GARCH Option Pricing

Time Series Analysis

The historian is a prophet in reverse.

— Friedrich von Schlegel (1772–1829)

GARCH Option Pricing

• Options can be priced when the underlying asset’s return follows a GARCH process.

• Let S_t denote the asset price at date t.

• Let h²_t be the conditional variance of the return over the period [t, t + 1) given the information at date t.

– “One day” is merely a convenient term for any elapsed time Δt.

aARCH (autoregressive conditional heteroskedastic) is due to Engle (1982), co-winner of the 2003 Nobel Prize in Economic Sciences. GARCH (generalized ARCH) is due to Bollerslev (1986) and Taylor (1986). A Bloomberg quant said to me on Feb 29, 2008, that GARCH is seldom used in trading.

GARCH Option Pricing (continued)

• Adopt the following risk-neutral process for the price dynamics:^a

ln S_t+1

S_t = r − h²_t

2 + h_t_t+1, (120) where

h²_t+1 = β₀ + β₁h²_t + β₂h²_t(_t+1 − c)², (121)

_t+1 ∼ N(0, 1) given information at date t, r = daily riskless return,

c ≥ 0.

aDuan (1995).

GARCH Option Pricing (continued)

• The five unknown parameters of the model are c, h₀, β₀, β₁, and β₂.

• It is postulated that β₀, β₁, β₂ ≥ 0 to make the conditional variance positive.

• There are other inequalities to satisfy (see text).

• The above process is called the nonlinear asymmetric GARCH (or NGARCH) model.

GARCH Option Pricing (continued)

• It captures the volatility clustering in asset returns first noted by Mandelbrot (1963).^a

– When c = 0, a large _t+1 results in a large h_t+1,

which in turns tends to yield a large h_t+2, and so on.

• It also captures the negative correlation between the asset return and changes in its (conditional) volatility.^b

– For c > 0, a positive _t+1 (good news) tends to decrease h_t+1, whereas a negative _t+1 (bad news) tends to do the opposite.

a“. . . large changes tend to be followed by large changes—of either sign—and small changes tend to be followed by small changes . . . ”

bNoted by Black (1976): Volatility tends to rise in response to “bad news” and fall in response to “good news.”

GARCH Option Pricing (continued)

• With y_t = ln^Δ S_t denoting the logarithmic price, the model becomes

y_t+1 = y_t + r − h²_t

2 + h_t_t+1. (122)

• The pair (y_t, h²_t) completely describes the current state.

GARCH Option Pricing (concluded)

• The conditional mean and variance of y_t+1 are clearly E[ y_t+1 | y_t, h²_t ] = y_t + r − h²_t

2 , (123) Var[y_t+1 | y_t, h²_t ] = h²_t. (124)

• Finally, given (y_t, h²_t), the correlation between y_t+1 and h_t+1 equals

−√ 2c

2 + 4c² , which is negative for c > 0.

GARCH Model: Inferences

• Suppose the parameters c, h₀, β₀, β₁, and β₂ are given.

• Then we can recover h₁, h₂, . . . , h_n and ₁, ₂, . . . , _n from the prices

S₀, S₁, . . . , S_n

under the GARCH model (120) on p. 936.

• This property is useful in statistical inferences.

The Ritchken-Trevor (RT) Algorithm

• The GARCH model is a continuous-state model.

• To approximate it, we turn to trees with discrete states.

• Path dependence in GARCH makes the tree for asset prices explode exponentially (why?).

• We need to mitigate this combinatorial explosion.

aRitchken & Trevor (1999).

The RT Algorithm (continued)

• Partition a day into n periods.

• Three states follow each state (y_t, h²_t) 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. 722).

• These 2n + 1 values must approximate the distribution of (y_t+1, h²_t+1).

• So the conditional moments (123)–(124) at date t + 1 on p. 940 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 h_t 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

γ_n =^Δ √γ n.

The RT Algorithm (continued)

• The jump size will be some integer multiple η of γ_n.

• We call η the jump parameter (see next page).

• Obviously, the magnitude of η grows with h_t.

• 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 of logarithmic price y_t+1.

The RT Algorithm (continued)

• The probabilities for the up, middle, and down branches are

p_u = h²_t

2η²γ² + r − (h²_t/2) 2ηγ√

n , (125)

p_m = 1 − h²_t

η²γ² , (126)

p_d = h²_t

2η²γ² − r − (h²_t/2) 2ηγ√

n . (127)

The RT Algorithm (continued)

• It can be shown that:

– The trinomial model takes on 2n + 1 values at date t + 1 for y_t+1 .

– These values have a matching mean for y_t+1 . – These values have an asymptotically matching

variance for y_t+1 .

• The central limit theorem guarantees convergence as n increases.^a

aAssume 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. 950).

• The resulting model is multinomial with 2n + 1 branches from any state (y_t, h²_t).

• 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. 401.

yt

6?^ηγn

-

 _{1 day}

This heptanomial tree is the outcome of the trinomial tree on p. 946 after its intermediate nodes are removed.

