# The Black-Karasinski Model

## Full text

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### The Black-Karasinski Model

a

• The BK model stipulates that the short rate follows d ln r = κ(t)(θ(t) − ln r) dt + σ(t) dW.

• This explicitly mean-reverting model depends on time through κ( · ), θ( · ), and σ( · ).

• The BK model hence has one more degree of freedom than the BDT model.

• The speed of mean reversion κ(t) and the short rate volatility σ(t) are independent.

aBlack and Karasinski (1991).

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### The Black-Karasinski Model: Discrete Time

• The discrete-time version of the BK model has the same representation as the BDT model.

• To maintain a combining binomial tree, however, requires some manipulations.

• The next plot illustrates the ideas in which t2 ≡ t1 + Δt1,

t3 ≡ t2 + Δt2.

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 ln rd(t2)

 

ln r(t1) ln rdu(t3) = ln rud(t3)

 

ln ru(t2)



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### The Black-Karasinski Model: Discrete Time (continued)

• Note that

lnrd(t2) = lnr(t1) +κ(t1)(θ(t1) − ln r(t1)) Δt1 − σ(t1)

Δt1 , lnru(t2) = lnr(t1) +κ(t1)(θ(t1) − ln r(t1)) Δt1 + σ(t1)

Δt1 .

• To make sure an up move followed by a down move coincides with a down move followed by an up move,

ln rd(t2) + κ(t2)(θ(t2) − ln rd(t2)) Δt2 + σ(t2)

Δt2 ,

= ln ru(t2) + κ(t2)(θ(t2) − ln ru(t2)) Δt2 − σ(t2)

Δt2 .

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### The Black-Karasinski Model: Discrete Time (continued)

• They imply

κ(t2) = 1 − (σ(t2)/σ(t1))

Δt2/Δt1

Δt2 .

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• So from Δt0, we can calculate the Δt1 that satisﬁes the combining condition and then iterate.

– t0 → Δt0 → t1 → Δt1 → t2 → Δt2 → · · · → T (roughly).a

aAs κ(t), θ(t), σ(t) are independent of r, the Δtis will not depend on r.

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### The Black-Karasinski Model: Discrete Time (concluded)

• Unequal durations Δti are often necessary to ensure a combining tree.a

aAmin (1991); Chen (R98922127) (2011); Lok (D99922028) and Lyuu (2016).

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### Problems with Lognormal Models in General

• Lognormal models such as BDT and BK share the problem that Eπ[ M (t) ] = ∞ for any ﬁnite t if they model the continuously compounded rate.a

• So periodically compounded rates should be modeled.b

• Another issue is computational.

• Lognormal models usually do not admit of analytical solutions to even basic ﬁxed-income securities.

• As a result, to price short-dated derivatives on long-term bonds, the tree has to be built over the life of the

underlying asset instead of the life of the derivative.

aHogan and Weintraub (1993).

bSandmann and Sondermann (1993).

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### Problems with Lognormal Models in General (concluded)

• This problem can be somewhat mitigated by adopting diﬀerent time steps.a

– Use a ﬁne time step up to the maturity of the short-dated derivative.

– Use a coarse time step beyond the maturity.

• A down side of this procedure is that it has to be tailor-made for each derivative.

• Finally, empirically, interest rates do not follow the lognormal distribution.

aHull and White (1993).

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### The Extended Vasicek Model

a

• Hull and White proposed models that extend the Vasicek model and the CIR model.

• They are called the extended Vasicek model and the extended CIR model.

• The extended Vasicek model adds time dependence to the original Vasicek model,

dr = (θ(t) − a(t) r) dt + σ(t) dW.

• Like the Ho-Lee model, this is a normal model.

• The inclusion of θ(t) allows for an exact ﬁt to the current spot rate curve.

aHull and White (1990).

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### The Extended Vasicek Model (concluded)

• Function σ(t) deﬁnes the short rate volatility, and a(t) determines the shape of the volatility structure.

