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

## Full text

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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. 1099).

– So it is harder to program.

• 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. (140) on p. 1103 that helps derive θ(ti) later.

• 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. 1112).

– The number of nodes above the r0-line, jmax, and that below the line, jmin, will be picked so that the probabilities (139) on p. 1100 are positive for all nodes.

– The tree’s branches and probabilities are in place.

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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)

• 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 ∆r = σ√

3∆t 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 = j∆r.

• Hence the probabilities in Eqs. (139) on p. 1100 use η ≡ −aj∆r∆t + (j − k) ∆r.

• 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(∆t)2 − 2aj∆t(j − k) + (j − k)2 − aj∆t + (j − k)

2 (142),

p2(i, j)

= 2

3 [

a2j2(∆t)2 − 2aj∆t(j − k) + (j − k)2 ]

, (143)

p3(i, j)

= 1

6 + a2j2(∆t)2 − 2aj∆t(j − k) + (j − k)2 + aj∆t− (j − k)

2 (144).

• 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 < ja∆t <

√2

3 (145)

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 < ja∆t < 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 336 ≈ 0.184, a good choice is

jmax = ⌈0.184/(a∆t)⌉.

• 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 at time ti.

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

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

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

e−r(0,ti+1)(i+1) ∆t = ∑

j

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

• Hence

βi = r(0, ti+1)(i + 1) ∆t + ln

j Q(i, j) e−rj∆t

∆t ,

and the short rate at node (i, j) equals βi + rj.

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

• 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 ∆t = 1 (year).

• Immediately, ∆r = 0.0173205 and jmax = 2.

• The plot on p. 1122 illustrates the 3-period trinomial tree after phase one.

• For example, the branching probabilities for node E are calculated by Eqs. (142)–(144) on p. 1115 with j = 2 and k = 1.

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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,

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−∆r + Q(1, 0) + Q(1,−1) e∆r ]

= 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×∆r + Q(2, 1) e−∆r + Q(2, 0) +Q(2,−1) e∆r + Q(2,−2) e2×∆r ]

= 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. 1128 plots βi for i = 0, 1, . . . , 29.

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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 Model

a

• This influential model is a forward rate model.

• It is also a popular model.

• The HJM model specifies the initial forward rate curve and the forward rate volatility structure, which 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 (HJM) (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.

• The lender — the mortgagee — can foreclose the loan by seizing the property if the borrower — the mortgagor — defaults, that is, fails to make the contractual payments.

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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 text).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.

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### 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, making the size and the timing of the cash flows unpredictable.

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### 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.

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Rule for distribution of cash flows: pro rata Loan 2

Loan 10 Loan 1

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

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### 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.

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

• 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.”

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### 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.

• CMOs were the first successful attempt to alter

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

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### 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. 1145.

• 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.

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### CMO Cash Flows without Prepayments

Interest Principal Remaining principal

Month A B Total A B Total A B Total

500,000 500,000 1,000,000

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

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

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

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

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

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

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### 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. (6) on p. 48 times (0.95)i.

• This done for all the months, the total interest payment at any month is the remaining principal of the previous

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### 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 1149 tabulates the cash flows of the same two-tranche CMO under 5% SMM.

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

\$204,421, can be verified as follows.

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### Another Example (concluded)

• The scheduled remaining principal is \$837,452 from p. 1145.

• The remaining principal is hence

837452 × 0.95 = 795579, which makes the total principal payment 1000000 − 795579 = 204421.

• 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.

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### CMO Cash Flows with Prepayments

Interest Principal Remaining principal

Month A B Total A B Total A B Total

500,000 500,000 1,000,000

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

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

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

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

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

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

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

Month-i numbers reflect the ith monthly payment.

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### 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. 1144 and p. 1146:

– 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.

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### 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.

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### Prepayments

• The prepayment option sets MBSs apart from other fixed-income securities.

• The exercise of options on most securities is expected to be “rational.”

• This kind of “rationality” is weakened when it comes to the homeowner’s decision to prepay.

• For example, even when the prevailing mortgage rate exceeds the mortgage’s loan rate, some loans are

prepaid.

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### Prepayment Risk

• Prepayment risk is the uncertainty in the amount and timing of the principal prepayments in the pool of

mortgages that collateralize the security.