The RT Algorithm (continued)

• A node with logarithmic price y_t + ηγ_n at date t + 1 follows the current node at date t with price y_t, 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!

j_u!j_m!j_d! p^j_u^up^j_m^mp^j_d^d,

with j_u, j_m, j_d ≥ 0, n = j_u + j_m + j_d, and  = j_u − j_d.

The RT Algorithm (continued)

• A particularly simple way to calculate the P ()s starts by noting that^a

(p_ux + p_m + p_dx⁻¹)ⁿ =

n

=−n

P () x^.

(128) – Convince yourself that this trick does the

“accounting” correctly.

• So we expand (p_ux + p_m + p_dx⁻¹)ⁿ and retrieve the probabilities by reading off the coefficients.

• It can be computed in O(n²) time, if not less.

aC. Wu (R90723065) (2003); Lyuu & C. Wu (R90723065) (2003, 2005).

The RT Algorithm (continued)

• The updating rule (121) on p. 936 must be modified to account for the adoption of the discrete-state model.

• The logarithmic price y_t + ηγ_n at date t + 1 following state (y_t, h²_t) is associated with this variance:

h²_t+1 = β₀ + β₁h²_t + β₂h²_t(^_t+1 − c)², (129) – Above, the z-score

^_t+1 = ηγ_n − (r − h²_t/2)

h_t ,  = 0, ±1, ±2, . . . , ±n, is a discrete random variable with 2n + 1 values.

The RT Algorithm (continued)

• Different conditional variances h²_t may require different η so that the probabilities calculated by

Eqs. (125)–(127) on p. 947 lie between 0 and 1.

• This implies varying jump sizes.

• The necessary requirement p_m ≥ 0 implies η ≥ h_t/γ.

• Hence we try

η =  h_t/γ ,  h_t/γ  + 1,  h_t/γ  + 2, . . . until valid probabilities are obtained or until their nonexistence is confirmed.

The RT Algorithm (continued)

• The sufficient and necessary condition for valid probabilities to exist is^a

| r − (h²_t/2) | 2ηγ√

n ≤ h²_t

2η²γ² ≤ min



1 − | r − (h²_t/2) | 2ηγ√

n , 1 2

 .

• The plot on p. 956 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. Wu (R90723065) (2003); Lyuu & C. 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. 956 such as nodes (2, 0) and (2, −1) have multiple jump sizes.

• The reason is path dependency of the model.

– Two paths can reach node (2, 0) from the root node, each with a different variance for the node.

– One variance results in η = 1.

– The other results in η = 2.

The RT Algorithm (concluded)

• The number of possible values of h²_t at a node can be exponential.

– Because each path brings a different variance h²_t.

• To address this problem, we record only the maximum and minimum h²_t at each node.^a

• Therefore, each node on the tree contains only two states (y_t, h²_max) and (y_t, h²_min).

• Each of (y_t, h²_max) and (y_t, h²_min) carries its own η and set of 2n + 1 branching probabilities.

aCakici & Topyan (2000). But see p. 993 for a potential problem.

Negative Aspects of the Ritchken-Trevor Algorithm

• 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. 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.

– γ_n = 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 h²_max(i, j) denote the maximum variance at node (i, j).

• Let h²_min(i, j) denote the minimum variance at node (i, j).

• Initially, h²_max(0, 0) = h²_min(0, 0) = h²₀.

• The resulting 3-day tree is depicted on p. 963.

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

y_t

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

Numerical Examples (continued)

• A top number inside a gray box refers to the minimum variance h²_min for the node.

• A bottom number inside a gray box refers to the maximum variance h²_max for the node.

• Variances are multiplied by 100,000 for readability.

• The top number inside a white box refers to the η for h²_min.

• The bottom number inside a white box refers to the η for h²_max.

Numerical Examples (continued)

• Let us see how the numbers are calculated.

• Start with the root node, node (0, 0).

• Try η = 1 in Eqs. (125)–(127) on p. 947 first to obtain p_u = 0.4974,

p_m = 0,

p_d = 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 (129) on p. 953 with  = 1 and h²_t = 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. (125)–(127) on p. 947 first to obtain

p_u = 0.1237, p_m = 0.7499, p_d = 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 (129) on p. 953.

• Carry out similar calculations for node (1, −1) with

 = −1 in updating rule (129).

• 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 (129) on p. 953 with  = 0 and h²_t = h²(1, 0).

• The result is h²_t+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 (129) on p. 953 with  = 1 and h²_t = h²(1, −1).

• The result is h²_t+1 = 0.000109603.

• We hence record

h²_min(2, 0) = 0.000101269, h²_max(2, 0) = 0.000109603.

Numerical Examples (continued)

• Consider state h²_max(2, 0) first.

• Because  h_max(2, 0)/γ  = 2, we first try η = 2 in Eqs. (125)–(127) on p. 947 to obtain

p_u = 0.1237, p_m = 0.7500, p_d = 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 h²_min(2, 0).