• Many European-style securities can be evaluated analytically.

• Eﬃcient numerical procedures can be developed for American-style securities.

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### The Hull-White Model

• The Hull-White model is the following special case, dr = (θ(t) − ar) dt + σ dW.

• When the current term structure is matched,a θ(t) = ∂f (0, t)

∂t + af (0, t) + σ2 2a

1 − e−2at .

aHull and White (1993).

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### The Extended CIR Model

• In the extended CIR model the short rate follows dr = (θ(t) − a(t) r) dt + σ(t)√

r dW.

• The functions θ(t), a(t), and σ(t) are implied from market observables.

• With constant parameters, there exist analytical solutions to a small set of interest rate-sensitive securities.

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### The Hull-White Model: Calibration

a

• We describe a trinomial forward induction scheme to calibrate the Hull-White model given a and σ.

• As with the Ho-Lee model, the set of achievable short rates is evenly spaced.

• Let r0 be the annualized, continuously compounded short rate at time zero.

• Every short rate on the tree takes on a value r0 + jΔr

for some integer j.

aHull and White (1993).

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### The Hull-White Model: Calibration (continued)

• Time increments on the tree are also equally spaced at Δt apart.

• Hence nodes are located at times iΔt for i = 0, 1, 2, . . . .

• We shall refer to the node on the tree with ti ≡ iΔt,

rj ≡ r0 + jΔr, as the (i, j) node.

• The short rate at node (i, j), which equals rj, is eﬀective for the time period [ ti, ti+1).

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### The Hull-White Model: Calibration (continued)

• Use

μi,j ≡ θ(ti) − arj (152) to denote the drift ratea of the short rate as seen from node (i, j).

• The three distinct possibilities for node (i, j) with three branches incident from it are displayed on p. 1109.

• The middle branch may be an increase of Δr, no change, or a decrease of Δr.

aOr, the annualized expected change.

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### The Hull-White Model: Calibration (continued)

(i, j)

(i + 1, j + 2)

*(i + 1, j + 1)

-(i + 1, j)

(i, j)

*(i + 1, j + 1)

-(i + 1, j) j(i + 1, j − 1)

(i, j) -(i + 1, j) j(i + 1, j − 1)

R(i + 1, j − 2)

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### The Hull-White Model: Calibration (continued)

• The upper and the lower branches bracket the middle branch.

• Deﬁne

p1(i, j) the probability of following the upper branch from node (i, j) p2(i, j) the probability of following the middle branch from node (i, j) p3(i, j) the probability of following the lower branch from node (i, j)

• The root of the tree is set to the current short rate r0.

• Inductively, the drift μi,j at node (i, j) is a function of (the still unknown) θ(ti).

– Both describe the expected change from node (i, j).

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### The Hull-White Model: Calibration (continued)

• Once θ(ti) is available, μi,j can be derived via Eq. (152) on p. 1108.

• This in turn determines the branching scheme at every node (i, j) for each j, as we will see shortly.

• The value of θ(ti) must thus be made consistent with the spot rate r(0, ti+2).a

aNot r(0, ti+1)!

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### The Hull-White Model: Calibration (continued)

• The branches emanating from node (i, j) with their probabilitiesa must be chosen to be consistent with μi,j and σ.

• This is done by letting the middle node be as closest to the current short rate rj plus the drift μi,jΔt.b

ap1(i, j), p2(i, j), and p3(i, j).

bA predecessor to Lyuu and Wu’s (R90723065) (2003, 2005) mean- tracking idea, which is the precursor of the binomial-trinomial tree of Dai (B82506025, R86526008, D8852600) and Lyuu (2006, 2008, 2010).

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### The Hull-White Model: Calibration (continued)

• Let k be the number among { j − 1, j, j + 1 } that

makes the short rate reached by the middle branch, rk, closest to

rj + μi,jΔt.