• This risk can be divided into contraction risk and extension risk.

• Contraction risk is the risk of having to reinvest the

prepayments at a rate lower than the coupon rate when interest rates decline.

• Extension risk is due to the slowdown of prepayments when interest rates climb, making the investor earn the security’s lower coupon rate rather than the market’s higher rate.

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### Prepayment Risk (concluded)

• Prepayments can be in whole or in part.

– The former is called liquidation.

– The latter is called curtailment.

• The holder of a pass-through security is exposed to the total prepayment risk associated with the underlying pool of mortgage loans.

• The CMO is designed to alter the distribution of that risk among the investors.

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### Other Risks

• Investors in mortgages are exposed to at least three other risks.

– Interest rate risk is inherent in any fixed-income security.

– Credit risk is the risk of loss from default.

∗ For privately insured mortgage, the risk is related to the credit rating of the company that insures the mortgage.

– Liquidity risk is the risk of loss if the investment must be sold quickly.

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### Prepayment: Causes

Prepayments have at least five components.

Home sale (“housing turnover”). The sale of a home generally leads to the prepayment of mortgage because of the full payment of the remaining principal.

Refinancing. Mortgagors can refinance their home

mortgage at a lower mortgage rate. This is the most volatile component of prepayment and constitutes the bulk of it when prepayments are extremely high.

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### Prepayment: Causes (concluded)

Default. Caused by foreclosure and subsequent liquidation of a mortgage. Relatively minor in most cases.

Curtailment. As the extra payment above the scheduled payment, curtailment applies to the principal and

shortens the maturity of fixed-rate loans. Its contribution to prepayments is minor.

Full payoﬀ (liquidation). There is evidence that many mortgagors pay oﬀ their mortgage completely when it is very seasoned and the remaining balance is small. Full payoﬀ can also be due to natural disasters.

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### Prepayment: Characteristics

• Prepayments usually increase as the mortgage ages — first at an increasing rate and then at a decreasing rate.

• They are higher in the spring and summer and lower in the fall and winter.

• They vary by the geographic locations of the underlying properties.

• They increase when interest rates drop but with a time lag.

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### Prepayment: Characteristics (continued)

• If prepayments were higher for some time because of high refinancing rates, they tend to slow down.

– Perhaps, homeowners who do not prepay when rates have been low for a prolonged time tend never to prepay.

• Plot on p. 1160 illustrates the typical price/yield curves of the Treasury and pass-through.

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0.05 0.1 0.15 0.2 0.25 0.3Interest rate 50

100 150 200

Price

The cusp

Treasury MBS

Price compression occurs as yields fall through a threshold.

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### Prepayment: Characteristics (concluded)

• As yields fall and the pass-through’s price moves above a certain price, it flattens and then follows a downward slope.

• This phenomenon is called the price compression of premium-priced MBSs.

• It demonstrates the negative convexity of such securities.

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## Analysis of Mortgage-Backed Securities

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Oh, well, if you cannot measure, measure anyhow.

— Frank H. Knight (1885–1972)

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### Uniqueness of MBS

• Compared with other fixed-income securities, the MBS is unique in two respects.

• Its cash flow consists of principal and interest (P&I).

• The cash flow may vary because of prepayments in the underlying mortgages.

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### Time Line

-

Time 0 Time 1 Time 2 Time 3 Time 4 Month 1 Month 2 Month 3 Month 4

• Mortgage payments are paid in arrears.

• A payment for month i occurs at time i, that is, end of month i.

• The end of a month will be identified with the beginning of the coming month.

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### Cash Flow Analysis

• A traditional mortgage has a fixed term, a fixed interest rate, and a fixed monthly payment.

• Page 1167 illustrates the scheduled P&I for a 30-year, 6% mortgage with an initial balance of \$100,000.

• Page 1168 depicts how the remaining principal balance decreases over time.

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### Scheduled Principal and Interest Payments

50 100 150 200 250 300 350 Month 100

200 300 400 500 600

Principal Interest

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### Scheduled Remaining Principal Balances

50 100 150 200 250 300 350 Month 20

40 60 80 100

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### Cash Flow Analysis (continued)

• In the early years, the P&I consists mostly of interest.

