MTH6154 Financial Mathematics I Stochastic Interest Rates Contents 4 Stochastic Interest Rates 45 4.1 Fixed Interest Rate Model............................ 45 4.2 Varying Interest Rate Model........................... 46 4.2.1 Set-up.................................. 47 4.2.2 Expected P&L.............................. 47 4.2.3 Variance of P&L............................. 48 4.3 The Log-normal Distribution........................... 49 4.4 Repeated Deposits................................ 51 4.4.1 Set-up.................................. 51 4.4.2 Expected P&L.............................. 51 4.4.3 Variance of P&L............................. 52 4 Stochastic Interest Rates Financial contracts are often of a long term nature. In practice deposits or loans for a period longer than one year will be rolled after one year: i.e. the initial rate will only be valid for one year and after that, the deposit will be reinvested again at the then prevailing one year-interest rate. Likewise most mortgages (the so called floating rate mortgages are based on the LIBOR rate set every day at 11am London time and valid fora loan/deposit of a fixed period, typically half a year. If the interest rate at the start of year i is denoted by R i then in practice we only know R 0 whereas the rates R 1, R 2,... will only be known at the start of years 1, 2,.... It seems reasonable to take R 1, R 2,... to be random variables. The resulting framework is sometimes referred to as Stochastic interest rates which means that interest rates in the future are taken to be random. In these notes, P&L means profit and loss. 4.1 Fixed Interest Rate Model In this section we consider an elementary example which illustrates the use of probability to model interest rates. Suppose an investor invests amount P for n years starting now, before the interest rate is known. Immediately after the investment is made, the interest rate is chosen according to 45
some probability distribution and fixed for the remainder of the investment. Suppoe there are K posssible interest rates, so that Ω = {r 1, r 2,..., r K }. Let R be the random interest rate, with probability distribution P(R = r i = p i, i = 1, 2,..., K. (Alternatively, we might suppose that R is a continuous random variable, for example it could be uniformly distributed. The expected value of R is E(R = K p i r i. i=1 The expected value of the accumulated value, however is not P (1 + E(R n, but P E((1 + R n. This is not equal to P ((1 + E(R n. Using Jensen s inequality (which is not a topic of this module one may show that P E((1 + R n P (1 + E(R n. Examples 4.1. Suppose that P = 5000, n = 5, K = 3, and 0.06 with probability 0.2 r k = 0.08 with probability 0.7 0.10 with probability 0.1 Then, the expected accumulated value is 5000E((1 + R 5 = 5000(0.2 1.06 5 + 0.7 1.08 5 + 0.1 1.10 5 However the mean rate of interest is = 7, 286. E(R = (0.2(0.06 + (0.7(0.08 + (0.1(0.10 = 7.8%. The accumulated value at the mean rate of interest is 5000 1.078 5 = 7, 279, which is less than the expected accumulated value. 4.2 Varying Interest Rate Model In this section we will consider the case of a single deposit of 1 a time 0 over n years in a bank account with variable interest rates each year. 46
4.2.1 Set-up We will assume that the investment period is n years staring from now and that R i starts for the annual interest rate earned on the period between year i and year i + 1. All R i are random variables. The random variable S n has its own distribution which may be difficult to analyse. In order to make calculations possible we will assume that all interest rates R 0,..., R are independent. If they are, in addition, identically distributed, we will write ρ and γ 2 for their common mean and variance. Note that if we place 1 in a deposit, this will grow to S n = (1 + R 0 (1 + R 1 (1 + R in year n, while if we place P in a deposit, this will grow to P S n. In this section we will consider placing a single unit of cash at time 0 until time n. The next section will consider a fund that receives 1 every year. Remark 4.2. The assumptions above constitute a first approach to stochastic interest rate modelling. In more complex approaches we would not assume that interest rates are independent or identically distributed. The reason for this is: 1. The interest rates in ten years, R 10, has more time to vary and so we would expect its variance to be larger than say the variance of R 1. 