Course 4 May 2001 Solutions

COURSE 4 

MAY 2001 

MULTIPLE-CHOICE ANSWER KEY 

1 D 26 A 

2 B 27 E 

3 C 28 E 

4 E 29 D 

5 C 30 C 

6 C 31 A 

7 D 32 A 

8 E 33 A 

9 D 34 D 

10 A 35 E 

11 D 36 D 

12 E 37 B 

13 B 38 B 

14 C 39 E 

15 A 40 C 

16 C 

17 E 

18 A 

19 B 

20 E 

21 C 

22 E 

23 B 

24 B 

25 C 

Course 4: May 2001 - 41 - STOP

Course 4 Exam Solutions 

May 2001 

Item Number: 1 

Key: D 

Solution 

Because the autocorrelation function is zero starting with lag 2, this must be an MA(1) model. 

Then, 

−θ 

− .4 = ρ = , −.4 − .4 θ =−θ ,.4 θ − θ + .4 = 0. 

1 

2 2 

1 2 

1 1 1 1 

1+ 

θ1 

This quadratic equation has two roots, 0.5 and 2. Because the coefficient’s absolute value must 

be less than 1, only 0.5 is acceptable. 

C42001A


Key: B 

Solution 

e λ 1 

The posterior distribution is π( λ|2) 

2! 3 

with parameters 3 and 3/4. The variance is 3(3/4) 2 = 27/16. 

−λ 

2 

−λ/3 2 −4 λ/3 

∝ e ∝ λ e which is a gamma distribution 



Key: C 

Solution 

( 1− 3) + ( 2− 3) + ( 3− 3) + ( 4− 3) + ( 5−3) 

2 2 2 2 2 

2 

2 

= = = 

µ σ 

( 1− 3) + ( 2− 3) + ( 3− 3) + ( 4− 3) + ( 5−3) 

4 4 4 4 4 

µ 

4 

= = 6.8 

5 

µ 

4 

6.8 

γ 

2 

= = = 

4 1.7 

σ 4 

5 

2 



Key: E 

Solution 

The calculations for each possibility are: 

Censored at 1 Censored at 2 Censored at 3 Censored at 4 Censored at 5 

Time d Y S $ d Y S $ d Y S $ d Y S $ d Y S $ 

1 0 5 1 1 5 .80 1 5 .80 1 5 .80 1 5 .80 

2 1 4 .75 0 4 .80 1 4 .60 1 4 .60 1 4 .60 

3 1 3 .50 1 3 .53 0 3 .60 1 3 .40 1 3 .40 

4 1 2 .25 1 2 .27 1 2 .30 0 2 .40 1 2 .20 

5 1 1 0 1 1 0 1 1 0 1 1 0 0 1 .20 

This result may also be obtained by general reasoning. If the lapse occurs late, then the “death” 

times come earlier, leading to lower survival probabilities. If the lapse occurs early, then the 

“death” times come later, leading to higher survival probabilities. 



Key: C 

Solution 

This is testing H0: β4 = β5 

= 0, which is a joint test on several regression coefficients. 

The key formula is 5.21 (p.130 of Pindyck and Rubinfeld). 

2 

We are given: R UR = 094 . 

Also, we know: N = 42, k = 5 (total number of coefficients to be estimated) and q = 2 

Using 5.21, we obtain F = 

b094 . − 0915 . g / 2 

b1 − 094 . g / b42 −5g 

= 77 . . 



Key: C 

Solution 

The number of expected claims (e) is proportional to the number of exposure units (n). Let e = 

cn. Using Bühlmann credibility and partial credibility gives: 

25 1 25/ 25 

= = c = 

100 2 25/ c+ k 25+ 

ck 

Therefore ck = 25. 

When we have 100 expected claims, 

100/ c 100 100 

Z = = = = 0.80. 

