Stat Hartman Winter 2018 Midterm Exam 28 February 2018

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1 Stat 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 Total: 53

2 Stat Hartman Midterm Exam - Page 2 of February You are given that: i = 6 A x = p x = 88 A x:1 = 588 q x 1 = 1 a [1 pt] Calculate ä x ä x = 1 A x d = /16 = b [1 pt] Calculate A 1 x:1 A 1 x:1 = A x:1 A 1 x:1 = = 1368 c [2 pts] Calculate A x+1 A x = A 1 x:1 + 1E x A x = A x+1 A x+1 = 51487

3 Stat Hartman Midterm Exam - Page 3 of February 218 d [2 pts] Calculate A x 1 A x 1 = vq x 1 + vp x 1 A x = = 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) 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( (45118)) 46812( ) = 72425

4 Stat Hartman Midterm Exam - Page 4 of February [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 v n (p x p x+1 p x+n 1 ) = v p x ( 1 + v px v n 1 n 1p x+1 ) = 1 E x ä x+1:n

5 Stat Hartman Midterm Exam - Page 5 of February 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) = b [1 pt] The present value of the annuity benefit if the annuitant dies at age 673 ( ) P V = 2 1 e 6(1) = 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 [( ) ] 1 e 6(t) = P [( = P 1 e 6(t)) ] 75 [ ] = P e 6(t) 25 = P [ 6(t) ln(25)] [ = P t ln(25) ] 6 = P [t ] = q 6 = p 6 ( = 1 5 = = 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 , 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

6 Stat Hartman Midterm Exam - Page 6 of February 218

7 Stat Hartman Midterm Exam - Page 7 of February 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 (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) b [2 pts] Calculate E[Z] E[Z] = ( ) + ( ) + ( ) + ( 478) = c [3 pts] Calculate V ar(z) E[Z 2 ] = ( ) + ( ) + ( ) + ( 2 478) = 49, V ar(z) = E[Z 2 ] E[Z] 2 = 49, (15746) 2 = 24,3487

8 Stat Hartman Midterm Exam - Page 8 of February 218 d [2 pts] Calculate P (Z < $15) P (Z < $15) = 478 from the PMF above

9 Stat Hartman Midterm Exam - Page 9 of February 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, , , , , , , 3223 a [1 pt] Show that 25 q 7225 = q 7225 = 1 25 p 7225 CF = 1 (p72 ) 25 CF = 1 ( l73 l 72 ) 25 CF = 1 ( 8, , CF = 8531 b [2 pts] Given that A (4) 7225 = 73341, find A(4) 725 A (4) 7225 = ( = c [2 pts] Find q 7125 ) ( 1 25q ) (8531) + ( 1 13 A (4) 725 = 7326 ) 13 ( 1 13 ) ) 25 25p 7225 A (4) 725 (1 8531) A (4) 725

10 Stat Hartman Midterm Exam - Page 1 of February q 7125 = 25 p q 7375 = 75 p 7125 p p 73 (1 75 p 7375 ) = 75 p 7125 p p 73 (1 25 p p 74 ) CF = (p71 ) 75 (p 72 ) (p 73 ) 75 (1 (p 73 ) 25 (p 74 ) 5 ) CF = 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 = / = 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, and 1, = 81, 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

11 Stat Hartman Midterm Exam - Page 11 of February 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) = 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) 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

Stat 475. Solutions to Homework Assignment 1

Stat 475. Solutions to Homework Assignment 1 Stat 475 Solutions to Homework Assignment. Jimmy recently purchased a house for he and his family to live in with a $3, 3-year mortgage. He is worried that should he die before the mortgage is paid, his