• Then it gradually shifts toward principal payment with the passage of time.

• However, the total P&I payment remains the same each month, hence the term level pay.

• In the absence of prepayments and servicing fees, identical characteristics hold for the pool’s P&I payments.

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### Cash Flow Analysis (continued)

• From Eq. (6) on p. 48 the remaining principal balance after the kth payment is

C 1 − (1 + r/m)−n+k

r/m . (146)

– C is the scheduled P&I payment of an n-month mortgage making m payments per year.

– r is the annual mortgage rate.

• For mortgages, m = 12.

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### Cash Flow Analysis (continued)

• The scheduled remaining principal balance after k

payments can be expressed as a portion of the original principal balance:

Balk ≡ 1 − (1 + r/m)k − 1 (1 + r/m)n − 1

= (1 + r/m)n − (1 + r/m)k

(1 + r/m)n − 1 . (147)

• This equation can be verified by dividing Eq. (146) (p. 1170) by the same equation with k = 0.

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### Cash Flow Analysis (continued)

• The remaining principal balance after k payments is RBk ≡ O × Balk,

where O will denote the original principal balance.

• The term factor denotes the portion of the remaining principal balance to its original principal balance.

• So Balk is the monthly factor when there are no prepayments.

• It is also known as the amortization factor.

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### Cash Flow Analysis (concluded)

• When the idea of factor is applied to a mortgage pool, it is called the paydown factor on the pool or simply the pool factor.

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### An Example

• The remaining balance of a 15-year mortgage with a 9%

mortgage rate after 54 months is

O × (1 + (0.09/12))180 − (1 + (0.09/12))54 (1 + (0.09/12))180 − 1

= O × 0.824866.

• In other words, roughly 82.49% of the original loan amount remains after 54 months.

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### P&I Analysis

• By the amortization principle, the tth interest payment equals

It ≡ RBt−1× r

m = O × r

m × (1 + r/m)n − (1 + r/m)t−1 (1 + r/m)n − 1 .

• The principal part of the tth monthly payment is Pt ≡ RBt−1 − RBt

= O × (r/m)(1 + r/m)t−1

(1 + r/m)n − 1 . (148)

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### P&I Analysis (concluded)

• The scheduled P&I payment at month t, or Pt + It, is (RBt−1 − RBt) + RBt−1 × r

m

= O ×

[ (r/m)(1 + r/m)n (1 + r/m)n − 1

]

, (149)

indeed a level pay independent of t.

• The term within the brackets, called the payment factor or annuity factor, is the monthly payment for each

dollar of mortgage.

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### An Example

• The mortgage on pp. 42ﬀ has a monthly payment of 250000 × (0.08/12) × (1 + (0.08/12))180

(1 + (0.08/12))180 − 1 = 2389.13 by Eq. (149) on p. 1176.

• This number agrees with the number derived earlier.

(71)

• We turn to ARM pricing as an interesting application of derivatives pricing and the analysis above.

• Consider a 3-year ARM with an interest rate that is 1%

above the 1-year T-bill rate at the beginning of the year.

• This 1% is called the margin.

• Assume this ARM carries annual, not monthly, payments.

• The T-bill rates follow the binomial process, in boldface, on p. 1179, and the risk-neutral probability is 0.5.

(72)

*

j

A

4.000%

5.000%

0.36721

*

j

B

3.526%

4.526%

0.53420 *

j

C

5.289%

6.289%

0.54765

-

1.0 D

2.895%

3.895%

1.03895

-

1.0 E

4.343%

5.343%

1.05343

-

1.0 F

6.514%

7.514%

1.07514

year 1 year 2 year 3

Stacked at each node are the T-bill rate, the mortgage rate, and the payment factor for a mortgage initiated at that node and ending at year 3 (based on the mortgage rate at the same node). The short rates are from p. 941.

(73)

• How much is the ARM worth to the issuer?

• Each new coupon rate at the reset date determines the level mortgage payment for the months until the next reset date as if the ARM were a fixed-rate loan with the new coupon rate and a maturity equal to that of the ARM.