100/ c+ k 100+ ck 100+ 

25 



Key: D 

Solution 

Following Example 2.71 in Loss Models (with no denominator because this problem has no 

deductible), the contribution to the likelihood function for a loss (x) below 1000 is 

1 x / θ 

f ( x) 

= θ − e 

− 

−1000/ 

θ 

while for observations censored at 1000 it is 1 − F(1000) 

= e . The 

likelihood function is: 

⎛ ⎞⎛ ⎞ 

j 

L( θ) = ⎜ θ e ⎟⎜ e ⎟= 

θ e 

62 100 

−1 − x / θ −1000/ θ −62 − (28,140+ 

38,000)/ θ 

∏ 

The logarithm and its derivative are: 

∏ 

⎝ j= 1 ⎠⎝ j= 

63 ⎠ 

l( θ) =−62ln( θ ) −66,140θ 

l′ ( θ) =− 62θ + 66,140θ 

−1 −2 

−1 

Setting the derivative equal to zero yields θ ˆ = 66,140/62= 1067 . 



Key: E 

Solution 

ˆ d 

1 1 1 

( 14 

i 

B ) = ∑ = + + = 4.65 

Qt ( ) 0.25+ 0.15+ 0.5 0.15+ 

0.5 0.5 

ti≤14 

i 



Key: D 

Solution 

ESSR 

− ESSUR 

F statistic = bN − kg b g 

q ESS 

N = 100 

k = 3 

q = 2 

Series I: F = 97 

Series II: F = 97 

Series III: F = 97 

b 

b 

UR 

g 

3552. 2− 

32338 . 

2 32338 . 

b 

b 

g 

13005 . − 11318 . 

2 11318 . 

b 

b 

g 

g 

g 

237. 0− 

2111 . 

2 2111 . 

b 

g 

g 

= 4. 78, 

Fail to reject 

= 7. 23, 

Reject 

= 595 . , Reject 



Key: A 

Solution 

= E N A E X A = b2 / 3gb990g 

= 660 

= E N B E X B = b4 / 

276g 

= 368 

c = h = c = h c = h = b gb g = = 

c h c h c 1 2 h b gb gb g 

PcA500h 12/ 28 3/ 7 and Pc B500h 

4 / 7 

E L A 

E L B 

P L 500 A P N 1A P X 500 A 4 / 9 1/ 3 4 / 27 12/ 

81 

P L = 500 B = P N = 2 B P X = X = 250 B = 4 / 9 2 / 3 2 / 3 = 16 / 81 

⇒ = = = 

( ) ( ) 

E⎡⎣L2 L1 = 500⎤⎦ 

= 3/7660+ 4/7368= 

493 



Key: D 

Solution 

2 2 

( ) ( ) [ ] 2 

VHM = ⎡ 

⎣ 

1/2660 + 1/2368 ⎤ 

⎦ 

− 514 = 285,512− 264,196= 

21,316 

Total variance = 296,962 

EPV = Total variance – VHM = 275,646 

EPV 

1 

k = = 12.9314⇒ Z = = 0.07178 

VHM 

1+ 

k 

Bühlmann credibility premium Z ( Z) 

= 500 + 514 1− = 513 



Key: E 

Solution 

The distribution function is ( ) 

yields the K-S statistic: 

F 

1 

x 

−3 

∫ 2(1 ) 1 . From it, the following table 

0 

2 

F x = + t dt = − 

( 1+ 

x) 

x .1 .2 .5 1.0 1.3 

F x 

b g .174 .306 .556 .75 .811 

x − 

0 .2 .4 .6 .8 

bx 

g .2 .4 .6 .8 1.0 

n 

n + 

The K-S statistic is 0.189. 

F 

Max diff. .174 .106 .156 .15 .189 



Key: B 

Solution 

2 

2 

∑c i h ∑c 2i 

2h 

Divide the first equation by Y − Y X − X 

2 

2 

∑cYi 

− Yh ∑cX3i 

− X3h 

to get the equations: 

and the second equation by 

r 

YX 

2 

sX 

s 

2 X3 

= $ 2 + $ 3 rX X = $ * 

2 + $ * 

β β β β 3r 

s s 

Y 

Y 

X X 

2 3 2 3 

r 

YX 

3 

sX 

s 

2 

X3 

= β $ rX X + β $ = β $ r 

2 3 

X2X 

+ β $ 

3 

s 

s 

* * 

2 3 2 3 

Y 

Y 

Solve these equations for $ * 

rYX − r . . . 

2 YX r 

3 X 04 09 06 

2X 

− 

3 

β 2 = 

= 

2 2 

1− 

r 1− 

0. 

6 

X X 

2 3 

b gb g 

b g 

= −0. 

22. 