– But note that μi,j is still not computed yet.

• Then the three nodes following node (i, j) are nodes (i + 1, k + 1), (i + 1, k), (i + 1, k − 1).

• See p. 1114 for a possible geometry.

• The resulting tree combines because of the constant jump sizes to reach k from j.

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(0, 0)

* - j

(1, 1)

* - j

(1, 0) 

*

(1, −1) -

* - j

* - j

* - j

* - j

- j R

* - j

* - j

* - j

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

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

Δt

6

?Δr

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### The Hull-White Model: Calibration (continued)

• The probabilities for moving along these branches are functions of μi,j, σ, j, and k:

p1(i, j) = σ2Δt + η2

2(Δr)2 + η

2Δr (153)

p2(i, j) = 1 − σ2Δt + η2

(Δr)2 (153)

p3(i, j) = σ2Δt + η2

2(Δr)2 η

2Δr (153) where

η ≡ μi,jΔt + (j − k) Δr.

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### The Hull-White Model: Calibration (continued)

• As trinomial tree algorithms are but explicit methods in disguise,a certain relations must hold for Δr and Δt to guarantee stability.

• It can be shown that their values must satisfy σ√

3Δt

2 ≤ Δr ≤ 2σ√ Δt

for the probabilities to lie between zero and one.

– For example, Δr can be set to σ

3Δt .b

• Now it only remains to determine θ(ti).

aRecall p. 761.

bHull and White (1988).

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### The Hull-White Model: Calibration (continued)

• At this point at time ti,

r(0, t1), r(0, t2), . . . , r(0, ti+1) have already been matched.

• Let Q(i, j) be the state price at node (i, j).

• By construction, the state prices Q(i, j) for all j are known by now.

• We begin with state price Q(0, 0) = 1.

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### The Hull-White Model: Calibration (continued)

• Let ˆr(i) refer to the short rate value at time ti.

• The value at time zero of a zero-coupon bond maturing at time ti+2 is then

e−r(0,ti+2)(i+2) Δt

= 

j

Q(i, j) e−rjΔt Eπ



e−ˆr(i+1) Δt ˆr(i) = rj



.(154)

• The right-hand side represents the value of \$1 obtained by holding a zero-coupon bond until time ti+1 and then reinvesting the proceeds at that time at the prevailing short rate ˆr(i + 1), which is stochastic.

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### The Hull-White Model: Calibration (continued)

• The expectation in Eq. (154) can be approximated by Eπ



e−ˆr(i+1) Δt  ˆr(i) = rj



≈ e−rjΔt

1 − μi,j(Δt)2 + σ2(Δt)3 2

. (155) – This solves the chicken-egg problem!

• Substitute Eq. (155) into Eq. (154) and replace μi,j

with θ(ti) − arj to obtain

θ(ti) ≈

j Q(i, j) e−2rjΔt 1 +arj(Δt)2 + σ2(Δt)3/2

− e−r(0,ti+2)(i+2) Δt t)2

j Q(i, j) e−2rjΔt .

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### The Hull-White Model: Calibration (continued)

• For the Hull-White model, the expectation in Eq. (155) is actually known analytically by Eq. (25) on p. 162:

Eπ



e−ˆr(i+1) Δt ˆr(i) = rj



= e−rjΔt+(−θ(ti)+arj2Δt/2)(Δt)2.

• Therefore, alternatively,

θ(ti) = r(0, ti+2)(i + 2)

Δt +σ2Δt

2 +ln

j Q(i, j) e−2rjΔt+arj(Δt)2

(Δt)2 .

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### The Hull-White Model: Calibration (concluded)

• With θ(ti) in hand, we can compute μi,j,a the

probabilities,b and ﬁnally the state prices at time ti+1:

Q(i + 1, j)

= 

(i, j) is connected to (i + 1, j) with probability pj∗

pje−rj∗ΔtQ(i, j).

• There are at most 5 choices for j (why?).

