The HullWhite 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 σr^{b}.
• But it has at least two shortcomings.
• First, the resulting trinomial tree is irregular (p. 1010).
– So higher complexity in programming.
• The second shortcoming is again a consequence of the tree’s irregular shape.
The HullWhite Model: Calibration with Irregular Trinomial Trees (concluded)
• Recall that the algorithm figured out θ(t_{i}) that matches the spot rate r(0, t_{i+2}) in order to determine the
branching schemes for the nodes at time t_{i}.
• But without those branches, the tree was not specified, and backward induction on the tree was not possible.
• To avoid this dilemma, the algorithm turned to the
continuoustime model to evaluate Eq. (122) on p. 1014 that helps derive θ(t_{i}) later.
• The resulting θ(t_{i}) hence might not yield a tree that matches the spot rates exactly.
The HullWhite Model: Calibration with Regular Trinomial Trees
^{a}• We will simplify the previous algorithm to exploit the fact that the HullWhite model has a constant diffusion term σ.
• The resulting trinomial tree will be regular.
• All the θ(t_{i}) terms can be chosen by backward induction to match the spot rates exactly.
• The tree is constructed in two phases.
aHull and White (1994).
The HullWhite Model: Calibration with Regular Trinomial Trees (continued)
• In the first phase, a tree is built for the θ(t) = 0 case, which is an OrnsteinUhlenbeck process:
dr = −ar dt + σ dW, r(0) = 0.
– The tree is daggershaped (p. 1024).
– The number of nodes above the r_{0}line, j_{max}, and that below the line, j_{min}, will be picked so that the probabilities (121) on p. 1011 are positive for all nodes.
– The tree’s branches and probabilities are in place.
The HullWhite Model: Calibration with Regular Trinomial Trees (concluded)
• Phase two fits the term structure.
– Backward induction is applied to calculate the β_{i} to add to the short rates on the tree at time t_{i} so that the spot rate r(0, t_{i+1}) is matched.
*  j
(0, 0) r_{0}
*  j
(1, 1)
*  j
(1, 0)
* 
(1, −1) j
*  j
*  j
*  j
*  j µ
* 
 j R
*  j
*  j
*  j
*  j µ
* 
 j R
*  j
*  j
*  j
*  j µ
* 
 j R
*  j
*  j
*  j
*  j µ
*  
¾
∆t
6?^{∆r}
The short rate at node (0, 0) equals r_{0} = 0; here j_{max} = 3 and j_{min} = 2.
The HullWhite Model: Calibration
• Set ∆r = σ√
3∆t and assume that a > 0.
• Node (i, j) is a top node if j = j_{max} and a bottom node if j = −j_{min}.
• Because the root of the tree has a short rate of r_{0} = 0, phase one adopts r_{j} = j∆r.
• Hence the probabilities in Eqs. (121) on p. 1011 use η ≡ −aj∆r∆t + (j − k) ∆r.
The HullWhite 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 , (124)
p2(i, j) = 2 3 −
h
a2j2(∆t)2 − 2aj∆t(j − k) + (j − k)2 i
, (125)
p3(i, j) = 1
6 + a2j2(∆t)2 − 2aj∆t(j − k) + (j − k)2 + aj∆t − (j − k)
2 . (126)
The HullWhite 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, p_{1}(i, j) = p_{3}(i, −j).
The HullWhite Model: Calibration (continued)
• The inequalities
3 − √ 6
3 < ja∆t <
r2
3 (127)
ensure that all the branching probabilities are positive in the upper half of the tree, that is, j > 0 (verify this).
• Similarly, the inequalities
− r2
3 < ja∆t < −3 − √ 6 3
ensure that the probabilities are positive in the lower half of the tree, that is, j < 0.
The HullWhite Model: Calibration (continued)
• To further make the tree symmetric across the r_{0}line, we let j_{min} = j_{max}.
