The RT Algorithm (continued)
• Partition a day into n periods.
• Three states follow each state (yt, h2t) after a period.
• As the trinomial model combines, each state at date t is followed by 2n + 1 states at date t + 1 (recall p. 682).
• These 2n + 1 values must approximate the distribution of (yt+1, h2t+1).
• So the conditional moments (115)–(116) at date t + 1 on p. 880 must be matched by the trinomial model to guarantee convergence to the continuous-state model.
The RT Algorithm (continued)
• It remains to pick the jump size and the three branching probabilities.
• The role of σ in the Black-Scholes option pricing model is played by ht in the GARCH model.
• As a jump size proportional to σ/√
n is picked in the BOPM, a comparable magnitude will be chosen here.
• Define γ =Δ h0, though other multiples of h0 are possible, and
γn =Δ √γ .
The RT Algorithm (continued)
• The jump size will be some integer multiple η of γn.
• We call η the jump parameter (see next page).
• Obviously, the magnitude of η grows with ht.
• The middle branch does not change the underlying asset’s price.
(0, 0) yt
(1, 1) (1, 0)
(1, −1)
6?ηγn
-
1 day
The seven values on the right approximate the distribution
The RT Algorithm (continued)
• The probabilities for the up, middle, and down branches are
pu = h2t
2η2γ2 + r − (h2t/2) 2ηγ√
n , (117)
pm = 1 − h2t
η2γ2 , (118)
pd = h2t
2η2γ2 − r − (h2t/2) 2ηγ√
n . (119)
The RT Algorithm (continued)
• It can be shown that:
– The trinomial model takes on 2n + 1 values at date t + 1 for yt+1 .
– These values have a matching mean for yt+1 . – These values have an asymptotically matching
variance for yt+1 .
• The central limit theorem guarantees convergence as n increases (if the probabilities are valid).
The RT Algorithm (continued)
• We can dispense with the intermediate nodes between dates to create a (2n + 1)-nomial tree (p. 890).
• The resulting model is multinomial with 2n + 1 branches from any state (yt, h2t).
• There are two reasons behind this manipulation.
– Interdate nodes are created merely to approximate the continuous-state model after one day.
– Keeping the interdate nodes results in a tree that can be n times larger.a
aContrast it with the case on p. 383.
yt
6?ηγn
-
1 day
This heptanomial tree is the outcome of the trinomial tree
The RT Algorithm (continued)
• A node with logarithmic price yt + ηγn at date t + 1 follows the current node at date t with price yt, where
−n ≤ ≤ n.
• To reach that price in n periods, the number of up moves must exceed that of down moves by exactly .
• The probability that this happens is P () =Δ
ju,jm,jd
n!
ju!jm!jd! pjuupjmmpjdd,
with ju, jm, jd ≥ 0, n = ju + jm + jd, and = ju − jd.
The RT Algorithm (continued)
• A particularly simple way to calculate the P ()s starts by noting that
(pux + pm + pdx−1)n =
n
=−n
P () x.
(120) – Convince yourself that this trick does the
“accounting” correctly.
• So we expand (pux + pm + pdx−1)n and retrieve the probabilities by reading off the coefficients.
The RT Algorithm (continued)
• The updating rule (113) on p. 877 must be modified to account for the adoption of the discrete-state model.
• The logarithmic price yt + ηγn at date t + 1 following state (yt, h2t) is associated with this variance:
h2t+1 = β0 + β1h2t + β2h2t(t+1 − c)2, (121) – Above,
t+1 = ηγn − (r − h2t/2)
ht , = 0, ±1, ±2, . . . , ±n, is a discrete random variable with 2n + 1 values.
The RT Algorithm (continued)
• Different conditional variances h2t may require different η so that the probabilities calculated by
Eqs. (117)–(119) on p. 887 lie between 0 and 1.
• This implies varying jump sizes.
• The necessary requirement pm ≥ 0 implies η ≥ ht/γ.