Key: C 

Solution 

σ 

2 

H 

di 

15 20 13 

20 = ∑ = + + = 0.0144 

2 2 2 

Y 100 65 40 

( ) 

ti 

≤20 

σ H b20g = 0. 0144 = 012 . 

i 



Key: A 

Solution 

=zτ 

µ $ S $ btgdt 

τ 

0 

b g b g b g b g 

= 10 . × 1 + 085 . × 7 + 05885 . × 9 + 03972 . × 8 = 1542 . 

z 

D 

L V$ $ S$ 

di 

µ = τ N M 

2 

τ 

∑ btgdt 

ti 

i= 

1 

QP Yib Yi − dig 

O 

2 2 2 

= 14424 . × 00018 . + 8474 . × 00068 . + 3178 . × 00120 . 

= 0.9840 

The 95% confidence interval is: 

15.42± 1.96× 

0.9840 

b 

135 . , 17. 

4 

g 



Key: C 

Solution 

The Weibull density function is 

.5 

.5 ( x / θ ) 

θ − − 

f ( x) = .5( x ) e . Therefore the likelihood function is 

The logarithm and its derivative are: 

10 

.5 

−.5 

−( x j/ θ) 

L( θ) = ∏.5( xjθ) 

e 

j= 

1 

−.5 

10 

10 −5 

∏ xj 

θ 

j= 

1 

⎛ ⎞ 

= (.5) 

e 

∑ 

⎜ ⎟ 

⎝ ⎠ 

∝ θ 

−.5 

−5 −488.97θ 

e 

. 

−.5 10 .5 

−θ 

x 

j = 1 

j 

l( θ) =−5lnθ−488.97θ 

l′ =− + 

−.5 

−1 −1.5 

( θ) 5θ 244.485 θ . 

Setting the derivative equal to zero yields 

2 

ˆ (244.485/5) 2391. 

θ = = 



Key: E 

Solution 

The estimated variance of the forecast errors is the sum of the squares of the error terms divided 

by T − p − q. In this case, after 100 observations, the sum of the squares of the error terms must 

equal 98, because the sum divided by b100 −1−1g, or 98, is 1.0. 

The 101 st observation introduces a new error term equal to 188−197, or − 9. The square of that 

term is 81. Adding 81 to the previous sum of 98 gives a new total of 179. Dividing 179 by 

b101−1−1g, or 99, gives a new estimated variance of 1.8. 



Key: A 

Solution 

− 5 .2 6 1.2 

The posterior distribution is ( |0) e λ − 

(.5)5 e λ − 

(.5).2e λ − 

π λ 

2.5 e λ − 

∝ ⎡ 

⎣ 

+ ⎤ 

⎦ 

= + .1 e 

λ . The 

∞ 

−6 1.2 

normalizing constant can be obtained from 2.5 e λ − λ 

∫ 

⎡ + .1 e ⎤dλ 

= .5 

0 ⎣ 

⎦ and therefore the exact 

−6λ 

−1.2λ 

posterior density is π( λ|0) = 5 e + .2e 

. The expected number of claims in the next year is 

∞ 

−6λ 

−1.2λ 

5 5 5 

the posterior mean, E( Λ |0) = ∫ λ[5 e + .2 e ] dλ 

= + = = .278. 

0 

36 36 18 



Key: B 

Solution 

There are 365 observations, so the expected count for k accidents is 

produces the following table: 

− .6 k 

e (.6) 

365pk 

= 365 which 

k! 

No. of accidents Observed Expected Chi-square 

0 209 200.32 0.38 

1 111 120.19 0.70 

2 33 36.06 0.26 

3 7 7.21 1.51** 

4 3 1.08 

5 2 0.14* 

*This number is 365 minus the sum of the other expected counts. 

**The last three cells must be grouped to get the expected count above 5. The calculation for the 

test statistic is (12 – 8.43) 2 /8.43 = 1.51. The total of the chi-square numbers is the test statistic of 

2.85. 



Key: E 

Solution 

The likelihood ratio test statistic is ( − 382.4+ 385.9) 

⋅ 2= 7. 