• As ^{3−}_{3}^{√}^{6} ≈ 0.184, a good choice is
j_{max} = d0.184/(a∆t)e.
• Phase two computes the β_{i}s to fit the spot rates.
• We begin with state price Q(0, 0) = 1.
• Inductively, suppose that spot rates r(0, t_{1}), r(0, t_{2}), . . . , r(0, t_{i}) have already been matched at time t_{i}.
The HullWhite Model: Calibration (continued)
• By construction, the state prices Q(i, j) for all j are known by now.
• The value of a zerocoupon bond maturing at time t_{i+1} equals
e^{−r(0,t}^{i+1}^{)(i+1) ∆t} = X
j
Q(i, j) e^{−(β}^{i}^{+r}^{j}^{)∆t} by riskneutral valuation.
• Hence
β_{i} = r(0, t_{i+1})(i + 1) ∆t + lnP
j Q(i, j) e^{−r}^{j}^{∆t}
∆t ,
The HullWhite Model: Calibration (concluded)
• The state prices at time t_{i+1},
Q(i + 1, j), −j_{max} ≤ j ≤ j_{max}, can now be calculated as before.
• The total running time is O(nj_{max}).
• The space requirement is O(n).
A Numerical Example
• Assume a = 0.1, σ = 0.01, and ∆t = 1 (year).
• Immediately, ∆r = 0.0173205 and j_{max} = 2.
• The plot on p. 1033 illustrates the 3period trinomial tree after phase one.
• For example, the branching probabilities for node E are calculated by Eqs. (124)–(126) on p. 1026 with j = 2 and k = 1.
*  j
A
*  j
B *
 j
C *
 j
D
 j R
E *
 j
F *
 j
G *
 j
H µ
*
I 
Node A, C, G B, F E D, H I
r (%) 0.00000 1.73205 3.46410 −1.73205 −3.46410 p_{1} 0.16667 0.12167 0.88667 0.22167 0.08667 p_{2} 0.66667 0.65667 0.02667 0.65667 0.02667 p_{3} 0.16667 0.22167 0.08667 0.12167 0.88667
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.
A Numerical Example (continued)
• Now,
β_{0} = r(0, 1) + ln Q(0, 0) e^{−r}^{0} = r(0, 1) = 3.82365%.
• Hence the short rate at node A equals β_{0} + r_{0} = 3.82365%.
• The state prices at year one are calculated as Q(1, 1) = p_{1}(0, 0) e^{−(β}^{0}^{+r}^{0}^{)} = 0.160414, Q(1, 0) = p_{2}(0, 0) e^{−(β}^{0}^{+r}^{0}^{)} = 0.641657, Q(1, −1) = p_{3}(0, 0) e^{−(β}^{0}^{+r}^{0}^{)} = 0.160414.
A Numerical Example (continued)
• The 2year rate spot rate r(0, 2) is matched by picking
β_{1} = r(0, 2)×2+ln h
Q(1, 1) e^{−∆r} + Q(1, 0) + Q(1, −1) e^{∆r} i
= 5.20459%.
• Hence the short rates at nodes B, C, and D equal β_{1} + r_{j},
where j = 1, 0, −1, respectively.
• They are found to be 6.93664%, 5.20459%, and 3.47254%.
A Numerical Example (continued)
• The state prices at year two are calculated as
Q(2, 2) = p_{1}(1, 1) e^{−(β}^{1}^{+r}^{1}^{)}Q(1, 1) = 0.018209,
Q(2, 1) = p_{2}(1, 1) e^{−(β}^{1}^{+r}^{1}^{)}Q(1, 1) + p_{1}(1, 0) e^{−(β}^{1}^{+r}^{0}^{)}Q(1, 0)
= 0.199799,
Q(2, 0) = p_{3}(1, 1) e^{−(β}^{1}^{+r}^{1}^{)}Q(1, 1) + p_{2}(1, 0) e^{−(β}^{1}^{+r}^{0}^{)}Q(1, 0) +p_{1}(1, −1) e^{−(β}^{1}^{+r}^{−1}^{)}Q(1, −1) = 0.473597,
Q(2, −1) = p_{3}(1, 0) e^{−(β}^{1}^{+r}^{0}^{)}Q(1, 0) + p_{2}(1, −1) e^{−(β}^{1}^{+r}^{−1}^{)}Q(1, −1)
= 0.203263,
Q(2, −2) = p_{3}(1, −1) e^{−(β}^{1}^{+r}^{−1}^{)}Q(1, −1) = 0.018851.