• Hence we try
η = ht/γ , ht/γ + 1, ht/γ + 2, . . . until valid probabilities are obtained or until their
The RT Algorithm (continued)
• The sufficient and necessary condition for valid probabilities to exist isa
| r − (h2t/2) | 2ηγ√
n ≤ h2t
2η2γ2 ≤ min
1 − | r − (h2t/2) | 2ηγ√
n , 1 2
.
• The plot on p. 896 uses n = 1 to illustrate our points for a 3-day model.
• For example, node (1, 1) of date 1 and node (2, 3) of date 2 pick η = 2.
aC. N. Wu (R90723065) (2003); Lyuu & C. N. Wu (R90723065) (2003, 2005).
y0
(1, 1)
(2, 3)
(2, 0)
(2, −1)
6?γn = γ1
-
3 days
The RT Algorithm (continued)
• The topology of the tree is not a standard combining multinomial tree.
• For example, a few nodes on p. 896 such as nodes (2, 0) and (2, −1) have multiple jump sizes.
• The reason is the path dependence of the model.
– Two paths can reach node (2, 0) from the root node, each with a different variance for the node.
– One of the variances results in η = 1.
– The other results in η = 2.
The RT Algorithm (concluded)
• The number of possible values of h2t at a node can be exponential.
– Because each path brings a different variance h2t.
• To address this problem, we record only the maximum and minimum h2t at each node.a
• Therefore, each node on the tree contains only two states (yt, h2max) and (yt, h2min).
• Each of (yt, h2max) and (yt, h2min) carries its own η and set of 2n + 1 branching probabilities.
Negative Aspects of the Ritchken-Trevor Algorithm
a• A small n may yield inaccurate option prices.
• But the tree will grow exponentially if n is large enough.
– Specifically, n > (1 − β1)/β2 when r = c = 0.
• A large n has another serious problem: The tree cannot grow beyond a certain date.
• Thus the choice of n may be quite limited in practice.
• The RT algorithm can be modified to be free of shortened maturity and exponential complexity.b
aLyuu & C. N. Wu (R90723065) (2003, 2005).
bIts size is only O(n2) if n ≤ (
(1 − β1)/β2 − c)2!
Numerical Examples
• Assume
– S0 = 100, y0 = lnS0 = 4.60517.
– r = 0.
– n = 1.
– h20 = 0.0001096, γ = h0 = 0.010469.
– γn = γ/√
n = 0.010469.
– β0 = 0.000006575, β1 = 0.9, β2 = 0.04, and c = 0.
Numerical Examples (continued)
• A daily variance of 0.0001096 corresponds to an annual volatility of
√365 × 0.0001096 ≈ 20%.
• Let h2(i, j) denote the variance at node (i, j).
• Initially, h2(0, 0) = h20 = 0.0001096.
Numerical Examples (continued)
• Let h2max(i, j) denote the maximum variance at node (i, j).
• Let h2min(i, j) denote the minimum variance at node (i, j).
• Initially, h2max(0, 0) = h2min(0, 0) = h20.
• The resulting three-day tree is depicted on p. 903.
13.4809 13.4809
12.2883 12.2883
11.7170 11.7170
12.2846 10.5733
10.9645 10.9645
10.5697 10.5256
13.4644 10.1305
10.9600 10.9600
10.5215 10.5215
10.9603 10.1269
10.6042 09.7717
10.9553 10.9553
12.2700 10.5173
11.7005 10.1231
10.9511 10.9511
12.2662 10.5135
13.4438 10.9473
yt
4.60517 4.61564 4.62611 4.63658 4.64705 4.65752
4.59470
4.58423
4.57376 0.01047
1 1
2 2
1 1
1 1
2 2
1 1
2 1
2 1
1 1
• A top number inside a gray box refers to the minimum variance h2min for the node.
• A bottom number inside a gray box refers to the maximum variance h2max for the node.
• Variances are multiplied by 100,000 for readability.
• A top number inside a white box refers to the η corresponding to h2min.
• A bottom number inside a white box refers to the η corresponding to h2max.