Key: C 

Solution 

8 

2 

20 

2 

⎛Yt 

−α⎞ ⎛Yt 

−α 

⎞ 

∑ ∑ to be a minimum. Setting the derivative equal to 

We need for S( α) 

= ⎜ ⎟ + ⎜ ⎟ 

t= 1⎝ 0.4 ⎠ t= 

9 ⎝ 0.6 ⎠ 

8 20 

1 1 

zero produces the equation S′ ( α) = 2( Yt 

− α) + 2( Yt 

− α) = 0. 

.4 t= 1 .6 t= 

9 

produces the equation: 

∑ ∑ Multiplying by 0.6 

0 = 3(8Y 

− 8 α) + 2(12Y 

−12 α) 

1 2 

0 = 24Y 

+ 24Y 

−48α 

1 2 

αˆ = .5 Y + .5Y 

1 2 



Key: E 

Solution 

Ο=4 

E = (.29 − .27) + (.35 − .27) + (.41 − .27) + 7.465 ( − .27) 

= 1.605 

b 

Chi-square statistic = 4 − 1605 . / 1605 . = 357 . 

The 0.05 level of significance is 3.84, so the answer is (E). 

g 

2 



Key: B 

Solution 

The Bühlmann-Straub credibility factor is 

n 

w + 

n + 

a 

v 

m 

which goes to 

n 

n + 

w 

a 

as m goes to infinity. 



Key: B 

Solution 

With a lognormal error component, the linear model should be for the logarithm of the 

observation. A model that conforms to the description is 

lnY = α + α D + β X + ε . 

* * * * 

1 2 1 

Exponentiating both sides yields 

* * * * 

α1 α2 D β1 

X ε 

Y = e e e e 

and then defining an unstarred quantity as its starred version exponentiated, we have 

Y 

= αα β ε 

D X 

1 2 1 

. 

Note that when D is 1, the value of Y is multiplied by 

2 

α and so the hypothesis to test is if this 

value is equal to 0.8. 



Key: C 

Solution 

The function of interest is f ( αβ , ) = α+ 10 β. 

The partial derivatives are 1 and 10 and so the 

variance can be estimated as 

⎡αˆ 2 

⎤⎡1 

⎤ 

σˆ 

[ 110] 

Var ˆ 

f 

= ⎢ ˆ⎥⎢ 10 

⎥ 

⎣β 

⎦⎣ ⎦ 

⎡ 0.00055 −0.00010⎤⎡ 1⎤ 

= [ 110] 

⎢ 

0.00010 0.00002 

⎥⎢ 

10 

⎥ 

⎣− 

⎦⎣ ⎦ 

= 0.00055− 2× 0.00010× 10+ 0.00002× 

100 

= 0.00055 

and so the standard deviation is the square root, or 0.02345. 



Key: A 

Solution 

Following Klein and Moeschberger with the time scale shifted by 1997: 

T i L i R i d i Y i P[ L< l | L< 

3] 

i 

0 0 2 6 

1 0 2 5 

2 0 2 4 15 (99/160) ⋅ 0 = 0 

0 1 1 3 

1 1 1 2 16 (9/10) ⋅ (11/16) = 99/160 

0 2 0 1 10 9/10 

9 99 

P[ L= 1| L< 3] = P[ L< 2| L< 3] − P[ L< 1| L< 3] 

= − = 0.28125 

10 160 



Key: E 

Solution 

From formula 15.41: 

yt 

825 784 710 918 

zt 

= , z6 = , z18 = , z30 = , z42 

= 

y% 

843 804 740 905 

t 

From formula 15.42: 

~z 6 = 

1 

4bz6 + z18 + z30 + z42g 

~ z 

1 . . . . . 

6 = 4 

09786 + 09751+ 09595+ 10144 = 09819 

b 

Adjustment factor = 12/11.9607 =1.0033 

g 

z 

y 

6 

a 

42 

= .9819*1.0033= 

0.9851 

918 

= = 932 

0.9851 



Key: E 

Solution 

Die/Spinner Prior probability Probability of 

getting a 12 

Posterior 

Probability 

AX 1/4 3/4× 1/2= 3/8 1/4 

AY 1/4 3/4× 1= 3/4 1/2 

BX 1/4 1/4× 1/2= 1/8 1/12 

BY 1/4 1/4× 1= 1/4 1/6 

Total 1 3/2 1 

Die/Spinner Expected Value Posterior Probability Expected Value × 

posterior probability 

AX (3/4)(1/2)(12+c) 1/4 1.125+3/32c 

AY (3/4)(12) 1/2 4.5 

BX (1/4)(1/2)(12+c) 1/12 0.125+1/96c 

BY (1/4)(12) 1/6 0.5 

Total 1 6.25+10/96c 

Because the expected value is 10, 6.25+10/96c = 10, so c = 36. 