A Numerical Example (concluded)
• The 3year rate spot rate r(0, 3) is matched by picking
β_{2} = r(0, 3) × 3 + ln h
Q(2, 2) e^{−2×∆r} + Q(2, 1) e^{−∆r} + Q(2, 0) +Q(2, −1) e^{∆r} + Q(2, −2) e^{2×∆r}
i
= 6.25359%.
• Hence the short rates at nodes E, F, G, H, and I equal β_{2} + r_{j}, 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. 1039 plots β_{i} for i = 0, 1, . . . , 29.
<HDU +L/
E_{L}+/
Introduction to MortgageBacked Securities
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
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.
MortgageBacked Securities
• A mortgagebacked security (MBS) is a bond backed by an undivided interest in a pool of mortgages.
• MBSs traditionally enjoy high returns, wide ranges of products, high credit quality, and liquidity.
• The mortgage market has witnessed tremendous innovations in product design.
MortgageBacked 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).
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 passthrough security (MPTS).
2. Collateralized mortgage obligation (CMO).
3. Stripped mortgagebacked security (SMBS).
Problems Investing in Mortgages
• The mortgage sector is one of the largest in the debt market (see text).
• 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.
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.
Mortgage PassThroughs
• 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 2^{1/2}% of the cash flow.
• Because of higher marketability, a passthrough is easier to sell than its individual loans.
Rule for distribution of cash flows: pro rata Loan 2
Loan 1
Passthrough: $1 million par pooled mortgage loans
Collateralized Mortgage Obligations (CMOs)
• A passthrough 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 passthroughs have a single maturity and are backed by individual mortgages.
• CMOs are multiplematurity, multiclass debt
instruments collateralized by passthroughs, stripped mortgagebacked securities, and whole loans.
Collateralized Mortgage Obligations (CMOs) (concluded)
• 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.”
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
An Example
• Consider a twotranche sequentialpay CMO backed by
$1,000,000 of mortgages with a 12% coupon and six months to maturity.
• The cash flow pattern for each tranche with zero
prepayment and zero servicing fee is shown on p. 1054.
• The calculation can be carried out first for the Total columns, which make up the amortization schedule, before the cash flow is allocated.
• Tranche A is retired after four months, and tranche B starts principal paydown at the end of month four.
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. Monthi numbers reflect the ith monthly payment.
Another Example
• When prepayments are present, the calculation is 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. (4) on p. 44 times (0.95)^{i}.
• This done for all the months, the total interest payment at any month is the remaining principal of the previous
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 1058 tabulates the cash flows of the same twotranche CMO under 5% SMM.
• For instance, the total principal payment at month one,
$204,421, can be verified as follows.
Another Example (concluded)
• The scheduled remaining principal is $837,452 from p. 1054.
• 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 three months, and tranche B starts principal paydown at the end of month three.
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
Monthi numbers reflect the ith monthly payment.
Stripped MortgageBacked Securities (SMBSs)
• They were created in February 1987 when Fannie Mae issued its Trust 1 stripped MBS.
• The principal and interest are divided between the PO strip and the IO strip.
• In the scenarios on p. 1053 and p. 1055:
– 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.
Stripped MortgageBacked Securities (SMBSs) (concluded)
• These new instruments allow investors to better exploit anticipated changes in interest rates.
• The collateral for an SMBS is a passthrough,.