• Because  h_min(2, 0)/γ  = 1, we first try η = 1 in Eqs. (125)–(127) on p. 947 to obtain

p_u = 0.4596, p_m = 0.0760, p_d = 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 (129) on p. 953 with

 = −1 and h²_t = h²(1, 1).^a

• The result is h²_t+1 = 0.0001227.

aNote that it is not  = −2. The reason is that h(1, 1) has η = 2 (p.

967).

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 (129) on p. 953 with

 = −1 and h²_t = h²(1, 0).

• The result is h²_t+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 (129) on p. 953 with  = 0 and h²_t = h²(1, −1).

• The result is h²_t+1 = 0.000105173.

• We hence record

h²_min(2, −1) = 0.000105173, h²_max(2, −1) = 0.0001227.

Numerical Examples (continued)

• Consider state h²_max(2, −1).

• Because  h_max(2, −1)/γ  = 2, we first try η = 2 in Eqs. (125)–(127) on p. 947 to obtain

p_u = 0.1385, p_m = 0.7201, p_d = 0.1414.

• As they are valid probabilities, the three branches from node (2, −1) with the maximum variance use double jumps.

Numerical Examples (continued)

• Next, consider state h²_min(2, −1).

• Because  h_min(2, −1)/γ  = 1, we first try η = 1 in Eqs. (125)–(127) on p. 947 to obtain

p_u = 0.4773, p_m = 0.0404, p_d = 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

• Recall the problems mentioned on p. 959.

• 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. Wu (R90723065) (2003, 2005).

25 50 75 100 125 150 175 Date 5000

10000 15000 20000 25000

Dotted line: n = 3; dashed line: n = 4; solid line: n = 5.

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 h²_max and h²_min.

• We now increase that number to K equally spaced variances between h²_max and h²_min 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. 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. 963.^a

• In general, the kth variance at node (i, j) is h²_min(i, j)+k h²_max(i, j) − h²_min(i, j)

K − 1 , k = 0, 1, . . . , K−1.

• Each interpolated variance’s jump parameter and branching probabilities can be computed as before.

aRepeated on p. 983.

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

y_t

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. 441, where we dealt with Asian options.

Numerical Examples

• We next use the tree on p. 983 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. 986 with a call price of 0.66346.

– The branching probabilities needed in backward induction can be found on p. 987.

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

0.00000 0.00000

S_t

100.00000 101.05240 102.11587 103.19054 104.27652 105.37392

98.95856

97.92797

96.90811

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 0.4773 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. 986.

• A gray line means the updated variance falls strictly between h²_max and h²_min.

• 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

0.1387 × 5.37392 + 0.7197 × 3.19054 + 0.1416 × 1.05240 = 3.19054.

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. 987.

Numerical Examples (continued)

• The option price corresponding to the minimum variance is 0 (p. 987).

• 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 x × 0 + (1 − x) × 0.14573 = 0.00362.

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

• This act tends to reduce the dynamic range of the variance.

aCakici & Topyan (2000).

Numerical Examples (concluded)

• Worse, 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.

– This is because 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. Wu (R90723065) (2005).

bY. Chang (B89704039, R93922034) (2006).

Complexities of GARCH Models

• The RT algorithm explodes exponentially if n is big enough (p. 959).

• 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. Wu (R90723065) (2003, 2005).

bSimilar to, but earlier than, the binomial-trinomial tree on pp. 745ff.

Complexities of GARCH Models (concluded)

Model 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. C. Chen (R95723051) (2008); Y. C. 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.

– Like the CRR tree for pricing options.

• The limiting distribution of the short rate at any future time is hence lognormal.

aDerman (2004), “complexity without calibration is pointless.”

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. 1004).

• In the figure on p. 1004, 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.

– r_h: 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 is attributed to Salomon Brothers.^a

aTuckman (2002).

Binomial Interest Rate Tree (continued)

• The short rate r can go to r_h 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^a σ = 1

2

√1

Δt ln

r_h r_



. (130)

• Above, σ is annualized, whereas r_ and r_h are period based.

aSee Exercise 23.2.3 in text.

Binomial Interest Rate Tree (continued)

• Note that

r_h

r_ = e^2σ√Δt.

• Thus greater volatility, hence uncertainty, leads to larger r_h/r_ and wider ranges of possible short rates.

• The ratio r_h/r_ may depend on time if the volatility is a function of time.

• Note that r_h/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 rates^a for period j, r_j, r_jv_j, r_jv_j², . . . , r_jv_j^j−1,

where

v_j =^Δ e^2σj√Δt (131) is the multiplicative ratio for the rates in period j (see figure on next page).

• We shall call r_j the baseline rates.

• The subscript j in σ_j above is meant to emphasize that the short rate volatility may be time dependent.

aNot j + 1.

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) e^{σ(t) W (t)}. (132) – The (percent) short rate volatility σ(t) is a

deterministic function of time.

• The expected value of r(t) equals μ(t) e^σ(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 achieves mean reversion to some long-term mean.

Memory Issues

• Path independency: The term structure at any node is independent of the path taken to reach it.

• So only the baseline rates r_i and the multiplicative ratios v_i 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.