Numerical Examples (continued)
• Let us see how the numbers are calculated.
• Start with the root node, node (0, 0).
• Try η = 1 in Eqs. (117)–(119) on p. 887 first to obtain pu = 0.4974,
pm = 0,
pd = 0.5026.
• As they are valid probabilities, the three branches from the root node use single jumps.
Numerical Examples (continued)
• Move on to node (1, 1).
• It has one predecessor node—node (0, 0)—and it takes an up move to reach the current node.
• So apply updating rule (121) on p. 893 with = 1 and h2t = h2(0, 0).
• The result is h2(1, 1) = 0.000109645.
Numerical Examples (continued)
• Because h(1, 1)/γ = 2, we try η = 2 in Eqs. (117)–(119) on p. 887 first to obtain
pu = 0.1237, pm = 0.7499, pd = 0.1264.
• As they are valid probabilities, the three branches from node (1, 1) use double jumps.
Numerical Examples (continued)
• Carry out similar calculations for node (1, 0) with
= 0 in updating rule (121) on p. 893.
• Carry out similar calculations for node (1, −1) with
= −1 in updating rule (121).
• Single jump η = 1 works for both nodes.
• The resulting variances are
h2(1, 0) = 0.000105215, h2(1, −1) = 0.000109553.
Numerical Examples (continued)
• Node (2, 0) has 2 predecessor nodes, (1, 0) and (1, −1).
• Both have to be considered in deriving the variances.
• Let us start with node (1, 0).
• Because it takes a middle move to reach the current
node, we apply updating rule (121) on p. 893 with = 0 and h2t = h2(1, 0).
• The result is h2t+1 = 0.000101269.
Numerical Examples (continued)
• Now move on to the other predecessor node (1, −1).
• Because it takes an up move to reach the current node, apply updating rule (121) on p. 893 with = 1 and h2t = h2(1, −1).
• The result is h2t+1 = 0.000109603.
• We hence record
h2min(2, 0) = 0.000101269, h2max(2, 0) = 0.000109603.
Numerical Examples (continued)
• Consider state h2max(2, 0) first.
• Because hmax(2, 0)/γ = 2, we first try η = 2 in Eqs. (117)–(119) on p. 887 to obtain
pu = 0.1237, pm = 0.7500, pd = 0.1263.
• As they are valid probabilities, the three branches from node (2, 0) with the maximum variance use double
jumps.
Numerical Examples (continued)
• Now consider state h2min(2, 0).
• Because hmin(2, 0)/γ = 1, we first try η = 1 in Eqs. (117)–(119) on p. 887 to obtain
pu = 0.4596, pm = 0.0760, pd = 0.4644.
• As they are valid probabilities, the three branches from node (2, 0) with the minimum variance use single jumps.
Numerical Examples (continued)
• Node (2, −1) has 3 predecessor nodes.
• Start with node (1, 1).
• Because it takes one down move to reach the current node, we apply updating rule (121) on p. 893 with
= −1 and h2t = h2(1, 1).a
• The result is h2t+1 = 0.0001227.
aNote that it is not = −2.
Numerical Examples (continued)
• Now move on to predecessor node (1, 0).
• Because it also takes a down move to reach the current node, we apply updating rule (121) on p. 893 with
= −1 and h2t = h2(1, 0).
• The result is h2t+1 = 0.000105609.
Numerical Examples (continued)
• Finally, consider predecessor node (1, −1).
• Because it takes a middle move to reach the current
node, we apply updating rule (121) on p. 893 with = 0 and h2t = h2(1, −1).
• The result is h2t+1 = 0.000105173.
• We hence record
h2min(2, −1) = 0.000105173, h2max(2, −1) = 0.0001227.
Numerical Examples (continued)
• Consider state h2max(2, −1).
• Because hmax(2, −1)/γ = 2, we first try η = 2 in Eqs. (117)–(119) on p. 887 to obtain
pu = 0.1385, pm = 0.7201, pd = 0.1414.