• The total running time is O(n2).

• The space requirement is O(n) (why?).

aSee Eq. (152) on p. 1108.

bSee Eqs. (153) on p. 1115.

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### Comments on the Hull-White Model

• One can try diﬀerent values of a and σ for each option.

• Or have an a value common to all options but use a diﬀerent σ value for each option.

• Either approach can match all the option prices exactly.

• But suppose the demand is for a single set of parameters that replicate all option prices.

• Then the Hull-White model can be calibrated to all the observed option prices by choosing a and σ that

minimize the mean-squared pricing error.a

aHull and White (1995).

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### The Hull-White Model: Calibration with Irregular Trinomial Trees

• The previous calibration algorithm is quite general.

• For example, it can be modiﬁed to apply to cases where the diﬀusion term has the form σrb.

• But it has at least two shortcomings.

• First, the resulting trinomial tree is irregular (p. 1114).

– So it is harder to program (for nonprogrammers).

• The second shortcoming is again a consequence of the tree’s irregular shape.

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### The Hull-White Model: Calibration with Irregular Trinomial Trees (concluded)

• Recall that the algorithm ﬁgured out θ(ti) that matches the spot rate r(0, ti+2) in order to determine the

branching schemes for the nodes at time ti.

• But without those branches, the tree was not speciﬁed, and backward induction on the tree was not possible.

• To avoid this chicken-egg dilemma, the algorithm turned to the continuous-time model to evaluate Eq. (154) on p. 1118 that helps derive θ(ti).

• The resulting θ(ti) hence might not yield a tree that matches the spot rates exactly.

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### The Hull-White Model: Calibration with Regular Trinomial Trees

a

• The next, simpler algorithm exploits the fact that the Hull-White model has a constant diﬀusion term σ.

• The resulting trinomial tree will be regular.

• All the θ(ti) terms can be chosen by backward induction to match the spot rates exactly.

• The tree is constructed in two phases.

aHull and White (1994).

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### The Hull-White Model: Calibration with Regular Trinomial Trees (continued)

• In the ﬁrst phase, a tree is built for the θ(t) = 0 case, which is an Ornstein-Uhlenbeck process:

dr = −ar dt + σ dW, r(0) = 0.

– The tree is dagger-shaped (preview p. 1127).

– The number of nodes above the r0-line is jmax, and that below the line is jmin.

– They will be picked so that the probabilities (153) on p. 1115 are positive for all nodes.

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(0, 0) r0

* - j

(1, 1)

* - j

(1, 0)

* -

(1, −1) j

* - j

* - j

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* - j



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* - j

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

Δt

6?Δr

The short rate at node (0, 0) equals r0 = 0; here jmax = 3 and jmin = 2.

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### The Hull-White Model: Calibration with Regular Trinomial Trees (concluded)

• The tree’s branches and probabilities are now in place.

• Phase two ﬁts the term structure.

– Backward induction is applied to calculate the βi to add to the short rates on the tree at time ti so that the spot rate r(0, ti+1) is matched.a

aContrast this with the previous algorithm, where it was the spot rate r(0, ti+2) that is matched!

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### The Hull-White Model: Calibration

• Set Δr = σ√

3Δt and assume that a > 0.

• Node (i, j) is a top node if j = jmax and a bottom node if j = −jmin.

• Because the root of the tree has a short rate of r0 = 0, phase one adopts rj = jΔr.

• Hence the probabilities in Eqs. (153) on p. 1115 use η ≡ −ajΔrΔt + (j − k) Δr.

• Recall that k denotes the middle branch.

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### The Hull-White Model: Calibration (continued)

• The probabilities become

p1(i, j)

= 1

6

+ a2j2(Δt)2 − 2ajΔt(j − k) + (j − k)2 − ajΔt + (j − k)

2 , (156)

p2(i, j)

= 2

3 

a2 j2(Δt)2 − 2ajΔt(j − k) + (j − k)2 

, (157)

p3(i, j)

= 1 6

+ a2j2(Δt)2 − 2ajΔt(j − k) + (j − k)2 + ajΔt − (j − k)

2 . (158)

• p1: up move; p2: ﬂat move; p3: down move.

