Solution 2.1. We determine the accumulation function/factor and use it as follows.

Transcription

1 Applied solutions The time value of money: Chapter questions Solution.. We determine the accumulation function/factor and use it as follows. a. The accumulation factor is A(t) =. t. b. The accumulation is given by A() =. = $.. c. The accumulation is given by A(. ) =.. = $.. d. The accumulation is given by A() + A(.) A( ) = $. Solution.. We determine the accumulation function/factor and use it as follows. a. The accumulation factor is A(t, T )=. T t which leads to a discount factor of (T t) =. A(t, T ) b. The present value is given by. = $9.9. c. The present value is given by,.. = $,.9. d. The present value is given by,. +,.. = $,.. Solution.. The accumulation function/factor is given by A(t, T )=. T t. a. The accumulated value is given by (A(, ) + A(, ) + A(, )) = = $,. b. The present value is given by A(, ) + A(, ) + A(, ) + A(, ) + + A(, ) A(, ) = $.

2 c. The present value is given by A(, ) + A(, ) + A(, ) + A(, ) + A(, ) + A(, ) + A(, ) + A(, ) + A(, 9) + = $,. A(, ) d. The accumulated value is given by A, + A, + A, + A, + A, + A, 9 + A, + A, + A, + A, + A, + A (, ) = $. Solution.. We have cashflows of, at t =, + at t = and + at t =. a. The equation of value at t = is obtained as, = ( + i) + ( + i) b. The net present value is given by NPV(i) =, + ( + i) + ( + i) which is such that i. NPV(%) = $. ii. NPV(%) = $. iii. NPV(%) = $. c. The internal yield is given by the value of i that corresponds to a root of the equation of value. This is a quadratic in /( + i) and can be solved to give i =.9% % per annum. Note that the value must be positive as a total of $, is obtained from a $, investment. Note also that the root of the EoV is also i such that NPV(i) = ; i % is consistent with the results of part b. d. The investment yields only % per annum and the borrowing costs exceed this. The investment should not then go ahead unless borrowing can be obtained at a rate lower than %.

3 Solution.. The cash flows are given by -$P at t = $ at t = $ at t = $ at t = $ at t = a. The fair price is determined by the equation of value. That is, P = ( + i) + ( + i) + ( + i) + ( + i) b. We therefore have P (%) = $.9 c. We need to solve = ( + i) + ( + i) + ( + i) + ( + i) and can do this computationally to determine that i =.% per annum. d. Both investments are for the same term ( years) and so the decision is determined by the expected return only. Since.% > % we should invest in the first asset at this price. The force of interest: Chapter questions Solution.. We are required to use the principle of consistency and partition the accumulation function between periods with di erent i. a. The present value is given by, A(, ) =, A(, )A(, )A(, )A(, ) =,.... = $. b. The accumulated value is given by, A(,.) =, A(, )A(, )A(, )A(,.) =,..... =$,.

4 c. The accumulated value is given by A(, 9) + A(, 9) +, A(, 9) = {A(, )A(, ) + A(, )} A(, )A(, 9) +, A(, 9) = ,. =$,. d. The present value is given by A(, ) + A(, ) + = + A(, ) A(, )A(, ) + A(, )A(, )A(, )A(, ) =$. Solution.. In each case we begin by converting the nominal rate to an e ective compounding rate. a. % per annum converted half yearly is equivalent to.% per months. The accumulated value is therefore. = $.. b. % per annum converted every years is equivalent to % per years. The present value is therefore. = $.. c. % per annum converted monthly is equivalent to.% per month. The accumulated value is therefore. +. = $.. d. % per annum converted half yearly is equivalent to.% per months. The required value is therefore.. =.9. Solution.. Let the annual nominal rate converted over a period of length p be i(p),thee ective rate over that time interval is the i(p) p. The principle of consistency can then be used then demonstrate that +i = p + i(p) p where i =.%. a. Our expression with p = leads to i () =.% per annum. b. Our expression with p = leads to i () =.% per annum.