• As they are valid probabilities, the three branches from node (2, −1) with the maximum variance use double
Numerical Examples (continued)
• Next, consider state h2min(2, −1).
• Because hmin(2, −1)/γ = 1, we first try η = 1 in Eqs. (117)–(119) on p. 887 to obtain
pu = 0.4773, pm = 0.0404, pd = 0.4823.
• As they are valid probabilities, the three branches from node (2, −1) with the minimum variance use single
jumps.
Numerical Examples (concluded)
• Other nodes at dates 2 and 3 can be handled similarly.
• In general, if a node has k predecessor nodes, then up to 2k variances will be calculated using the updating rule.
– This is because each predecessor node keeps two variance numbers.
• But only the maximum and minimum variances will be kept.
Negative Aspects of the RT Algorithm Revisited
a• Recall the problems mentioned on p. 899.
• In our case, combinatorial explosion occurs when n > 1 − β1
β2 = 1 − 0.9
0.04 = 2.5 (see the next plot).
• Suppose we are willing to accept the exponential running time and pick n = 100 to seek accuracy.
• But the problem of shortened maturity forces the tree to stop at date 9!
aLyuu & C. N. Wu (R90723065) (2003, 2005).
25 50 75 100 125 150 175 Date 5000
10000 15000 20000 25000
Backward Induction on the RT Tree
• After the RT tree is constructed, it can be used to price options by backward induction.
• Recall that each node keeps two variances h2max and h2min.
• We now increase that number to K equally spaced variances between h2max and h2min at each node.
• Besides the minimum and maximum variances, the other K − 2 variances in between are linearly interpolated.a
aIn practice, log-linear interpolation works better (Lyuu & C. N. Wu (R90723065), 2005). Log-cubic interpolation works even better (C. Liu (R92922123), 2005).
Backward Induction on the RT Tree (continued)
• For example, if K = 3, then a variance of 10.5436 × 10−6
will be added between the maximum and minimum variances at node (2, 0) on p. 903.a
• In general, the kth variance at node (i, j) is h2min(i, j)+k h2max(i, j) − h2min(i, j)
K − 1 , k = 0, 1, . . . , K−1.
• Each interpolated variance’s jump parameter and branching probabilities can be computed as before.
13.4809 13.4809
12.2883 12.2883
11.7170 11.7170
12.2846 10.5733
10.9645 10.9645
10.5697 10.5256
13.4644 10.1305
10.9600 10.9600
10.5215 10.5215
10.9603 10.1269
10.6042 09.7717
10.9553 10.9553
12.2700 10.5173
11.7005 10.1231
10.9511 10.9511
12.2662 10.5135
13.4438 10.9473
yt
4.60517 4.61564 4.62611 4.63658 4.64705 4.65752
4.59470
4.58423
4.57376 0.01047
1 1
2 2
1 1
1 1
2 2
1 1
2 1
2 1
1 1
Backward Induction on the RT Tree (concluded)
• Suppose a variance falls between two of the K variances during backward induction.
• Linear interpolation of the option prices corresponding to the two bracketing variances will be used as the
approximate option price.
• The above ideas are reminiscent of the ones on p. 409, where we dealt with Asian options.
Numerical Examples
• We next use the tree on p. 923 to price a European call option with a strike price of 100 and expiring at date 3.
• Recall that the riskless interest rate is zero.
• Assume K = 2; hence there are no interpolated variances.
• The pricing tree is shown on p. 926 with a call price of 0.66346.
– The branching probabilities needed in backward induction can be found on p. 927.