2. Interest rates are not generally independent: if rates become low then we expect future rates to become low as well. I.e. the correlation ρ Ri,R i+1 is in general positive. 4.2.2 Expected P&L As stated in the previous section, the final P&L of the investment under consideration is the random variable: S n = (1 + R 0 (1 + R 1 (1 + R = (1 + R i. In order to better understand the performance of this investment it seems sensible to calculate the mean and variance of this quantity. We start calculating the mean ( E (S n = E (1 + R i = E (1 + R i If the R i sre i.i.d., then we have E (S n = (1 + ρ = (1 + ρ n as the variables are independent. Therefore E(P S n = P (1 + ρ n. 47
4.2.3 Variance of P&L Next we calculate the variance which tells us how likely it is for the investment to depart from its expectation. We start by calculating the second moment E(S 2 n. In the calculation we will use that for any random variable, X, we have E(X 2 = Var(X + (E (X 2 (1 which follows immediately from the definition of variance. To calculate the second moment we separate out (1 + R 0 out as this term is not random: E ( ( Sn 2 = E (1 + R i 2 = E ( (1 + R i 2 = E ( 1 + 2R i + Ri 2 ( = 1 + 2E(Ri + E(Ri 2 ( = 1 + 2E(Ri + Var(R i + (E(R i 2 ( = (1 + E(Ri 2 + Var(R i as the variables are independent If the R i are i.i.d., then we have E(S 2 n = ( (1 + ρ 2 + γ 2 n. We can now use this to calculate the variance: Var(S n = E(S 2 n E(S n 2 = ( (1 + ρ 2 + γ 2 n (1 + ρ 2n (2 Therefore, Var(P S n = P 2 [( (1 + ρ 2 + γ 2 n (1 + ρ 2n ]. Remark 4.3. As a sanity-check, let us look at the formula above in the case the variance of the interest rates is zero, γ 2 = 0. We would expect that the variance of the value of our deposit should also be zero. This is indeed the case, as from (2: ( (1 + ρ 2 + γ 2 n (1 + ρ 2n = ( (1 + ρ 2 n (1 + ρ 2n = 0. As another sanity-check, note that if the variance of the interest rates goes up, γ 2, then so does the variance of our P&L. This can be seen by just visually inspecting (2 or by checking that the derivative with respect to γ is positive. 48
Examples 4.4. Suppose that the R i all have the distribution given in Example 4.1 and that we invest the principle 5000 for n = 5 years. We will find the mean and variance of the accumulated value of the investment. The accumulated value is 5000S n. We found that ρ = E(R = 0.078 and know that the expected accumulated value is We must find γ 2. Now, E(5000S n = 5000(1 + ρ 5 = 7286. γ 2 = E(R 2 ρ 2 = 0.2 0.06 2 + 0.7 0.08 2 + 0.1 0.10 2 ρ 2 = 0.0062 0.078 2 = 0.000116. Thus the variance of the accumulated value is Var(5000S n = 25, 000, 000 [ (1.078 2 + 0.000116 5 (1.078 10] = 26448.73. 4.3 The Log-normal Distribution There is one distribution for the 1+R i for which the distribution of S n can be stated precisely: the log-normal distribution. A random variable Y is log-normally distributed with parameters µ and σ 2, i.e. Y LogNormal(µ, σ 2, if log Y N(µ, σ 2. We can write Y as Y = e X, where X N(µ, σ 2. We will use transformation of random variables to determine the p.d.f. f Y (y. Theorem 4.5. If Y LogNormal(µ, σ 2, then { f Y (y = 1 2πσy exp (log y µ2 ( 2σ 2 if y > 0; 0 if y 0. Proof. We apply the Transformation of Variables formula with f X (x = 1 exp ( x2 2πσ 2σ 2 and g(x = e x. The function g takes the interval (, to the interval (0,. It has inverse g 1 (y = log y. The rest is a simple calculation. 49
If we suppose that Y 1 LogNormal(µ 1, σ 2 1 and Y 1 LogNormal(µ 1, σ 2 1 are independent, then Y 1 = e X 1 and Y 2 = e X 2 for independent X 1 N(µ 1, σ 2 1 and Y 1 N(µ 1, σ 2 1. Therefore, Y 1 Y 2 = e X 1 e X 2 = e X 1+X 2, and since it follows that X 1 + X 2 N(µ 1 + µ 1, σ 2 1 + σ 2 2, Y 1 Y 2 LogNormal(µ 1 + µ 2, σ 2 1 + σ 2 2. By extending the previous results to n variables we have shown Theorem 4.6. If the random variables 1+R i are i.i.d. with distribution 1+R i LogNormal(µ, σ 2, then S n = (1 + R i LogNormal(nµ, nσ 2. We state the cummulative distribution function of S n explicitly: P(S n x = P(e X 1+ +X n x = P(X 1 + + X n log x ( X1 + + X n nµ = P nσ ( log x nµ = Φ. nσ log x nµ nσ Examples 4.7. Find the upper and lower quartiles for the accumulated value at the end of 5 years of an initial investment of 1000, using the varying interest rate model and assuming that the annual growth rate has a log-normal distribution with parameters µ = 0.075 and σ 2 = 0.025 2. By definition, the accumulated amount X = 1000S 5 will exceed the upper quartile u with probability % 25, i.e. 0.75 = P(X u = P(1000S 5 u = P(S 5 u/1000. Therefore, ( log(u/1000 5µ 0.75 = Φ 5σ From the normal distribution tables we find that Φ(0.6745 = 0.75. So we must have log(u/1000 5µ 5σ = 0.6745, that is, ( u = 1000 exp 5µ + 0.6745σ 5 = 1510.90. Similarly, the lower quartile is l = 1000 exp ( 5µ 0.6745σ 5 = 1401.15. 50