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### The Hull-White Model: Calibration (continued)

• The dagger shape dictates this:

– Let k = j − 1 if node (i, j) is a top node.

– Let k = j + 1 if node (i, j) is a bottom node.

– Let k = j for the rest of the nodes.

• Note that the probabilities are identical for nodes (i, j) with the same j.

• Furthermore, p1(i, j) = p3(i, −j).

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### The Hull-White Model: Calibration (continued)

• The inequalities

3 6

3 < jaΔt <

2

3 (159)

ensure that all the branching probabilities are positive in the upper half of the tree, that is, j > 0 (verify this).

• Similarly, the inequalities

2

3 < jaΔt < −3 6 3

ensure that the probabilities are positive in the lower half of the tree, that is, j < 0.

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### The Hull-White Model: Calibration (continued)

• To further make the tree symmetric across the r0-line, we let jmin = jmax.

• As

3 6

3 ≈ 0.184, a good choice is

jmax = 0.184/(aΔt) .

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### The Hull-White Model: Calibration (continued)

• Phase two computes the βis to ﬁt the spot rates.

• We begin with state price Q(0, 0) = 1.

• Inductively, suppose that spot rates

r(0, t1), r(0, t2), . . . , r(0, ti) have already been matched as of time ti.

• By construction, the state prices Q(i, j) for all j are known by now.

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### The Hull-White Model: Calibration (continued)

• The value of a zero-coupon bond maturing at time ti+1 equals

e−r(0,ti+1)(i+1) Δt = 

j

Q(i, j) e−(βi+rj)Δt by risk-neutral valuation.

• Hence

βi = r(0, ti+1)(i + 1) Δt + ln

j Q(i, j) e−rjΔt

Δt .

(160)

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### The Hull-White Model: Calibration (concluded)

• The short rate at node (i, j) now equals βi + rj.

• The state prices at time ti+1,

Q(i + 1, j), − min(i + 1, jmax) ≤ j ≤ min(i + 1, jmax), can now be calculated as before.

• The total running time is O(njmax).

• The space requirement is O(n).

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### A Numerical Example

• Assume a = 0.1, σ = 0.01, and Δt = 1 (year).

• Immediately, Δr = 0.0173205 and jmax = 2.

• The plot on p. 1138 illustrates the 3-period trinomial tree after phase one.

• For example, the branching probabilities for node E are calculated by Eqs. (156)–(158) on p. 1130 with j = 2 and k = 1.

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* - j

A

* - j

B *

- j

C *

- j

D

- j R

E *

- j

F *

- j

G *

- j

H 

* -

I

Node A, C, G B, F E D, H I

r (%) 0.00000 1.73205 3.46410 −1.73205 −3.46410 p1 0.16667 0.12167 0.88667 0.22167 0.08667 p2 0.66667 0.65667 0.02667 0.65667 0.02667 p3 0.16667 0.22167 0.08667 0.12167 0.88667

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### A Numerical Example (continued)

• Suppose that phase two is to ﬁt the spot rate curve 0.08 − 0.05 × e−0.18×t.

• The annualized continuously compounded spot rates are r(0, 1) = 3.82365%, r(0, 2) = 4.51162%, r(0, 3) = 5.08626%.

• Start with state price Q(0, 0) = 1 at node A.

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### A Numerical Example (continued)

• Now, by Eq. (160) on p. 1135,

β0 = r(0, 1) + ln Q(0, 0) e−r0 = r(0, 1) = 3.82365%.

• Hence the short rate at node A equals β0 + r0 = 3.82365%.