5.37392 5.37392
3.19054 3.19054
3.19054 3.19054
2.11587 2.11587
1.20241 1.20241
1.05240 1.05240
1.05240 1.05240
0.66346 0.66346
0.52360 0.52360
0.26172 0.48366
0.00000 0.00000
0.13012 0.13012
0.14573 0.00000
0.00000 0.00000
0.00000 0.00000
0.00000 0.00000
St
100.00000 101.05240 102.11587 103.19054 104.27652 105.37392
98.95856
97.92797
1 1
2 2
1 1
1 1
2 2
1 1
2 1
2 1
1 1
h2[i][j][0]
η[i][j][0]
rb[i][0]
10.9600 10.9600
10.9553 10.9553
10.5215 10.5215
10.9645 10.9645
10.9511 10.9511
12.2700 10.5173
10.9603 10.1269
10.5697 10.5256
12.2883 12.2883
13.4438 10.9473
12.2662 10.5135
11.7005 10.1231
10.6042 09.7717
13.4644 10.1305
12.2846 10.5733
11.7170 11.7170
13.4809 13.4809
h2[0][ ][ ]
1 1
1 1
1 1
2 2
1 1
2 1
2 1
1 1
2 2
η[0][ ][ ]
p[0][ ][ ][ ] 0
0
1
−1
3
−2
5
−3 rb[0][ ]
0.4974 0.4974 0.0000 0.0000 0.5026 0.5026
0.4972 0.4972 0.0004 0.0004 0.5024 0.5024 0.4775 0.4775 0.0400 0.0400 0.4825 0.4825 0.1237 0.1237 0.7499 0.7499 0.1264 0.1264
0.4970 0.4970 0.0008 0.0008 0.5022 0.5022 0.4773 0.1385 0.0404 0.7201 0.4823 0.1414 0.4596 0.1237 0.0760 0.7500 0.4644 0.1263 0.4777 0.4797 0.0396 0.0356 0.4827 0.4847 0.1387 0.1387 0.7197 0.7197 0.1416 0.1416
h2[1][ ][ ]
h2[2][ ][ ]
h2[3][ ][ ]
η[1][ ][ ]
η[2][ ][ ]
p[1][ ][ ][ ]
p[2][ ][ ][ ] rb[i][1]
h2[i][j][1]
η[i][j][1]
p[i][j][0][1] p[i][j][1][1]
p[i][j][0][0] p[i][j][1][0]
p[i][j][0][−1] p[i][j][1][−1]
rb[1][ ] rb[2][ ] rb[3][ ]
j
3 2 1 0
−1
−2
3 2 1 0
−1
−2 j
5 4 3 2
0
j
−3
−2
−1 1
Numerical Examples (continued)
• Let us derive some of the numbers on p. 926.
• A gray line means the updated variance falls strictly between h2max and h2min.
• The option price for a terminal node at date 3 equals max(S3 − 100, 0), independent of the variance level.
• Now move on to nodes at date 2.
• The option price at node (2, 3) depends on those at nodes (3, 5), (3, 3), and (3, 1).
• It therefore equals
Numerical Examples (continued)
• Option prices for other nodes at date 2 can be computed similarly.
• For node (1, 1), the option price for both variances is
0.1237 × 3.19054 + 0.7499 × 1.05240 + 0.1264 × 0.14573 = 1.20241.
• Node (1, 0) is most interesting.
• We knew that a down move from it gives a variance of 0.000105609.
• This number falls between the minimum variance
0.000105173 and the maximum variance 0.0001227 at node (2, −1) on p. 923.
Numerical Examples (continued)
• The option price corresponding to the minimum variance is 0 (p. 927).
• The option price corresponding to the maximum variance is 0.14573.
• The equation
x × 0.000105173 + (1 − x) × 0.0001227 = 0.000105609 is satisfied by x = 0.9751.
• So the option for the down state is approximated by
Numerical Examples (continued)
• The up move leads to the state with option price 1.05240.
• The middle move leads to the state with option price 0.48366.
• The option price at node (1, 0) is finally calculated as
0.4775 × 1.05240 + 0.0400 × 0.48366 + 0.4825 × 0.00362 = 0.52360.
Numerical Examples (continued)
• A variance following an interpolated variance may exceed the maximum variance or be exceeded by the minimum variance.
• When this happens, the option price corresponding to the maximum or minimum variance will be used during backward induction.a
aCakici & Topyan (2000).
Numerical Examples (concluded)
• But an interpolated variance may choose a branch that goes into a node that is not reached in forward
induction.a
• In this case, the algorithm fails.
