Stat 475 - Hartman Winter 218 Midterm Exam 28 February 218 Name: This exam contains 11 pages (including this cover page) and 6 problems Check to see if any pages are missing You may only use an SOA-approved calculator and a pencil or pen on this exam You are required to show your work on each problem on this exam Grade calculation errors: If I made an arithmetic mistake (I miscounted your total points) please come and see me and I will fix it Regrade requests: I make every effort to grade your test (and those of your classmates) fairly If you feel I graded a portion of your test too harshly, please write an explanation on the back of the test and turn it into me by Wednesday March 15th in class Please note that to maintain fairness your entire test will be regraded, potentially resulting in a lower overall grade Problem Points Score 1 11 2 4 3 1 4 11 5 1 6 7 Total: 53
Stat 475 - Hartman Midterm Exam - Page 2 of 11 28 February 218 1 You are given that: i = 6 A x = 3691 1p x = 88 A x:1 = 588 q x 1 = 1 a [1 pt] Calculate ä x ä x = 1 A x d = 1 3691 6/16 = 111459 b [1 pt] Calculate A 1 x:1 A 1 x:1 = A x:1 A 1 x:1 = 588 4512 = 1368 c [2 pts] Calculate A x+1 A x = A 1 x:1 + 1E x A x+1 3691 = 1368 + 45118A x+1 A x+1 = 51487
Stat 475 - Hartman Midterm Exam - Page 3 of 11 28 February 218 d [2 pts] Calculate A x 1 A x 1 = vq x 1 + vp x 1 A x = 1 99 3691 + 16 16 = 3542 e [2 pts] Under UDD, calculate Āx:1 Ā x:1 UDD = i δ A1 x:1 + A 1 x:1 UDD = 6 (1368) + 4512 ln(16) UDD = 5926 f [3 pts] Under UDD, calculate ä (12) x:1 ä (12) x:1 = α(12)ä x:1 β(12)(1 1E x ) = α(12)(ä x 1 E x ä x+1 ) β(12)(1 1 E x ) = 128(111459 85763(45118)) 46812(1 45118) = 72425
Stat 475 - Hartman Midterm Exam - Page 4 of 11 28 February 218 2 [4 pts] Consider the equation a x:n = ä x+1:n Is this a valid relationship? If so, prove it If not, determine a relationship between a x:n and ä x+1:n that is indeed valid (no proof required, just write the relationship and simplify as far as possible) It may help to draw a picture a x:n = n v k kp x k=1 = v p x + v 2 2p x + + v n np x = v p x + v 2 p x p x+1 + + v n (p x p x+1 p x+n 1 ) = v p x ( 1 + v px+1 + + v n 1 n 1p x+1 ) = 1 E x ä x+1:n
Stat 475 - Hartman Midterm Exam - Page 5 of 11 28 February 218 3 A whole life annuity pays a continuous benefit to (6) at a rate of 2 per year The first 1 years of payments are guaranteed to be made (ie, this is a life-and-1-year-certain continuous annuity) You are given: { ( 5 t ) 2 (i) t p 6 = 5 for t 5 for t > 5 (ii) δ = 6 Let Y denote the present value of this annuity benefit Calculate: a [1 pt] The present value of the annuity benefit if the annuitant dies at age 782 ( ) P V = 2 1 e 6(182) = 2214851 6 b [1 pt] The present value of the annuity benefit if the annuitant dies at age 673 ( ) P V = 2 1 e 6(1) = 153961 6 c [3 pts] The probability that the present value of the annuity benefit will be at most 25, that is, P (Y 25) Let t be the future lifetime of (6) Then [ ( ) ] 1 e 6(t) P (P V 25) = P 2 25 6 [( ) ] 1 e 6(t) = P 125 6 [( = P 1 e 6(t)) ] 75 [ ] = P e 6(t) 25 = P [ 6(t) ln(25)] [ = P t ln(25) ] 6 = P [t 231491] = 231491 q 6 = 1 231491 p 6 ( 5 231491 = 1 5 = 1 2893 = 717 d [2 pts] The probability that the present value of the annuity benefit will be at most 15, that is, P (Y 15) Since the first 1 years are guaranteed, the minimum present value of this annuity is 153961, so the probability that the PV will be less than or equal to 15 is e [3 pts] Draw a sketch of the CDF (cumulative distribution function) of Y Be sure to mark the values corresponding to any discontinuities, limiting values, and asymptotes ) 2
Stat 475 - Hartman Midterm Exam - Page 6 of 11 28 February 218
