The Accumulation Of Annual Investments With Independent Rates Of Return

The next application of the stochastic approach to the theory of interest which will be considered is the accumulation of annual investments. There are two examples of this application which may be of particular interest: first, an investor may wish to make a series of annual investments (this may be an insurance company wishing to invest annual premiums from an insurance policy); second, an investor may purchase a security, such as a bond, and wish to accumulate the coupon payments from the bond until the maturity date. The coupon payments will then form a series of annual (or semi-annual) investments invested at an unknown rate of interest. We will again assume that the values taken by the rate of interest are independent of each other.

The accumulation of a series of annual investments of one unit per annum, for n years, will be denoted by Pn.

The annual investments can be seen merely as a series of single investments, invested at the beginning of each of the next n years.

If the accumulation of (n — 1) annual premiums is considered, then Pn — 1 = (1 + ii)(1 + i2) ... (1 + in — 1)

The relationship between Pn and Pn — 1 is Pn = (1 + in)(1 + Pn — 1). That is, the accumulation of n annual premiums is equal to the accumulation of (n — 1) annual premiums, further accumulated for one year plus an additional investment of one unit accumulated over the nth year. Therefore, E[Pn] = E[(1 + In)(1 + Pn — 1)] using this relationship. Pn — 1 is dependent only on the values taken by the random variables (I1), (I2), ..., (In — ^ which have all been assumed to be independent of the value taken by the random variable In. Therefore, using Probability Result 5:

This relationship can now be used to determine successive values of E[Pn] Again, to simplify the notation, E[It] will be defined as ¿t,

E [P1] = E [1 + I1] = (1 + ¿1) E [P2] = E [1 + I2]E [1 + P1]

= (1 + ¿2)(1 + 1 + ¿1) = (1 + ¿0(1 + ¿2) + (1 + ¿2) E [P3] = E [1 + Is]E [1 + P2]

= (1 + ¿s)[1 + (1 + ¿1)(1 + ¿2) + (1 + ¿2)] = (1 + ¿1)(1 + ¿2)(1 + ¿3) + (1 + ¿2)(1 + ¿3) + (1 + ¿3)

In general:

E [Pn] = E [1 + In]E [1 + Pn — 1] = (1 + ¿n)E [1 + Pn — 1]

= (1 + ¿n)[1 + (1 + ¿n — 1)(1 + ¿n — 2) ... (1 + ¿1)

+ (1 + ¿n — 1)(1 + ¿n — 2) ... (1 + ¿2) +'" + (1 + ¿n — 1)] = (1 + ¿n)(1 + ¿n — 1) ... (1 + ¿1) + (1 + ¿n)(1 + ¿n — 1) ... (1 + ¿2)

+ ••• + (1 + ¿n)(1 + ¿n — 1) + (1 + ¿n) (28)

It may be regarded as more desirable to keep the expression in the more general form

It is also of interest to derive the second moment of the accumulation of annual investments

E[P2] = E[(1 + /„)2(1 + P„ _ 1)2] = E[(1 + /„)2]E[(1 + P„ _ i)2] (due to independence)

Again, the recurrence relationship can be used to find an expression for E[P2] in successive years; defining st2 to be the variance of the rate of interest in year t

Similarly,

= E [1 + 2/2 + 122]E [1 + 2P1 + p2] = (1 + 2^2 + s22 + ¿2 )(1 + 2E [P1] + + ¿2)

where E[P1] is calculated as above

In general:

= (1 + 2a„ + s2 + a2)[1 + 2E [P„ _ 1] + E [P2_ 1]]

This recurrence relationship can be used to calculate successive values of the second moment of P„.

The variance of P„ can then be found, in the usual way, by subtracting the square of the first moment. Therefore, var(P„) = (1 + 2t„ + s„2 + l 2){1 + 2E [P„ _ 1] + E [P2_ 1]} - (1 + l„)2{E[1 + P„ _ 1]}2

Once again, convenient results can be obtained if it is assumed that E[It] = a for all values of t and that st2 = s2 for all values of t: that is if we assume that the first and second moments of the distribution of It are the same for all t. Equation (9.28) then becomes

E [P„] = (1 + a) " + (1 + a) 1 + ... + (1 + a) In general, from equations (9.29) and (9.33) it can be seen that

E[P«] = (1 + a)E[1 + P„ _ 1] = (calculated at rate of interest a)

Therefore, E[P„] is equal to the accumulation of a series of annual investments calculated at the expected rate of interest. This is analogous to the result, derived in the previous section that E[A«] = (1 + a)«. This result could also be derived by summing the expected accumulations of a series of annual premiums.

E[P2] = (1 + 2l + s2 + l2){1 + 2E[Pn_ 1] + E[Pn_ i]2} (9.35)

var(P„) = (1 + 2l + s2 + l2){1 + 2E[Pn_ 1] + E[P2_ 1]} - (E[Pn])2 (9.36)

Clearly, unless computer methods are used, the calculation of second moments and variances of accumulations of annual investments will become difficult for large values of n.

Example 9.17

An investment manager purchases a bond on 1 January 2001 which has a maturity value of £100 and pays a coupon of £5 per £100 nominal at the end of every year. The bond is redeemed on 31 December 2006 and the investment manager wishes to invest the coupon payments, on deposit, until the bond is redeemed. It is assumed that the rate of interest at which the coupon payments can be reinvested is a random variable and the rate of interest in any one year is independent of the rate of interest in any other year.

(i) Find the mean value of the total accumulated investment on 31 December 2006 if the rate of interest has an expected value of 8% in 2002, 7% in 2003, 6% in 2004, and 5% in 2005 and 2006.