• The RT algorithm does not have this problem as all interpolated variances are involved in the
forward-induction phase.
• It may be hard to calculate the implied β1 and β2 from option prices.b
aLyuu & C. N. Wu (R90723065) (2005).
bY. Chang (R93922034) (2006).
Complexities of GARCH Models
a• The RT algorithm explodes exponentially if n is big enough (p. 899).
• The mean-tracking tree of Lyuu and Wu (2005) makes sure explosion does not happen if n is not too large.b
• The next page summarizes the situations for many GARCH option pricing models.
– Our earlier treatment is for NGARCH only.
aLyuu & C. N. Wu (R90723065) (2003, 2005).
bSimilar to, but earlier than, the binomial-trinomial tree on pp. 703ff.
Complexities of GARCH Models (concluded)
aModel Explosion Non-explosion
NGARCH β1 + β2n > 1 β1 + β2(√
n + λ + c)2 ≤ 1 LGARCH β1 + β2n > 1 β1 + β2(√
n + λ)2 ≤ 1 AGARCH β1 + β2n > 1 β1 + β2(√
n + λ)2 ≤ 1 GJR-GARCH β1 + β2n > 1 β1 + (β2 + β3)(√
n + λ)2 ≤ 1 TS-GARCH β1 + β2√
n > 1 β1 + β2(λ +√
n) ≤ 1 TGARCH β1 + β2√
n > 1 β1 + (β2 + β3)(λ + √
n) ≤ 1 Heston-Nandi β1 + β2(c − 12)2 > 1 β1 + β2c2 ≤ 1
& c ≤ 12
VGARCH β1 + (β2/4) > 1 β1 ≤ 1
aY. Chen (R95723051) (2008); Y. Chen (R95723051), Lyuu, & Wen (D94922003) (2012).
Introduction to Term Structure Modeling
The fox often ran to the hole by which they had come in, to find out if his body was still thin enough to slip through it.
— Grimm’s Fairy Tales
And the worst thing you can have is models and spreadsheets.
— Warren Buffet, May 3, 2008
Outline
• Use the binomial interest rate tree to model stochastic term structure.
– Illustrates the basic ideas underlying future models.
– Applications are generic in that pricing and hedging methodologies can be easily adapted to other models.
• Although the idea is similar to the earlier one used in option pricing, the current task is more complicated.
– The evolution of an entire term structure, not just a single stock price, is to be modeled.
– Interest rates of various maturities cannot evolve arbitrarily, or arbitrage profits may occur.
Issues
• A stochastic interest rate model performs two tasks.
– Provides a stochastic process that defines future term structures without arbitrage profits.
– “Consistent” with the observed term structures.
History
• The methodology was founded by Merton (1970).
• Modern interest rate modeling is often traced to 1977 when Vasicek and Cox, Ingersoll, and Ross developed simultaneously their influential models.
• Early models have fitting problems because they may not price today’s benchmark bonds correctly.
• An alternative approach pioneered by Ho and Lee (1986) makes fitting the market yield curve mandatory.
• Models based on such a paradigm are called (somewhat misleadingly) arbitrage-free or no-arbitrage models.
Binomial Interest Rate Tree
• Goal is to construct a no-arbitrage interest rate tree consistent with the yields and/or yield volatilities of zero-coupon bonds of all maturities.
– This procedure is called calibration.a
• Pick a binomial tree model in which the logarithm of the future short rate obeys the binomial distribution.
– Exactly like the CRR tree.
• The limiting distribution of the short rate at any future time is hence lognormal.
Binomial Interest Rate Tree (continued)
• A binomial tree of future short rates is constructed.
• Every short rate is followed by two short rates in the following period (p. 944).
• In the figure on p. 944, node A coincides with the start of period j during which the short rate r is in effect.
• At the conclusion of period j, a new short rate goes into effect for period j + 1.
r
* r
0.5
j rh
0.5 A
B
C
period j − 1 period j period j + 1 time j − 1 time j
Binomial Interest Rate Tree (continued)
• This may take one of two possible values:
– r: the “low” short-rate outcome at node B.
