 17.51%; 25.81 − 12.5
P = (28.14)(1.065) − 10(28.14)(0.02) − 12.5(0.05) = 24.06 1.065
r = 24.06 − 28.14 = −26.09%; 28.14 − 12.5
P = (24.06)(1.085) − 10(24.06)(0.02) − 12.5(0.07) = 20.80 1.085
r = 20.80 − 24.06 = −28.20%. 24.06 − 12.5
= =
Net gain after 6 years
principal left in the pool at the end of Year 6 − initial investment − borrowed amount 20.80 − 12.5 − 7.5 = 0.80.
If invested in bank of amount 7.5 billion (without borrowing to gain leverage), then the net gain = 7.5(1.06)(1.055)(1.05)(1.045)(1.04)(1.06) − 7.5 = 2.64.
Note the significant mark-to-market losses in the bond portfolio when the interest rates increased by 2% in two consecutive years (Years 5 and 6).
9. Recall the formula:
−λ(T )T 1 − e−[y(T )−y∗(T )]T = Q(T ) = 1 − R ,
1 − e
where y(T ) − y∗(T ) is the credit spread over [0, T ]. Using the approximation: 1 − e−x ≈ x
when x is small, we deduce that
λ(T)≈ y(T)−y∗(T).
1−R
The average default intensity over 3 years is approximately given by
λ(3) ≈ 0.0050 = 0.0071 = 0.71% 1 − 0.3
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per year. Similarly, the average default intensity over 5 years is approximately given by λ(5) ≈ 0.0060 = 0.0086 = 0.86%
1 − 0.3
per year. Let λ(3;5) denote the default intensity per year in years 4 and 5. Using the
relation: λ(3)×3+λ(3;5)×(5−3)=λ(5)×5,weobtain
λ(3;5) ≈ 0.86×5−0.71×3 = 0.0107 = 1.07%.
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10. This question illustrates the use of two risky bonds with different maturities to find the term structure of unconditional default probabilities over [0, 3] and [3, 5]. The table for the first bond is
The market price of the bond is 98.35 and the risk-free value is 101.23. It follows that Q1 is given by
178.31Q1 = 101.23 − 98.35 The table for the second bond is
The market price of the bond is 96.24 and the risk-free value is 101.97. It follows that 180.56Q1 + 108.53Q2 = 101.97 − 96.24,
from which we get Q2 = 0.0260. The bond prices therefore imply an unconditional prob- ability of default of 1.61% per year for the first three years and 2.60% for the next two years.
11. We assume:
Recovery rate = 0.2
Interest rate = 6%
Notional = 1
Loss given default = (1− recovery rate) × notional = 0.8
so that Q1 = 0.0161.
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Examples of some sample calculations
• Accrual payment at Year 1.5 = 0.015 × 12 × 12 × 0.9139s = 0.00343s.
• P V of expected payment at Year 1.5 = 0.015 × 12 × 0.8 × 0.9139 = 0.005484.
The spread of the CDS with recovery rate 0.2 = 0.04799/3.3711 = 1.4237%. The spread of the binary CDS = 0.05999/3.3711 = 1.7796% = 1.4237% × 1.0 .
0.8
12. Note that holding the risky corporate bond plus CDS generates the same cash flow as holding the riskfree bond and selling the bond upon default of Company X. The current data show that the credit spread of the risky bond is 250bps = 9.5% − 7% while the annual premium of the CDS is only 150bps. Hence, it is advantageous to hold the risky bond and hedge the default at the cost of 150bps.
The breakeven CDS premium is seen to be 250bps. If the CDS premium increases to 300bps, we simply do the opposite: short selling the risky bond plus CDS and long the riskfree bond.
The arbitrage may not be perfect since there are at least two potential risks embedded in the transactions.
(a) There exists potential counterparty risk of the CDS Protection Seller.
(b) There exists uncertainty in the sale price of the riskfree bond when it is liquidated
upon the default of X.
13. Barings Fall
The downfall of Barings PLC was due to a single derivative trader, 28-year-old Nicholas Leeson. The loss was caused by a large exposure to the Japanese stock market, which was achieved through the futures market. Barings’ notional positions in stock index futures on the Nikkei 225 on the Singapore and Osaka exchanges added up to a staggering $ 7 billion. As the Japanese market fell more than 15 percent in the first two months of 1995, Barings Futures suffered huge losses. As losses mounted, Leeson increased the size of the position, in a stubborn belief that he was right. There were allegations that senior bank executives were aware of the risks involved and has approved cash transfers of $ 1 billion to help Leeson make margin calls. In a case where a trader is taking unauthorized positions, the real question is the strength of an investment house’s internal controls and the external controls done by exchanges and regulators.