Key: D 

Solution 

Summing the ranks of the A’s (being careful to average for ties) gives 

(1+2+6+6+8.5+8.5+10+11+13.5+20) = 86.5 (or 123.5 working with B’s). 

* 

E R = 105, Var R = 175,so r =± 1.398giving p= 

0.162. 

( ) ( ) 

H0 H0 



Key: C 

Solution 

The maximum likelihood estimate of the mean, θ, is the sample mean, 367.9. 

2 

Two years of inflation at 5% inflates this scale parameter toθ = 1.05 (367.9) = 405.6. 

The expected amount paid per loss is 

E( X ∧1000) −EX 

( ∧100) 

= − − + 

−1000/405.6 −100/405.6 

405.6(1 e 1 e ) 

= 283. 



Key: A 

Solution 

L 

( â) 

β1 β 2 

β 

⎛ 

3 

⎛ e ⎞⎛ e ⎞⎛e 

⎞⎞ 

= ⎜⎜ β1 β2 β ⎟⎜ 

3 β2 

β ⎟⎜ 

3 β ⎟ 

3 

e + e + e e + e e ⎟ 

⎝⎝ ⎠⎝ ⎠⎝ ⎠⎠ 



Key: A 

Solution 

2 3 

∑∑ 

i= 

1 

( − ) 

mij xij xi 

i= 1 j= 

1 

2020 

vˆ = = = 505 

2 

2+ 

2 

2 

∑ 

( 3−1) 

2 

( − ) −ˆ ( 2−1) 

m x x v 

2 

4800− 

505 4295 

i i 

i= 

1 

ˆ = = = = 114.5333 

2 2 2 

1 2 

m− 

m 

100− ( 1/100) ( 25 + 75 ) 37.5 

∑ i 

m i= 

1 

a 

∑ 

ˆ vˆ 

k = = 4.4092 

aˆ 

25 

Zˆ 1 

= = 0.850074 

25 + kˆ 

75 

and Zˆ 2 

= = 0.944475 

75+ 

kˆ 

ZX ˆ 

Zˆ 

+ Zˆ 

X 

+ Zˆ 

1 1 2 2 

ˆ = = 105.4208 

µ 

1 2 

Bühlmann credibility premium Zˆ ( ) ( Z ˆ 

)( µ ˆ ) 

= 97 + 1− = 98.26 

1 1 



Key: A 

Solution 

According to a statement on page 159 of Pindyck and Rubinfeld, the estimator is unbiased. 



Key: D 

Solution 

The observations are right and left truncated and the truncation depends upon the report year. 

For report year 1997 only claims settled at durations 1 and 2 can be observed, so the denominator 

must be the sum of those two probabilities. For 1998, only durations 0 and 1 can be observed 

and for 1999 only duration 0 can be observed. Calculation of the denominator probabilities is 

summarized below. 

Probabilities 

Year 

Year Settled 

Reported 1998 1999 

1997 b1− pgp 

1− p p 

1998 1− p 

Sum 

(Denominator) 

− p p + p 

2 

b g b1 g b1 

g 

b g b1− pgp 

b1 − p gb1+ 

pg 

b g b1− pg 

1999 1− p 

The likelihood function is: 

( ) 

( ) ( ) 

( 1+ 

p) 

11 

2 

( ) 

( ) ( ) 

( ) 

( ) ( ) 

( ) 

( ) ( ) 

( ) 

( ) 

3 1 5 2 4 

⎛ 1− p p ⎞⎛ 1− p p ⎞⎛ 1−p ⎞⎛ 1− p p ⎞ ⎛ 1− 

p ⎞ 

L( p) 

= ⎜ 1− p p 1+ p ⎟⎜ 1− p p 1+ p ⎟⎜ 1− p 1+ p ⎟⎜ 1− p 1+ p ⎟ ⎜ 1− 

p ⎟ 

⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠ 

3 1 5 2 

⎛ 1 ⎞⎛ p ⎞⎛ 1 ⎞⎛ p ⎞ 

= ⎜ 

1 p ⎟⎜ 

1 p ⎟⎜ 

1 p ⎟⎜ 

1 p ⎟ 

⎝ + ⎠⎝ + ⎠⎝ + ⎠⎝ + ⎠ 

3 

p 

= 

The loglikelihood is l( p) = 3ln( p) − 11ln( 1+ 

p) 

Take the derivative with respect to p to obtain the equation to solve: 

3 11 

− = 0 

p 1 + p 

b 

g 

The solution is p $ = 3 . 