• For example, for the interest rate tree on p. 1179, the scenario A → B → E will leave our three-year ARM

with a remaining principal at the end of the second year diﬀerent from that under the scenario A → C → E.

(74)

• This path dependency calls for care in algorithmic design to avoid exponential complexity.

• Attach to each node on the binomial tree the annual payment per \$1 of principal for a mortgage initiated at that node and ending at year 3.

– In other words, the payment factor.

• At node B, for example, the annual payment factor can be calculated by Eq. (149) on p. 1176 with r = 0.04526, m = 1, and n = 2 as

0.04526 × (1.04526)2

(1.04526)2 − 1 = 0.53420.

(75)

• The payment factors for other nodes on p. 1179 are calculated in the same manner.

• We now apply backward induction to price the ARM (see p. 1183).

• At each node on the tree, the net value of an ARM of value \$1 initiated at that node and ending at the end of the third year is calculated.

• For example, the value is zero at terminal nodes since the ARM is immediately repaid.

(76)

Y

A

0.0189916

Y

B

0.0144236

Y

C

0.0141396



0 D

0.0097186



0 E

0.0095837



0 F

0.0093884

year 1 year 2 year 3

(77)

• At node D, the value is 1.03895

1.02895 − 1 = 0.0097186,

which is simply the net present value of the payment 1.03895 next year.

– Recall that the issuer makes a loan of \$1 at D.

• The values at nodes E and F can be computed similarly.

(78)

• At node B, we first figure out the remaining principal balance after the payment one year hence as

1 − (0.53420 − 0.04526) = 0.51106,

because \$0.04526 of the payment of \$0.53426 constitutes the interest.

• The issuer will receive \$0.01 above the T-bill rate next year, and the value of the ARM is either \$0.0097186 or

\$0.0095837 per \$1, each with probability 0.5.

(79)

• The ARM’s value at node B thus equals 0.51106 × (0.0097186 + 0.0095837)/2 + 0.01

1.03526 = 0.0144236.

• The values at nodes C and A can be calculated similarly as

(1 − (0.54765 − 0.06289)) × (0.0095837 + 0.0093884)/2 + 0.01 1.05289

= 0.0141396

(1 − (0.36721 − 0.05)) × (0.0144236 + 0.0141396)/2 + 0.01 1.04

= 0.0189916,

(80)

• The value of the ARM to the issuer is hence \$0.0189916 per \$1 of loan amount.

• The above idea of scaling has wide applicability in pricing certain classes of path-dependent securities.a

aFor example, newly issued lookback options.

(81)

### More on ARMs

• ARMs are indexed to publicly available indices such as:

– libor

– The constant maturity Treasury rate (CMT) – The Cost of Funds Index (COFI).

• COFI is based on an average cost of funds.

• So it moves relatively sluggishly compared with libor.

• Since 1990, the need for securitization gradually shift in libor’s favor.a

aSee Morgenson (2012). The libor rate-fixing scandal broke in June 2012.

(82)

### More on ARMs (continued)

• If the ARM coupon reflects fully and instantaneously current market rates, then the ARM security will be priced close to par and refinancings rarely occur.

• In reality, adjustments are imperfect in many ways.

• At the reset date, a margin is added to the benchmark index to determine the new coupon.

(83)

### More on ARMs (concluded)

• ARMs often have periodic rate caps that limit the amount by which the coupon rate may increase or decrease at the reset date.

• They also have lifetime caps and floors.

• To attract borrowers, mortgage lenders usually oﬀer a below-market initial rate (the “teaser” rate).

• The reset interval, the time period between adjustments in the ARM coupon rate, is often annual, which is not frequent enough.

• But these terms are easy to incorporate into the pricing

(84)

### Expressing Prepayment Speeds

• The cash flow of a mortgage derivative is determined from that of the mortgage pool.

• The single most important factor complicating this endeavor is the unpredictability of prepayments.

• Recall that prepayment represents the principal payment made in excess of the scheduled principal amortization.

(85)

### Expressing Prepayment Speeds (concluded)

• Compare the amortization factor Balt of the pool with the reported factor to determine if prepayments have occurred.

• The amount by which the reported factor is exceeded by the amortization factor is the prepayment amount.