Daiwa’s Losses
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On September 26, 1995, Daiwa Bank announced that a 44-year-old trader in New York, Toshihide Igushi, has accumulated losses estimated at $ 1.1 billion (a similar magnitude to those that befell Barings). Igushi had concealed more than 30,000 trades over 11 years, starting in 1984, in US Treasury bonds. As the losses grew, the trader exceeded his position limits to make up for the losses. He eventually started selling, in the name of Daiwa, securities deposited by clients at the New York branch. The problem arose since Igushi had control of both the front and back offices. Japanese banks rely on a group spirit that acts as an internal safety mechanism. In overseas operations, such an approach can be fatal. This loss highlighted the poor risk-management policies of Japanese banks. The Federal Reserve Board in US had inspected Daiwa’s offices in November 1992 and November 1993. Daiwa failed to implement major changes and even reported that it deliberately hid records and temporarily removed bond traders in order to pass the 1992 inspection.
14. The delta indicates that when the value of the euro exchange rate increases by $0.01, the value of the bank’s position increases by 0.01 × 30, 000 = $300. The gamma indicates that when the euro exchange rate increases by $0.01, the delta of the portfolio decreases by 0.01 × 80, 000 = 800. For delta neutrality, 30, 000 euros should be shorted. When the exchange rate moves up to 0.93, we expect the delta of the portfolio to decrease by (0.93 − 0.90) × 80, 000 = 2, 400 so that it becomes 27, 600. To maintain delta neutrality, it is therefore necessary for the bank to unwind its short position by 2, 400 euros so that a net 27, 600 = 30, 000 − 2, 400 euros have been shorted. When a portfolio is delta neutral and has a negative gamma, a loss is experienced when there is a large movement in the underlying exchange rate since euro appreciates quite substantially from 0.90 to 0.93. We can conclude that the bank is likely to have lost money in its short position in euro.
15. The dynamic hedge should replicate a long position in the call. Due to the positive delta, this implies a long position of ∆ × 100 = 50 shares. If the delta falls, the position needs to be adjusted by selling (0.5 − 0.44) × 100 = 6 shares to maintain delta-neutral in the portfolio.
16. The answer is a. Since gamma is negative, we need to buy a call to increase the portfolio gamma back to zero. The number is 600/1.5 = 400 calls. This will increase the delta from zero to 400 × 0.75 = 300. Hence, we must sell 300 shares to bring back the delta to zero. Note that positions in shares have zero gamma so that the sales of 300 shares has no impact on gamma, so we do not need to be changed with change in gamma under change in stock position.
17. The answer is d. For a short-maturity at-the-money (ATM) option, the delta changes most drastically since it fluctuates between 0 and 1 within a short span of time. There- fore, the gamma has the highest value. Sensitivity to volatility, as measured by vega, is proportional to square root of maturity. Long gamma means that the portfolio is long options with high gamma, typically short-maturity ATM options. Short vega means that the portfolio is short options with high vega, typically long-maturity ATM options.
18. Portfolio insurance is a form of dynamic hedging that replicates a long position in a put option. Recall that a put option on stock index provides downside protection of a portfolio of stocks, where a well diversified portfolio of stocks can be proxied by the stock index. When the stock prices fall, the put delta drops in value and comes closer to −1; say, the put delta drops from −0.6 to −0.8. If the value of the portfolio decreases, one should sell the index futures in the amount that represents the change in the put delta.
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19. The answer is b. The optimal hedge ratio is ρσs/σf = 0.77×2.6/3.2 = 0.626. Taking into account the size of the position, the number of contracts to buy is 0.626 × 1, 000/25 = 25.04.
20. (a) $1 million.
(b) The expected shortfall is 0.9 × 10 + 0.1 × 1, or $9.1 million.
(c) There is a probability of 0.0092 = 0.000081 of a loss of $20 million, a probability of 2 × 0.009 × 0.991 = 0.017838 of a loss of $11 million, and a probability of 0.9912 = 0.982081 of loss of $2 million. The VaR when the confidence level is 99% is therefore $11 million.
(d) The expected shortfall is (0.000081 × 20 + 0.009919 × 11)/0.01 = $11.07 million. (e) Since 1+1 < 11, the subadditivity condition is not satisfied for VaR. Since 9.1+9.1 >
11.07, it is satisfied for expected shortfall.
(f) VaR is the loss that is not expected to be exceeded with a certain confidence level. Expected shortfall is the expected loss conditional on the loss being worse than the VaR level. Expected shortfall has the advantage that it always satisfies the subadditivity (diversification is good) condition.
21. (a)
The market-risk capital required is
10-day 99% VaR × multiplicative factor = 10×2.33×5×4 = 147.36 (million dollars).
(b) Assume there are 252 working days in one year. Then economic capital using a one-year time horizon and a 99.9% confidence limit is
√
252 × (3.09 × 5)2 + (2522 − 252)(3.09 × 5)2 × 0.05 = 902.81 (million dollars).
22. The worst move in the index value, which is αSσ√T = 1.645 × 3, 000(24%/√260)√10 =
232.3. Multiplying by the delta of 0.6 gives 139.
23. The correlation information stated in the question is summarized in the following table:
Correlations between losses:
MR, CR, and OR refer to market risk, credit risk, and operational risk; 1 and 2 refer to the business units.