• The state prices at year one are calculated as Q(1, 1) = p1(0, 0) e−(β0+r0) = 0.160414, Q(1, 0) = p2(0, 0) e−(β0+r0) = 0.641657, Q(1, −1) = p3(0, 0) e−(β0+r0) = 0.160414.

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### A Numerical Example (continued)

• The 2-year rate spot rate r(0, 2) is matched by picking

β1 = r(0, 2)×2+ln

Q(1, 1) e−Δr + Q(1, 0) + Q(1, −1) eΔr 

= 5.20459%.

• Hence the short rates at nodes B, C, and D equal β1 + rj,

where j = 1, 0, −1, respectively.

• They are found to be 6.93664%, 5.20459%, and 3.47254%.

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### A Numerical Example (continued)

• The state prices at year two are calculated as

Q(2, 2) = p1(1, 1) e−(β1+r1)Q(1, 1) = 0.018209,

Q(2, 1) = p2(1, 1) e−(β1+r1)Q(1, 1) + p1(1, 0) e−(β1+r0)Q(1, 0)

= 0.199799,

Q(2, 0) = p3(1, 1) e−(β1+r1)Q(1, 1) + p2(1, 0) e−(β1+r0)Q(1, 0) +p1(1, −1) e−(β1+r−1)Q(1, −1) = 0.473597,

Q(2, −1) = p3(1, 0) e−(β1+r0)Q(1, 0) + p2(1, −1) e−(β1+r−1)Q(1, −1)

= 0.203263,

Q(2, −2) = p3(1, −1) e−(β1+r−1)Q(1, −1) = 0.018851.

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### A Numerical Example (concluded)

• The 3-year rate spot rate r(0, 3) is matched by picking

β2 = r(0, 3) × 3 + ln

Q(2, 2) e−2×Δr + Q(2, 1) e−Δr + Q(2, 0) +Q(2, −1) eΔr + Q(2, −2) e2×Δr 

= 6.25359%.

• Hence the short rates at nodes E, F, G, H, and I equal β2 + rj, where j = 2, 1, 0, −1, −2, respectively.

• They are found to be 9.71769%, 7.98564%, 6.25359%, 4.52154%, and 2.78949%.

• The ﬁgure on p. 1144 plots βi for i = 0, 1, . . . , 29.

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      <HDU +L/







 -L+/

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### The (Whole) Yield Curve Approach

• We have seen several Markovian short rate models.

• The Markovian approach is computationally eﬃcient.

• But it is diﬃcult to model the behavior of yields and bond prices of diﬀerent maturities.

• The alternative yield curve approach regards the whole term structure as the state of a process and directly speciﬁes how it evolves.

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### The Heath-Jarrow-Morton (HJM) Model

a

• This inﬂuential model is a forward rate model.

• The HJM model speciﬁes the initial forward rate curve and the forward rate volatility structure.

– The volatility structure describes the volatility of each forward rate for a given maturity date.

• Like the Black-Scholes option pricing model, neither risk preference assumptions nor the drifts of forward rates are needed.

aHeath, Jarrow, and Morton (1992).

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### Introduction to Mortgage-Backed Securities

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Anyone stupid enough to promise to be responsible for a stranger’s debts deserves to have his own property held to guarantee payment.

— Proverbs 27:13

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### Mortgages

• A mortgage is a loan secured by the collateral of real estate property.

• Suppose the borrower (the mortgagor) defaults, that is, fails to make the contractual payments.

• The lender (the mortgagee) can foreclose the loan by seizing the property.

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### Mortgage-Backed Securities

• A mortgage-backed security (MBS) is a bond backed by an undivided interest in a pool of mortgages.a

• MBSs traditionally enjoy high returns, wide ranges of products, high credit quality, and liquidity.

• The mortgage market has witnessed tremendous innovations in product design.

aThey can be traced to 1880s (Levy (2012)).

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### Mortgage-Backed Securities (concluded)

• The complexity of the products and the prepayment

option require advanced models and software techniques.