(ii) Calculate the mean value of the total accumulated investment on 31 December 2006 if the rate of interest has an expected value of 6.5% in each of the years from 2002 to 2006 inclusive.

(iii) State an expression in terms of moments of P4 which would allow you to calculate the variance of the accumulated value of the investment, assuming that the variance of the interest rate is the same in each year and is equal to 0.01.

Answer

In this case, a series of five annual investments of £5 is being considered; if the expected value of these investments is found, the £100 maturity value plus the final £5 coupon payment can simply be added to this expected value to find the mean of the total, as the final payments are not random variables.

(i) Returning to equation (9.28) and using Probability Result 2:

E[5(P5)] = 5 x E[P5] = 5[(1 + L5)(1 + L4)(1 + lb)(1 + L2)(1 + L1)

+ (1 + L5)(1 + L4)(1 + L3)(1 + L2) + (1 + L5)(1 + L4)(1 + L3) + (1 + L5)(1 + L4) + (1 + L5)]

(where l1 is the rate of interest in 2002, l2 the rate of interest in 2003, etc.)

= 5[(1.05) 2(1.06)(1.07)(1.08) + (1.05)2(1.06)(1.07)

+ (1.05)2(1.06) + (1.05)2 + (1.05)] = 5[1.3505 + 1.2505 + 1.1687 + 1.1025 + 1.05]

The expected value of the final investment, at maturity, is therefore

This answer could also be found by using the recurrence relationship in equation (9.29) explicitly, i.e. E[Pn] = (1 + tn)E[1 + Pn _j], where in is E[z„], and this approach is illustrated below.

E [P1] =

1 + ¿1) = 1.08

E [P2] =

1 + ¿2)E [1 + P1] =

(1.07)(1 + 1.08)

= 2.2256

E [P3] =

1 + ¿3)E [1 + P2] =

(1.06)(1 + 2.2256)

= 3.41914

E [P4] =

1 + ¿4)E [1 + P3] =

(1.05)(1 + 3.41914)

= 4.64010

E [P5] =

1 + ¿5)E [1 + P4] =

(1.05)(1 + 4.64010)

= 5.9221.

Therefore, the expected accumulated value of the coupon payments is

5 x E[P5] = 5 x 5.9221 = 29.611 and the expected accumulated value of the whole investment is

(ii) If the interest rate has the same expected value in each year, equation (9.34) can be used to calculate the expected value of the accumulation of the coupons:

0.065

Therefore the expected accumulated value of the total investment is

(iii) It is now necessary to find an expression for the variance of a random variable (5 xp 5) plus a constant (105). Using Probability Result 3, this variance must equal 52 var(P5), as the addition of the constant does, not affect the variance. From equation (9.24)

E [P52] = (1 + 2i + s2 + 12)[1 + 2E [P4] + E [P42]]

where 1 is the expected value of the interest rate and s2 is the variance of the rate of interest (both of which are the same in all years). Inserting these values for the parameters, the following expression is obtained:

(1 + 2 x 0.065 + 0.01 + 0.0652)[1 + 2€n + E[P42]]

0.065

184 Investment Mathematics

Therefore,

E[P52] = 1.4423E[10.3872 + E[P42]] = 11.8853 + 1.4423 x E[P42]

Therefore var(Pt) = 11.8853 + 1.14423E[P42] - [E[P5]]2 (using Probability Result 1)

Thus the variance of the total investment is

25[1.14423E[P42] - 24.8834] = 28.60575E[P42] - 622.085

A further example of the application of the stochastic approach to financial mathematics arises from analysing the accumulation of a level annual deposit per year when the interest rate is assumed to be a random variable. This is illustrated below.

Example 9.18

In any year the yield on funds invested on deposit has a mean value i and standard deviation s and is independent of yields in all previous years. Money is invested only at the beginning of the year. Find the mean and variance, at the end of 2 years, of the accumulation of an investment of one unit of money per fl«n«m at the beginning of each year if i = 0.06 and s = 0.01.

Answer

Using equation (9.34), the expected value, at the end of 2 years, of the annual investments is

0.06

To obtain the variance of the deposit, it is required to establish successive values of E[P2]. In this case, as the deposit is for only 2 years, it is only necessary to calculate E[Pj2] and then this value is used to calculate E[P|] Using equation (9.35)

E[Pj2] = (1 + 2i + s2 + i2) = (1 + 2 x 0.06 + 0.012 + 0.062) = 1.1237

Therefore var(P2) = E[P22] - [E[P2]]2 = 4.7686 - 2.18362 = 0.000 49.

More complex applications of the stochastic approach to the theory of interest will be discussed in Chapter 17.

Annex 9.1 Properties of the expected value

(iv) Writing the joint probability {p(X =xi) and (Y = yj)} as P(xi, yj):

and for each yj Thus,

^^ y^ xiyj;p(xi)p(yj) since the variables are independent i = 1 j = 1

Annex 9.2 Properties of the variance

(i) For equation 9.15

(ii) For equation 9.16

var(X + c) = E[{(X + c) - (y + c)}2] = E[(X - M)2] = var(X)

(iii) For equation 9.18

var(cX + kY) = E[(cX + kY)2] - {E[cX + kY]}2 = E [c 2X2 + 2ckXY + k 2 Y2] - {c2E[X]2 + 2ckE[X]E[Y] + k2E[Y]2} and since X and Y are independent E[XY] = E[X]E[Y] so, var(cX + kY) = c2(E[X2] - E[X]2) + k2(E[Y2] - E[Y]2) = c 2 var(X) + k2 var(Y)

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