• CMOs and SMBSs are usually called derivative MBSs.
Prepayments
• The prepayment option sets MBSs apart from other fixedincome 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.
• Even when the prevailing mortgage rate exceeds the mortgage’s loan rate, some loans are prepaid.
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
Prepayment Risk (concluded)
• Prepayments can be in whole or in part.
– The former is called liquidation.
– The latter is called curtailment.
• Prepayments, however, need not always result in losses.
• The holder of a passthrough 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.
Other Risks
• Investors in mortgages are exposed to at least three other risks.
– Interest rate risk is inherent in any fixedincome 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.
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.
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 fixedrate 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.
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.
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. 1069 illustrates the typical price/yield curves of the Treasury and passthrough.
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.
Prepayment: Characteristics (concluded)
• As yields fall and the passthrough’s price moves above a certain price, it flattens and then follows a downward slope.
• This phenomenon is called the price compression of premiumpriced MBSs.
• It demonstrates the negative convexity of such securities.
Analysis of MortgageBacked Securities
Oh, well, if you cannot measure, measure anyhow.
— Frank H. Knight (1885–1972)
Uniqueness of MBS
• Compared with other fixedincome 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.
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.
Cash Flow Analysis
• A traditional mortgage has a fixed term, a fixed interest rate, and a fixed monthly payment.
• Page 1076 illustrates the scheduled P&I for a 30year, 6% mortgage with an initial balance of $100,000.
• Page 1077 depicts how the remaining principal balance decreases over time.
Scheduled Principal and Interest Payments
50 100 150 200 250 300 350 Month 100
200 300 400 500 600
Principal Interest
Scheduled Remaining Principal Balances
50 100 150 200 250 300 350 Month 20
40 60 80 100
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.
• Identical characteristics hold for the pool’s P&I
payments in the absence of prepayments and servicing fees.
Cash Flow Analysis (continued)
• From Eq. (4) on p. 44 the remaining principal balance after the kth payment is
C 1 − (1 + r/m)^{−n+k}
r/m . (128)
– C is the scheduled P&I payment of an nmonth mortgage making m payments per year.
– r is the annual mortgage rate.
• For mortgages, m = 12.
Cash Flow Analysis (continued)
• The remaining principal balance after k payments can be expressed as a portion of the original principal
balance:
Bal_{k} ≡ 1 − (1 + r/m)^{k} − 1 (1 + r/m)^{n} − 1
= (1 + r/m)^{n} − (1 + r/m)^{k}
(1 + r/m)^{n} − 1 . (129)
• This equation can be verified by dividing Eq. (128) (p. 1079) by Bal_{0}.
Cash Flow Analysis (continued)
• The remaining principal balance after k payments is RB_{k} ≡ O × Bal_{k},
where O will denote the original principal balance.
• The term factor denotes the portion of the remaining principal balance to its original principal balance
expressed as a decimal.
• So Bal_{k} is the monthly factor when there are no prepayments.
• It is also known as the amortization factor.
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.
An Example
• The remaining balance of a 15year 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.
P&I Analysis
• By the amortization principle, the tth interest payment equals
I_{t} ≡ RB_{t−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 P_{t} ≡ RB_{t−1} − RB_{t}
= O × (r/m)(1 + r/m)^{t−1}
(1 + r/m)^{n} − 1 . (130)
P&I Analysis (concluded)
• The scheduled P&I payment at month t, or P_{t} + I_{t}, is (RB_{t−1} − RB_{t}) + RB_{t−1} × r
m
= O ×
· (r/m)(1 + r/m)^{n} (1 + r/m)^{n} − 1
¸
, (131)
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.
An Example
• The mortgage on pp. 39ff has a monthly payment of 250000 × (0.08/12) × (1 + (0.08/12))^{180}
(1 + (0.08/12))^{180} − 1 = 2389.13 by Eq. (131) on p. 1085.
• This number agrees with the number derived earlier.