Stat 475 - Hartman Midterm Exam - Page 7 of 11 28 February 218 4 Let Z be the random variable representing the present value of benefits for a 3-year term insurance with death benefit payable at the end of the year of death issued to [82] You are given: (i) The death benefit is $4 if the insured dies in the first year, $35 if the insured dies in the second year, and $3 if the insured dies in the third year (ii) Mortality is given by the following select and ultimate mortality table with a 2 year select period [x] q [x] q [x]+1 q x+2 x + 2 81 22 212 22 83 82 28 218 228 84 83 216 224 236 85 84 221 234 242 86 (iii) i = 1% a [4 pts] Give the PMF (Probability Mass Function, in the form of a table of numbers) of Z z P (Z = z) 36364 28 28925 173 22539 141 478 b [2 pts] Calculate E[Z] E[Z] = (36364 28) + (28925 173) + (22539 141) + ( 478) = 15746 c [3 pts] Calculate V ar(z) E[Z 2 ] = (36364 2 28) + (28925 2 173) + (22539 2 141) + ( 2 478) = 49, 14172 V ar(z) = E[Z 2 ] E[Z] 2 = 49, 14172 (15746) 2 = 24,3487
Stat 475 - Hartman Midterm Exam - Page 8 of 11 28 February 218 d [2 pts] Calculate P (Z < $15) P (Z < $15) = 478 from the PMF above
Stat 475 - Hartman Midterm Exam - Page 9 of 11 28 February 218 5 Assume that mortality is given by the following life table excerpt, along with the constant force of mortality fractional age assumption Assume that interest is given by a nominal annual rate, convertible quarterly, of i (4) = 12% x l x 69 22, 7449 7 19, 32942 71 15, 46354 72 11, 59765 73 8, 11836 74 5, 35812 75 3, 3223 a [1 pt] Show that 25 q 7225 = 8531 25q 7225 = 1 25 p 7225 CF = 1 (p72 ) 25 CF = 1 ( l73 l 72 ) 25 CF = 1 ( 8, 11836 11, 59765 CF = 8531 b [2 pts] Given that A (4) 7225 = 73341, find A(4) 725 A (4) 7225 = ( 1 73341 = c [2 pts] Find 25 75 q 7125 ) ( 1 25q 7225 + 13 ) (8531) + ( 1 13 A (4) 725 = 7326 ) 13 ( 1 13 ) ) 25 25p 7225 A (4) 725 (1 8531) A (4) 725
Stat 475 - Hartman Midterm Exam - Page 1 of 11 28 February 218 25 75 q 7125 = 25 p 7125 75 q 7375 = 75 p 7125 p 72 75 p 73 (1 75 p 7375 ) = 75 p 7125 p 72 75 p 73 (1 25 p 7375 5 p 74 ) CF = (p71 ) 75 (p 72 ) (p 73 ) 75 (1 (p 73 ) 25 (p 74 ) 5 ) CF = 11992 Let Z represent the present value of a 5-year term life insurance with $1, death benefit payable at the end of the quarter of death, issued to (69) Using the mortality and interest assumptions described above: d [2 pts] Find P r[z > $] Z will be positive if and only if the death benefit is paid, which happens only when the insured dies within the 5 year term: P r[z > $] = 5 q 69 = 1 5 p 69 = 1 l 74 /l 69 = 1 535812/227449 = 76438 e [3 pts] Find P r[z > $82, ] You can find, either algebraically or through trial and error that ( ) 1 6 ( ) 1 7 1, = 83, 74843 and 1, = 81, 3915 13 13 so that Z > $82, if and only if the death benefit is paid at time 6/4 or sooner The probability of this happening is 15q 69 = 1 15 p 69 = 1 p 69 5 p 7 CF = 1 p69 (p 7 ) 5 CF = 23974
Stat 475 - Hartman Midterm Exam - Page 11 of 11 28 February 218 6 A person currently age 5 pledges that, upon her death, her estate will make a donation of $1,, to a particular university Assume that the force of interest is δ = 5 and the force of mortality is given by µ x = { 2 for 5 x 8 4 for x > 8 a [3 pts] Calculate 4 p 5 for this person In general, for this person, for t 3, we will have and for t > 3, we will have Then 4 p 5 = e 6 4(4) = 36788 tp 5 = e t 2 ds = e 2 t tp 5 = 3 p 5 t 3 p 8 = e 2(3) e t 3 4 ds = e 2(3) e 4(t 3) = e 6 4t b [4 pts] Calculate the EPV of this person s donation The EPV is 1,, Ā5 = 1,, = 1,, = 1,, = 1,, [ 3 [ 3 [ 3 [ 3 e δ t tp 5 µ 5+t dt e δ t tp 5 µ 5+t dt + 3 e 5 t e 2 t (2) dt + e t(5+2) (2) dt + ] e δ t tp 5 µ 5+t dt 3 3 ] e 5 t e 6 4t (4) dt ] e t(5+4) e 6 (4) dt ] = 1,, e t(7) (2) dt + e t(9) e 6 (4) dt 3 { [ = 1,, 2 ] 3 ] } 7 e t(7) + [ (4)e6 e t(9) 9 3 { = 1,, 2 7 e 3(7) + 2 7 e (7) } (4)e6 e ( )(9) + (4)e6 e (3)(9) 9 9 { = 1,, 2 7 e 3(7) + 2 } 7 + (4)e6 e (3)(9) 9 = 35,15181