5 c. Our expression with p =. leads to i (.) =.% per annum. d. Here we use. = e and obtain =.9% per annum. We note that i (.) >i>i () >i () > which is consistent with the definition = lim p! i(p). Solution.. We have =.% per annum. a. The accumulation factor is given by A(,t)=e t = e.t. b. The accumulation is given by A(, ) = e. = $.. c. The accumulation is given by A(,.) = e.. = $9.. d. The net accumulation is given by A(, ) + A(., ) A(9, ) =e. + e.. e. =$. Solution.. In each case we use the expression (t) = d ln A(,t) dt and perform the required derivative. a. A(,t)=e.t ) (t) =. b. A(,t)=e.t.t ) (t) =.t. c. A(,t)= t ) (t) = t d. A(,t)=.( + sin(t)) ) (t) = cos t +sin t

6 Approximating price sensitivities: Chapter questions Solution.. The e ective duration D(i ) is such that D(i )= dp (i ) ) P (i ) P (i ) ' D(i )P (i ) P (i ) di i i where P (i ) is the price of the asset at the previous interest rate and i a new interest rate. Using the data given to us, we see that P (i ) ' P (i ) { (i i )D(i )})P (i ) '. {.(i.)} a. Using i =.% we find P (i ) ' $.9. b. Using i =.% per annum we find P (i ) ' $.9. c. Using i =.% per annum we find P (i ) ' $.. d. Using i =.% per annum we find P (i ) ' $.. We see that an increase in the desired yield reduces the price that should be paid - this makes sense as the incomes are fixed. The accuracy of estimate decreases with increased i.. The smaller this value the more accurate the price will be. Solution.. The duration (i ) is such that (i )= dp ( ) ) P ( ) P ( ) ' ( )P ( ) P ( ) d where P ( ) is the price of the asset at the previous force of interest and Using the data given to us, we see that a new force of interest. P ( ) ' P ( ) { ( )D( )})P ( ) ' {.(.)} a. Using i =.9% ) =.% we find P ( ) ' $.. b. Using i =.% ) =.9% we find P ( ) ' $.. c. Using =.% we find P ( ) ' $..

7 d. Using =.% we find P ( ) ' $9.. Solution.. The cash flows are as follows. -$P at t = $ at t = $ at t = $ at t = a. The price of the asset is given by the equation of value P (i) = +i + ( + i) + ( + i) = e + e + e which is evaluated as follows. i. P (%) = $. ii. P (.%) = $. b. The duration of the asset is given by = P P = e + e + e e + e + e and so (i =.) =. years. c. We use the expression in Question to determine that P ( ) '. {.(ln. ln.)} = $. This estimate is close to the actual value of $. which reflects that there has been a very small change in and the approximation holds. Solution.. We consider $ nominal of the bond. The cash flows are therefore -$P at t = $. at t =.

8 $. at t =. $. at t =. $. at t = a. The equation of value for this system is P (i) =. ( + i). + ( + i) ( + i) + ( + i) =. e. + e e +e This is evaluated as follows. i. P (.%) ' $9. ii. P (.%) ' $. iii. P (.%) ' $. We note that the price decreases with increasing desired yield. This is necessary as the income cash flows are fixed and so an increased yield can only be obtained from a reduced price. b. Using the equation in Question, we have =..e. + e +...+e + e.(e. + e e ) + e and so (i =.) =.. c. We use the expression in Question to determine that P ( ) '. {.(ln( + i ) ln.)} We therefore obtain i. i =.% ) =.% and so P ( ) ' $9.. ii. i =.% ) =.% and so P ( ) ' $.. iii. i =.% ) =.% and so P ( ) ' $.. Solution.. The cash flows are

9 -$ at t = $ at t = $ at t = $x at t = The price is given by the equation of value (at t = ) = e + e + xe and the duration is such that. = e + e +xe These represent two simultaneous equations that can be solved using Wolfram Alpha to give x.9 and.. The unknown yield is therefore i.%. Annuities (Chapter ) questions Solution.. Recall that a n = (+i) n each of n years. i is the present value of a unit payment made at the end of a. $ paid at the end of every year for years has present value $a = $.. b. $ paid at the start of every year for years has present value $a = $9.. c. A -year payment stream with $ paid at the end of the first years and $ paid at the end of the second ten years can be considered as an immediate annuity and a deferred annuity. The present value is given by $ + (+i) a = $. d. We work in time units of -years and an e ective interest rate of + j =.. The present value at j% of $ paid at t =,,..., is therefore given by $a = $.. Solution.. The cash flows are -$. at t =