MR-1 CR-1 OR-1 MR-2 CR-2 OR-2
√
MR-1 1.0 CR-1 0.3 OR-1 0.0 MR-2 0.2 CR-2 0.0 OR-2 0.0
0.3 0.0 1.0 0.0 0.0 1.0 0.0 0.0 0.7 0.0 0.0 0.0
0.2 0.0 0.0 0.0 0.7 0.0 0.0 0.0 0.0 1.0 0.3 0.0 0.3 1.0 0.0 0.0 0.0 1.0
The total enterprise-wide economic capital is the square root of
102 +502 +302 +302 +502 +102 +2×0.3×10×30+2×0.2×10×50
+ 2 × 0.7 × 30 × 30 + 2 × 0.3 × 50 × 30 which is 97.67.
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24.
(a) Recall
m∫1 k m−k a−1 b−1 Ck 0 p ̃(1−p ̃) p ̃ (1−p ̃) dp ̃
25. 26.
Under the assumption of uniform default probability p, face value F and correlation coefficient ρ, the diversity score D(ρ) is reduced to
Total economic capital for Business Unit 1 is
√
102 +302 +502 +2×0.3×10×30 = 60.66.
Total economic capital for Business Unit 2 is
Hence, we have
√
502 +302 +102 +2×0.3×50×30 = 66.33.
P[M = k] = ∫ 1 xa−1(1 − x)b−1dx 0
.
using the definition of the beta distribution, we obtain
P[M =k]=cmβ(a+k,b+m−k).
k β(a,b) (b) The default correlation coefficient is given by
ρ(Xi , Xj ) = Var(p) = E[p](1 − E[p])
ab (a+b)2 (a+b+1)
.
From a a+b
= p, we deduce that
a = b = a + b. p 1−p
ρ(Xi,Xj) = 1 = 1 1+a 1+b
.
For fixed value of p, the default-event correlation decreases as a increases.
a a+b
(1 −
a ) a+b
D(ρ) = n2F2p(1 − p) = n(n − 1)F2ρp(1 − p) + nF2p(1 − p)
n . 1 + ρ(n − 1)
p 1−p
Note that D(ρ) is an increasing function in m since dD(ρ) = 1−ρ
>0.
= 1. ρ
On the other hand, we have
dn [1 + ρ(n − 1)]2
lim D(ρ) = 1
n→∞ n1 +ρ(1−n1)
Hence, we deduce that 1/ρ is an upper bond of D(ρ).
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27.
Given that Li|Λi = λi ∼ Pois(λi) and Li|Λ = λ, i = 1,2,…,m, are independent, by virtue of the property of Poisson distribution where both its mean and variance are equal to the Poisson parameter,
we have
E[Li] = E[E[Li|Λ]] = E[Λi]
var(Li) = var(E[Li|Λ]) + E[var(Li|Λ)]
= var(Λi) + E[Λi].
Also, by the tower rule in conditional probability and independence of Li|Λ and Lj|Λ,
we have
E[LiLj] = E[E[LiLj|Λ]]
= E[E[Li|Λ]E[Lj|Λ]]
= E[ΛiΛj]. Lastly, the correlation coefficient is given by
cov(Li,Lj) ρ(Li,Lj) = √ √
var(Li) var(Lj) E[ΛiΛj ] − E[Λi]E[Λj ]
=√√ var(Li) var(Lj)
28.
cov(Λi,Λj) =√√.
var(Λi) + E[Λi] var(Λj) + E[Λj]
(a) The m obligors all survive up to their respective times t1,t2,…,tm, provided that idiosyncratic shock i does not occur up to time ti, i = 1, 2, . . . , m, and the macro- economic shock does not occur up to max(t1, t2, . . . , tm). Since these shocks are independent, so the joint survival function is given by
S(t1, t2, . . . , tm) =
[∏m ]
exp(−λiti) exp(−λmax(t1, t2, . . . , tm))
i=1
= exp(−(λ1t1 + λ2t2 + ··· + λmtm + λmax(t1,t2,…,tm))).
(b) The two-dimensional marginal copula is given by
where Note that
so that where
Cτ(ui,uj) = S(0,…,0,ti,0,…,tj,0,…,0) =exp(−(λiti +λjtj +λmax(ti,tj)))
=exp(−(λi +λ)ti)exp(−(λj +λ)tj)min(eλti,eλtj) = Si (ti )Sj (tj )min(eλti , eλtj ),
ui = Si(ti) = e−(λi+λ)ti and uj = Sj(tj) = e−(λj+λ)tj .
[ ]−λ u−θi = e−(λi+λ)ti λi+λ
C τ (ui , uj ) = ui uj min(u−θi , u−θj ) = min(uj u1−θi , ui u1−θj ), ijij
θi=λ andθj=λ. λi + λ λj + λ
i
−θ [ ]−λ
= eλti, = eλtj ,
u j = e−(λj+λ)tj λj+λ j
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