– In fact, the mortgage market probably could not have operated eﬃciently without them.a

• They also consume lots of computing power.

• Our focus will be on residential mortgages.

• But the underlying principles are applicable to other types of assets.

aMerton (1994).

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### Types of MBSs

• An MBS is issued with pools of mortgage loans as the collateral.

• The cash ﬂows of the mortgages making up the pool naturally reﬂect upon those of the MBS.

• There are three basic types of MBSs:

1. Mortgage pass-through security (MPTS).

2. Collateralized mortgage obligation (CMO).

3. Stripped mortgage-backed security (SMBS).

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### Problems Investing in Mortgages

• The mortgage sector is one of the largest in the debt market (see p. 3 of the textbook).a

• Individual mortgages are unattractive for many investors.

• Often at hundreds of thousands of U.S. dollars or more, they demand too much investment.

• Most investors lack the resources and knowledge to assess the credit risk involved.

aThe outstanding balance was US\$8.1 trillion as of 2012 vs. the US Treasury’s US\$10.9 trillion according to SIFMA.

(61)

### Problems Investing in Mortgages (concluded)

• Recall that a traditional mortgage is ﬁxed rate, level payment, and fully amortized.

• So the percentage of principal and interest (P&I) varying from month to month, creating accounting headaches.

• Prepayment levels ﬂuctuate with a host of factors.

• That makes the size and the timing of the cash ﬂows unpredictable.

(62)

### Mortgage Pass-Throughs

a

• The simplest kind of MBS.

• Payments from the underlying mortgages are passed

from the mortgage holders through the servicing agency, after a fee is subtracted.

• They are distributed to the security holder on a pro rata basis.

– The holder of a \$25,000 certiﬁcate from a \$1 million pool is entitled to 21/2% (or 1/40th) of the cash ﬂow.

• Because of higher marketability, a pass-through is easier to sell than its individual loans.

aFirst issued by Ginnie Mae in 1970.

(63)

Rule for distribution of cash flows: pro rata Loan 2

Loan 10 Loan 1

Pass-through: \$1 million par pooled mortgage loans

(64)

### Collateralized Mortgage Obligations (CMOs)

• A pass-through exposes the investor to the total prepayment risk.

• Such risk is undesirable from an asset/liability perspective.

• To deal with prepayment uncertainty, CMOs were created.a

• Mortgage pass-throughs have a single maturity and are backed by individual mortgages.

aIn June 1983 by Freddie Mac with the help of First Boston, which was acquired by Credit Suisse in 1990.

(65)

### Collateralized Mortgage Obligations (CMOs) (continued)

• CMOs are multiple-maturity, multiclass debt

instruments collateralized by pass-throughs, stripped mortgage-backed securities, and whole loans.

• The total prepayment risk is now divided among classes of bonds called classes or tranches.a

• The principal, scheduled and prepaid, is allocated on a prioritized basis so as to redistribute the prepayment risk among the tranches in an unequal way.

aTranche is a French word for “slice.”

(66)

### Collateralized Mortgage Obligations (CMOs) (concluded)

• CMOs were the ﬁrst successful attempt to alter

mortgage cash ﬂows in a security form that attracts a wide range of investors

– The outstanding balance of agency CMOs was US\$1.1 trillion as of the ﬁrst quarter of 2015.a

aSIFMA (2015).

(67)

### Sequential Tranche Paydown

• In the sequential tranche paydown structure, Class A receives principal paydown and prepayments before

Class B, which in turn does it before Class C, and so on.

• Each tranche thus has a diﬀerent eﬀective maturity.

• Each tranche may even have a diﬀerent coupon rate.

(68)

### An Example

• Consider a two-tranche sequential-pay CMO backed by

\$1,000,000 of mortgages with a 12% coupon and 6 months to maturity.

• The cash ﬂow pattern for each tranche with zero

prepayment and zero servicing fee is shown on p. 1162.