10 $ at t =. $ at t = $ at t = We identify the regular cash flow as an -year annuity. a. The equation of value is given by. = a + ( + i) = + ( + i) i ( + i) b. Wolfram Alpha or Goal Seek can be used to determine that the equation of value is solved by i.9% per annum. The keen reader may wish to implement their own numerical method to solve this equation. Solution.. The cash flows of every half-year mean that it is sensible to work in units of half years. The dividend cash flow is then a perpetuity in arrears and the equation of value is P (j) = j where j is the -month e ective compounding rate. a. We have i = % per annum and so + j = p. which leads to.99. b. We have i =.% per annum and so + j = p. which leads to.. c. We have i = j = % per -months which leads to.. d. We have i =.% per quarter and so + j =. which leads to.. Solution.. The cash flows are -P at t = x at t = x( + r) at t =

11 . x( + r) n at t = n a. The equation of value is given by P =x +i + x +r ( + i) x( + r)n ( + i) " n = x # n +r +r +r +r +i i +i = x j% aat n +r where + j = +i +r. b. We evaluate the above expression at n =, x = $, r = % and i = % per annum and obtain P = $.. c. We evaluate the above expression at n =, x = $, r = % and i = % per annum and obtain P = $.. Note that j is negative. This reflects that the discount rate i is less than the inflation rate r. Each additional cash flow (as t increases) therefore adds an increasing amount to the overall present value. Solution.. The cash flows are -P at t = x at t = x + a at t =.. x +(n )a at t = n a. After some work, we find that the equation of value is given by P = x +i + x + a x +(n )a ( + i) ( + i) ( + i)an n( + i) n =(x a)a n + a i

12 b. We evaluate the above expression at n =, x = $, a = $ and i = % per annum and obtain P = $.. The force of interest as a function of time: Chapter questions Solution.. In each case we use the expression Z t A(,t)=exp (s)ds a. Using (t) =.t we find Z t A(,t)=exp.sds =exp.t b. Using (t) =.t.t we find Z t A(,t)=exp (.s..s)ds =exp t.t c. Using (t) =. t + t we find Z t t A(,t)=exp. s + s ds =exp. + t t d. Using (t) = e t we find Z t A(,t)=exp e s ds =exp e t Solution.. The accumultion factor between times t and t is given by Z t A(t,t )=exp ( cos t)ds = e ((t t)+sin(t) sin(t)) t a. The accumulated value at t = of $ deposited at t = is given by $A(, ) = $. b. The present value of $ due at t = is given by $ A(, ) = $9.. c. The accumulated value at t = of $ deposited at t = and $ deposited at t = is given

13 by $A(, ) + $A(, ) = $.99. d. The value at t = of $ due at t =, t = and t = is given by $ A(, ) + A(, ) + = $. A(, ) Solution.. The force of interest is the piecewise function ><., apple t< (t) = >:. +.(t ), t > and so the accumulation factor will also be a piecewise function. Using the principle of consistency, we write and so >< A(,t), apple t< A(,t)= >: A(, )A(,t), t > >< e.t, apple t< A(,t)= >: e +.t +.t, t > Solution.. The piecewise force of interest >< (t) = >:., apple t<., apple t< at + bt, t leads to a piecewise accumulation factor >< A(,t)= >: A(,t)=e.t apple t< A(, )A(,t)=e. e.(t ) = e.t. apple t< A(, )A(, )A(,t)=e. e a(t )+ b(t ) t We have two accumulations that can be used to form two simultaneous equations =e. e a( )+ b( ) =e. e a( )+ b( )

14 These can be solved (using Wolfram Alpha, say) to obtain a. and b.. Solution.. The accumulation factor is given by Z t. A(,t)=exp.s ds =exp t The payment stream is received continuously at rate (t). Consider a small element of time dt placed at time t, the amount of cash received in this element is (t)dt and this has present value e. t t dt These elements exist between t = and t =, leading to an overall present value of Z t e. t dt = $. Mortality: Chapter 9 questions Solution 9.. We use the standard notation t p x and t q x. a. The probability that a newborn lives to at least age is denoted p. b. The probability that an year old does not survive to retirement at age is denoted q. c. The probability that a year old dies before his 9th birthday is denoted q. d. The probability that a year old lives to her th birthday is denoted p. Solution 9.. We use the standard notation t p x and t q x. a. The probability that a newborn dies between his th and th birthday is denoted p. q. b. The probability that a year old will die aged is denoted p.q c. The probability that a year old will die between her th and th birthday is denoted p. q. d. The probability that a year old will die aged either, or is denoted p.q + p.q + p.q.