4.4 Repeated Deposits We next explore the properties of an investment where we place additional deposits every year. 4.4.1 Set-up Assume that at times 0,1,..., n 1 we place 1 in a deposit. Assume that interest rates are as in 4.2.1: R 0, R 1,..., R are independent random variables. At time n the accumulated value of the account will be denoted by A n. We can calculate this as: A n = (1 + R 0 (1 + R 1 (1 + R for the 1 deposited now +(1 + R 1 (1 + R for the 1 deposited at time 1. +(1 + R for the 1 deposited at time n 1 This sum can be re-written in a more compact form as the recursion A n = (A + 1(1 + R (3 which is meaningful for n 1 as long as we define A 1 = 0: for n = 1, A 1 stands for the value of our 1 investment at year 1 which is A 1 = 1 + R 0. Remarks 4.8. 1. This recursion can be interpreted financially as follows: the value of the fund at time n, A n, is equal to the value it had at time n 1, A, plus the 1 contribution of that year, all increased by the interest rate from n 1 to n, (1 + R. 2. Remember that A n is a random variable. As usual, in order to better understand it, we would like to know its variance and variance. 3. Using the recursion (3, A i can be written as a (complicated formula involving R 0, R 1,..., R i 1. For example A 1 = 1 + R 0 A 2 = (A 1 + 1(1 + R 1 = (R 0 + 2(R 1 + 1 A 3 = (A 2 + 1(1 + R 2 = ((R 0 + 2(R 1 + 1 + 1(R 2 + 1. 4. Note that this recursion implies that A i only depends on R 0, R 1,..., R i 1. As R i is independent of all these random variables we conclude that R i and A i are independent. This will be useful later when we calculate the expectation of formula (3. 4.4.2 Expected P&L To calculate the expectation of the value of our investment we apply the expectation operation to the recursion formula above. We would expect this might lead to a similar recursion formula for expectations. Taking expectations in (3 yields: E (A n = E ((A + 1(1 + R = E (1 + A + R + A R = 1 + E (A + E (R + E (A E (R as R and A = 1 + E (A + E(R + E (A E(R are independent = (1 + E (A (1 + E(R 51.
This is the sought-after recursion formula for the expectations: E (A n = (1 + E (A (1 + E(R which is valid for n 1. In general the resulting formulæ for E(A n are quite complicated. For n = 0, 1, 2, if the R i are i.i.d., then we find: E (A 1 = 1 + ρ E (A 2 = (1 + E (A 1 (1 + ρ = (2 + ρ(1 + ρ E (A 3 = (1 + E (A 2 (1 + ρ = (1 + (2 + ρ(1 + ρ (1 + ρ. Remark 4.9. As a quick sanity-check for the formulæ above, note that if the expected rate is zero, ρ = 0, then the expected value of our savings satisfies E(A n = E(A + 1, and as A 1 = 1 + ρ = 1 we get that E(A n = n. This is reasonable: if the expected interest rate is zero, then at time n the expected value of a fund where we place 1 every year is n. 4.4.3 Variance of P&L The calculation of variance, as in 4.2.3, will depend on the calculation of the second moment. To simplify notation a bit, we will assume that the R i are i.i.d. Will use the recursion formula (3. If we square it, we get: A 2 n = (A + 1 2 (1 + R 2 (4 valid for n > 1. Taking expectation will yield the second moment: E ( ( A 2 n = E (A + 1 2 (1 + R 2 = E (( ( 1 + 2A + A 2 1 + 2R + R 2 = E ( ( 1 + 2A + A 2 E 1 + 2R + R 2 as R and A are independent = ( 1 + 2E(A + E(A 2 ( 1 + 2E(R + E(R 2 = ( 1 + 2E(A + Var(A + E(A 2 ( 1 + 2E(R + Var(R + E(R 2 = ( (1 + E(A 2 + Var(A ( (1 + ρ 2 + γ 2. Therefore Var(A n = E ( A 2 n E (An 2 = ( (1 + E(A 2 + Var(A ( (1 + ρ 2 + γ 2 (1 + E(A 2 (1 + ρ 2 = (1 + E(A 2 γ 2 + Var(A ( (1 + ρ 2 + γ 2. This complicated formula tells us that we can calculate Var(A n once we know Var(A and the expectations in 4.4.2. For example: Var(A 1 = Var(1 + R 0 = γ 2 Var(A 2 = (1 + E(A 1 2 γ 2 + Var(A 1 ( (1 + ρ 2 + γ 2 = (2 + ρ 2 γ 2 + γ 2 ( (1 + ρ 2 + γ 2 Var(A 3 = (1 + E(A 2 2 γ 2 + Var(A 2 ( (1 + ρ 2 + γ 2 = (1 + (2 + ρ(1 + ρ 2 γ 2 + [ (2 + ρ 2 γ 2 + γ 2 ( (1 + ρ 2 + γ 2] ( (1 + ρ 2 + γ 2. These formulæ are increasingly unmanageable by hand but can be easily programmed in Excel or some other system. 52