– rh: the “high” short-rate outcome at node C.
• Each branch has a 50% chance of occurring in a risk-neutral economy.
• We require that the paths combine as the binomial process unfolds.
• This model can be traced to Salomon Brothers.a
aTuckman (2002).
Binomial Interest Rate Tree (continued)
• The short rate r can go to rh and r with equal
risk-neutral probability 1/2 in a period of length Δt.
• Hence the volatility of ln r after Δt time is σ = 1
2
√1
Δt ln
rh r
(122) (see Exercise 23.2.3 in text).
• Above, σ is annualized, whereas r and rh are period based.
Binomial Interest Rate Tree (continued)
• Note that
rh
r = e2σ√Δt.
• Thus greater volatility, hence uncertainty, leads to larger rh/r and wider ranges of possible short rates.
• The ratio rh/r may depend on time if the volatility is a function of time.
• Note that rh/r has nothing to do with the current short rate r if σ is independent of r.
Binomial Interest Rate Tree (continued)
• In general there are j possible ratesa for period j, rj, rjvj, rjvj2, . . . , rjvjj−1,
where
vj =Δ e2σj√Δt (123) is the multiplicative ratio for the rates in period j (see figure on next page).
• We shall call rj the baseline rates.
• The subscript j in σj above is meant to emphasize that
Baseline rates
A C
B B
C
C
D
D D H D
H L
H L! !
H L! ! H
H!
Binomial Interest Rate Tree (concluded)
• In the limit, the short rate follows
r(t) = μ(t) eσ(t) W (t). (124) – The (percent) short rate volatility σ(t) is a
deterministic function of time.
• The expected value of r(t) equals μ(t) eσ(t)2(t/2).
• Hence a declining short rate volatility is usually imposed to preclude the short rate from assuming implausibly high values.
• Incidentally, this is how the binomial interest rate tree
Memory Issues
• Path independency: The term structure at any node is independent of the path taken to reach it.
• So only the baseline rates ri and the multiplicative ratios vi need to be stored in computer memory.
• This takes up only O(n) space.a
• Storing the whole tree would take up O(n2) space.
– Daily interest rate movements for 30 years require roughly (30 × 365)2/2 ≈ 6 × 107 double-precision floating-point numbers (half a gigabyte!).
aThroughout, n denotes the depth of the tree.
Set Things in Motion
• The abstract process is now in place.
• We need the annualized rates of return of the riskless bonds that make up the benchmark yield curve and their volatilities.
• In the U.S., for example, the on-the-run yield curve
obtained by the most recently issued Treasury securities may be used as the benchmark curve.
Set Things in Motion (concluded)
• The term structure of (yield) volatilitiesa can be estimated from:
– Historical data (historical volatility).
– Or interest rate option prices such as cap prices (implied volatility).
• The binomial tree should be found that is consistent with both term structures.
• Here we focus on the term structure of interest rates.
aOr simply the volatility (term) structure.
Model Term Structures
• The model price is computed by backward induction.
• Refer back to the figure on p. 944.
• Given that the values at nodes B and C are PB and PC, respectively, the value at node A is then
PB + PC
2(1 + r) + cash flow at node A.
• We compute the values column by column without
explicitly expanding the binomial interest rate tree (see next page).
HL
A C
B
Cash flows:
B
C
C
D
D
D D
+
+ 2 2 H
= B
+ 2 2
HL
!
= B
H
HL
+ 2 2
HL
! "
? D
2
2
2!
2"
Term Structure Dynamics
• An n-period zero-coupon bond’s price can be computed by assigning $1 to every node at period n and then
applying backward induction.
• Repeating this step for n = 1, 2, . . . , one obtains the market discount function implied by the tree.
• The tree therefore determines a term structure.
• It also contains a term structure dynamics.
– Taking any node in the tree as the current state induces a binomial interest rate tree and, again, a