• The calculation can be carried out ﬁrst for the Total columns, which make up the amortization schedule.

• Then the cash ﬂow is allocated.

• Tranche A is retired after 4 months, and tranche B starts principal paydown at the end of month 4.

(69)

### CMO Cash Flows without Prepayments

Interest Principal Remaining principal

Month A B Total A B Total A B Tota

500,000 500,000 1,000,0

1 5,000 5,000 10,000 162,548 0 162,548 337,452 500,000 837,4

2 3,375 5,000 8,375 164,173 0 164,173 173,279 500,000 673,2

3 1,733 5,000 6,733 165,815 0 165,815 7,464 500,000 507,4

4 75 5,000 5,075 7,464 160,009 167,473 0 339,991 339,9

5 0 3,400 3,400 0 169,148 169,148 0 170,843 170,8

6 0 1,708 1,708 0 170,843 170,843 0 0

Total 10,183 25,108 35,291 500,000 500,000 1,000,000

The total monthly payment is \$172,548. Month-i numbers reﬂect the ith monthly payment.

(70)

### Another Example

• When prepayments are present, the calculation is only slightly more complex.

• Suppose the single monthly mortality (SMM) per month is 5%.

• This means the prepayment amount is 5% of the remaining principal.

• The remaining principal at month i after prepayment then equals the scheduled remaining principal as

computed by Eq. (7) on p. 46 times (0.95)i.

• This done for all the months, the interest payment at any month is the remaining principal of the previous month times 1%.

(71)

### Another Example (continued)

• The prepayment amount equals the remaining principal times 0.05/0.95.

– The division by 0.95 yields the remaining principal before prepayment.

• Page 1165 tabulates the cash ﬂows of the same two-tranche CMO under 5% SMM.

(72)

### Another Example (continued)

Interest Principal Remaining principal

Month A B Total A B Total A B Total

500,000 500,000 1,000,00

1 5,000 5,000 10,000 204,421 0 204,421 295,579 500,000 795,57

2 2,956 5,000 7,956 187,946 0 187,946 107,633 500,000 607,63

3 1,076 5,000 6,076 107,633 64,915 172,548 0 435,085 435,08

4 0 4,351 4,351 0 158,163 158,163 0 276,922 276,92

5 0 2,769 2,769 0 144,730 144,730 0 132,192 132,19

6 0 1,322 1,322 0 132,192 132,192 0 0

Total 9,032 23,442 32,474 500,000 500,000 1,000,000

Month-i numbers reﬂect the ith monthly payment.

(73)

### Another Example (continued)

• For instance, the total principal payment at month one,

\$204,421, can be veriﬁed as follows.

• The scheduled remaining principal is \$837,452 from p. 1162.

• The remaining principal is hence

837452 × 0.95 = 795579.

• That makes the total principal payment 1000000 − 795579 = 204421.

(74)

### Another Example (concluded)

• As tranche A’s remaining principal is \$500,000, all 204,421 dollars go to tranche A.

– Incidentally, the prepayment is

837452 × 5% = 41873.

• Tranche A is retired after 3 months, and tranche B starts principal paydown at the end of month 3.

(75)

### Stripped Mortgage-Backed Securities (SMBSs)

a

• The principal and interest are divided between the PO strip and the IO strip.

• In the scenarios on p. 1161 and p. 1163:

– The IO strip receives all the interest payments under the Interest/Total column.

– The PO strip receives all the principal payments under the Principal/Total column.

aThey were created in February 1987 when Fannie Mae issued its Trust 1 stripped MBS.

(76)

### Stripped Mortgage-Backed Securities (SMBSs) (concluded)

• These new instruments allow investors to better exploit anticipated changes in interest rates.a

• The collateral for an SMBS is a pass-through.

• CMOs and SMBSs are usually called derivative MBSs.

aSee p. 357 of the textbook.

Updating...

## References

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