8 



Key: E 

Solution 

( ) ( ) 

t d Y ∆ H % 4 −t 

H t K t* 

i 

( t) 

∆Vˆ 

⎡ ⎣ 

H% 

( t) 

⎤ t* = ⎦ b Kbt * g 

∆ % ˆ ( ) Kt (*) 

∆V ⎡H% 

t ⎤ 

⎛ ⎞ 

⎜ 

b 

⎣ ⎦ b 

⎟ ⎝ ⎠ 

1 1 9 0.1111 0.01235 1.0000 0.000 0.0000 0.000000 

3 2 8 0.2500 0.03125 0.3333 0.741 0.0617 0.001905 

5 1 5 0.2000 0.04000 −0.3333 0.741 0.0494 0.002439 

8 2 3 0.6667 0.22222 −1.3333 0.000 0.0000 0.000000 

10 1 1 1.0000 1.00000 −2.0000 0.000 0.0000 0.000000 

h ˆ(4) = 0.1111 

2 

σ [ hˆ 

(4)] = 0.00434 

2 

Lower confidence limit: 

1.96 0.00434/0.1111 

0.1111e − = 0.035 

Upper confidence limit: 

1.96 0.00434/0.1111 

0.1111e = 0.355 



Key: D 

Solution 

See page 555 of Pindyck and Rubinfeld. The variance is 1/T, not T, and the statement is then 

only true for large displacements. 



Key: B 

Solution 

The first step is to get the posterior distribution of P given no claims: 

π( p|0) ∝ f(0| p) π( p) 

= e 

−+ 1 0.01p 

(0.01) 

0.01p 

∝ e 

,0 100 

The normalizing constant can be found from 

∫ 

100 100 

0.01p 

0.01p 

0 

e dp = 100e = 100( e− 1) = 171.83. 

0 

The posterior density is 

Then, 

1 

π = < < 

171.83 

0.01p 

( p|0) e ,0 p 100. 

1 100 

171.83 171.83 

100 

0.01 p 

.5 

Pr( P > 50|0) = ∫ e dp = ( e− e ) = 0.622. 

50 



Key: B 

Solution 

µ ( p) = 1−0.01p 

µ = E[ µ ( p)] = E(1− 0.01 p) = 1− 0.01(50) = 0.5 

2 

100 

a = Var[ µ ( p)] = Var(1− 0.01 p) = 0.0001 Var( p) = 0.0001 = 1/12 

12 

vp ( ) = 1−0.01p 

v= Evp [ ( )] = 0.5 

k = v/ a= (1/2)/(1/12) = 6 

Z = 4/(4+ 6) = 0.4 

P = 0.4(5/4) + 0.6(0.5) = 0.8 

c 

The number of claims is Poisson, so the mean and variance are the Poisson parameter, 1 – 0.01p. 

The mean and variance of p come from the uniform distribution. 



Key: E 

Solution 

For a mixture, the mean and second raw moments are a mixture of the component means and 

second raw moments. Therefore, 

E( X) = p(1) + (1 − p)(10) = 10−9p 

2 

E( X ) = p(2) + (1 − p)(200) = 200−198p 

Var X p p p p 

2 2 

( ) = 200−198 −(10− 9 ) = 100−18 − 81 = 4. 

The only positive root of the quadratic equation is p = 0.983. 



Key: C 

Solution 

s 

2 

= 

∑ 

2 

ε$ 

i 967 

1934 

N − 2 

= 5 

= . 

2 

β$ 

s 

2 

s 1934 . 

= = = 0. 

0967 

2 

x 2000 

∑ 

i 

s 

β$ 

= 031 .

Course 4 May 2001 Solutions

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