(86)

### Single Monthly Mortality

• A SMM of ω means ω% of the scheduled remaining balance at the end of the month will prepay (recall p.

1146).

• In other words, the SMM is the percentage of the remaining balance that prepays for the month.

• Suppose the remaining principal balance of an MBS at the beginning of a month is \$50,000, the SMM is 0.5%, and the scheduled principal payment is \$70.

• Then the prepayment for the month is 0.005 × (50,000 − 70) ≈ 250 dollars.

(87)

### Single Monthly Mortality (concluded)

• If the same monthly prepayment speed s is maintained since the issuance of the pool, the remaining principal balance at month i will be RBi × (1 − s/100)i.

• It goes without saying that prepayment speeds must lie between 0% and 100%.

(88)

### An Example

• Take the mortgage on p. 1174.

• Its amortization factor at the 54th month is 0.824866.

• If the actual factor is 0.8, then the (implied) SMM for the initial period of 54 months is

100 × [

1

( 0.8 0.824866

)1/54 ]

= 0.0566677.

• In other words, roughly 0.057% of the remaining principal is prepaid per month.

(89)

### Conditional Prepayment Rate

• The conditional prepayment rate (CPR) is the annualized equivalent of a SMM,

CPR = 100 × [

1 (

1 SMM 100

)12 ] .

• Conversely,

SMM = 100 × [

1 (

1 CPR 100

)1/12 ] .

(90)

### Conditional Prepayment Rate (concluded)

• For example, the SMM of 0.0566677 on p. 1195 is equivalent to a CPR of

100 × [

1 (

1

(0.0566677 100

)12)]

= 0.677897.

• Roughly 0.68% of the remaining principal is prepaid annually.

• The figures on 1198 plot the principal and interest cash flows under various prepayment speeds.

• Observe that with accelerated prepayments, the principal cash flow is shifted forward in time.

(91)

50 100 150 200 250 300 350 Month 200

400 600 800 1000 1200 1400

2%

4%

6%

10%

15%

50 100 150 200 250 300 350 Month 100

200 300 400 500 600 700 800

15%

10%6%

4%

2%

Principal (left) and interest (right) cash flows at various CPRs. The 6% mortgage has 30 years to maturity and an original loan amount of \$100,000.

(92)

### PSA

• In 1985 the Public Securities Association (PSA) standardized a prepayment model.

• The PSA standard is expressed as a monthly series of CPRs.

– It reflects the increase in CPR that occurs as the pool seasons.

• At the time the PSA proposed its standard, a seasoned 30-year GNMA’s typical prepayment speed was ∼ 6%

CPR.

(93)

### PSA (continued)

• The PSA standard postulates the following prepayment speeds:

– The CPR is 0.2% for the first month.

– It increases thereafter by 0.2% per month until it reaches 6% per year for the 30th month.

– It then stays at 6% for the remaining years.

• The PSA benchmark is also referred to as 100 PSA.

• Other speeds are expressed as some percentage of PSA.

– 50 PSA means one-half the PSA CPRs.

– 150 PSA means one-and-a-half the PSA CPRs.

(94)

0 50 100 150 200 250 300 350 Mortgage age (month)

0 2 4 6 8 10

CPR (%)

100 PSA 150 PSA

50 PSA

(95)

### PSA (concluded)

• Mathematically,

CPR =

6% × PSA100 if the pool age exceeds 30 months

0.2% × m × PSA100 if the pool age m ≤ 30 months

• Conversely,

PSA =

100 × CPR6 if the pool age exceeds 30 months

100 × 0.2CPR×m if the pool age m ≤ 30 months

(96)

### Cash Flows at 50 and 100 PSAs

50 100 150 200 250 300 350 Month 100

200 300 400 500

Principal

Interest

50 100 150 200 250 300 350 Month 100

200 300 400 500

Principal

Interest

The 6% mortgage has 30 years to maturity and an original loan amount of \$100,000. The 100 PSA scenario is on the left, and the 50 PSA is on the right.

(97)

### Prepayment Vector

• The PSA tries to capture how prepayments vary with age.

• But it should be viewed as a market convention rather than a model.

• A vector of PSAs generated by a prepayment model should be used to describe the monthly prepayment speed through time.