15 Solution 9.. We use the life table as directed. Recall that tp x = l x+t l x and tq x = l x l x+t l x leading to p.q + p.q + p.q = l l l l l + l l l l l + l l l l l = l (l l + l l + l l ) =.% Solution 9.. The required probability is expressed in terms of -year quantities as p.p.p.p.q =p.p.p.p.( p ) =e.e.e.e. e =.% Solution 9.. The required probability is expressed as p. q =p.p.p.p.p.( p ) =p.p.p.p.p.( p.p ) = =.% Markov chains: Chapter questions Solution.. We have P = and note that the n-step transition matrix is given by P (n) = P n.

16 a. The -step transition matrix is P () = = b. The -step transition matrix is P () = P = P P = c. The -step transition matrix is P () = P = P P = d. The -step transition matrix is P () = P = P P = Note that the n-step matrices appear to be converging towards a stationary structure. That is, additional time steps will eventually make no di erence to the transition probabilities. Solution.. We refer to the no-claims discount model has states and associated -step transition probabilities shown in Figure.. a. The -step transition matrix is given by P =....

17 b. The -step transition matrix is given by P () = P = c. The -step transition matrix is given by P () = P = Solution.. The probability that a policyholder currently in State will be in State after time steps is given by element p () =. That is there is zero probability. The reader should also convince himself of this by looking at all possible groups of movements from State. Solution.. A stationary distribution is such that it is una ected by a further time step. That is, we require {n i } such that apple n n n n.... apple = n n n n This leads to the simultaneous equations n +.n = n ><.n = n.n = n >:.n + n = n This system is solved by any values of n and n as long as n ==n. This is to be expected as the

18 end states (States and ) are absorbing. Solution.. We refer to the no-claims discount model has states and associated -step transition probabilities shown in Figure. and begin by stating the -step transition matrix. P = The steady annual revenue is determined by the stationary distribution obtained from apple n n n n apple = n n n n Some e ort leads to the solution n = n n = n n which is subject to n + n + n + n =,. Therefore,, policyholders pay $ per annum, policyholders pay $9 per annum, policyholders pay $ per annum, policyholders pay $ per annum leading to a stationary revenue of $9,,. per annum. Continuous probability distributions: Chapter questions These questions all require the evaluation of di cult integrals to a reasonable degree of accuracy. The choice of the numerical scheme is left to the reader, however, given that accuracy is important, we use

19 Simpson s rule of order in all that follows. That is, Z b a f(x)dx ' h (f(a = x )+f(x )+f(x )+f(x )+f(x )+f(x )+f(b = x )) Solution.. We have P (a <X<b)= Z b a f X (x)dx = Z b a p e x dx and so each probability can be evaluated as required to obtain the following. Note that we can approximate by some finite number, say x =. a. P (X <.) = P ( <X<.) ' 9.% b. P (X >.) = P ( >X>.) '.% c. P (. <X<.) '.% d. P (. <X<.) ' 9.% Solution.. We are required to show that E[X] = Z x p e x dx = Of course this is true as the function is odd. However, we can evaluate it numerically to obtain E[X] ' Z x p e x dx = Solution.. We work with h Y (y) = cos y for y (, p ). a. We require h Y (y) > for y (, p ), which is true if >. Also, we require Z p cos y dy =

20 and this fixes. In particular = Z p cos y dy '. b. We note that P ( <Y <) = Z cos. y dy '.9% c. Using the same approach to part b., we find that P ( <Y < p ) '.%. d. Using the same approach to part b., we find that P ( p <Y < p ) '.%. Solution.. We have f X (x) = xe (x ) for x [, ]. a. We note that f X (x) for all x [, ] when >. The value of is obtained from Z xe (x ) dx = That is, =.. b. We require P ( <X<) and this is obtained from Z.xe (x ) dx '.% c. We require P ( <X<) and this is obtained from Z.xe (x ) dx '.% d. We obtain E[X] = Z.x e (x ) dx '. Solution.. We have f Y (y) =!y e (y ) for y [, ].

21 a. We note that f Y (y) for all y [, ] when!>. The value of! is obtained from Z!y e (y ) dy = That is,! '.. b. We obtain c. We obtain var[y ]= E[Y ]= Z Z.y e (y ) dy '..(y.) y e (y ) dy '.