• The monthly cash flows can be derived thereof.

(98)

### Prepayment Vector (continued)

• Similarly, the CPR should be seen purely as a measure of speed rather than a model.

• If one treats a single CPR number as the true

prepayment speed, that number will be called the constant prepayment rate.

• This simple model crashes with the empirical fact that pools with new production loans typically prepay at a slower rate than seasoned pools.

• A vector of CPRs should be preferred.

(99)

### Prepayment Vector (concluded)

• A CPR/SMM vector is easier to work with than a PSA vector because of the lack of dependence on the pool age.

• But they are all equivalent as a CPR vector can always be converted into an equivalent PSA vector and vice versa.

(100)

### Cash Flow Generation

• Each cash flow is composed of the principal payment, the interest payment, and the principal prepayment.

• Let Bk denote the actual remaining principal balance at month k.

• The pool’s actual remaining principal balance at time i − 1 is Bi−1.

(101)

### Cash Flow Generation (continued)

• The principal and interest payments at time i are

Pi ≡ Bi−1

(Bali−1 − Bali Bali−1

)

(150)

= Bi−1 r/m

(1 + r/m)n−i+1 − 1 (151) Ii ≡ Bi−1 r − α

m (152)

– α is the servicing spread (or servicing fee rate),

which consists of the servicing fee for the servicer as well as the guarantee fee.

(102)

### Cash Flow Generation (continued)

• The prepayment at time i is PPi = Bi−1 Bali

Bali−1 × SMMi.

– SMMi is the prepayment speed for month i.

• If the total principal payment from the pool is Pi + PPi, the remaining principal balance is

Bi = Bi−1 − Pi − PPi

= Bi−1 [

1

(Bali−1 − Bali Bali−1

)

Bali

Bali−1 × SMMi ]

= Bi−1 × Bali × (1 − SMMi)

Bali−1 . (153)

(103)

### Cash Flow Generation (continued)

• Equation (153) can be applied iteratively to yielda

Bi = RBi ×

i j=1

(1 − SMMj). (154)

• Define

bi

i j=1

(1 − SMMj).

aRBi is defined on p. 1172.

(104)

### Cash Flow Generation (continued)

• Then the scheduled P&I isa

Pi = bi−1Pi and Ii = bi−1Ii. (155) – Ii ≡ RBi−1 × (r − α)/m is the scheduled interest

payment.

• The scheduled cash flow and the bi determined by the prepayment vector are all that are needed to calculate the projected actual cash flows.

aPi and Ii are defined on p. 1175.

(105)

### Cash Flow Generation (concluded)

• If the servicing fees do not exist (that is, α = 0), the

projected monthly payment before prepayment at month i becomes

Pi + Ii = bi−1(Pi + Ii) = bi−1C. (156) – C is the scheduled monthly payment on the original

principal.

• See Figure 29.10 in the text for a linear-time algorithm for generating the mortgage pool’s cash flow.

(106)

### Cash Flows of Sequential-Pay CMOs

• Take a 3-tranche sequential-pay CMO backed by

\$3,000,000 of mortgages with a 12% coupon and 6 months to maturity.

• The 3 tranches are called A, B, and Z.

• All three tranches carry the same coupon rate of 12%.

(107)

### Cash Flows of Sequential-Pay CMOs (continued)

• The Z tranche consists of Z bonds.

– A Z bond receives no payments until all previous tranches are retired.

– Although a Z bond carries an explicit coupon rate, the owed interest is accrued and added to the

principal balance of that tranche.

– The Z bond thus protects earlier tranches from extension risk

• When a Z bond starts receiving cash payments, it becomes a pass-through instrument.

(108)

### Cash Flows of Sequential-Pay CMOs (continued)

• The Z tranche’s coupon cash flows are initially used to pay down the tranches preceding it.

• Its existence (as in the ABZ structure here) accelerates the principal repayments of the sequential-pay bonds.

• Assume the ensuing monthly interest rates are 1%, 0.9%, 1.1%, 1.2%, 1.1%, 1.0%.

• Assume that the SMMs are 5%, 6%, 5%, 4%, 5%, 6%.

• We want to calculate the cash flow and then the fair price of each tranche.

Updating...

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