fixed income analysis

Module 11: Assignment1
PIMCO video (2 points)
Provide an executive summary of this module’s video.
2
GSAM global fixed income commentary (2 points)
Provide a brief discussion of this week’s “chart of the week.”
3
Research questions (9 points)
1. With reference to the GS research report “An OAS Primer,” answer the following questions
(4 points):
• What is the difference between the on-the-run ad the off-the-run curves?
• How is the price of a currently callable Treasury security calculated?
• How is the OAS on an agency security calculated?
• What are the tree main factors driving prepayments in Goldman’s model?
2. With reference to the PIMCO research report “A New PIMCO GNMA MBS Prepayment
Model,” answer the following questions (5 points):
• What is the difference between Fannie Mae or Freddie Mac MBSs and GNMA MBSs?
• What is the difference between the mortgages in the typical GNMA loan pool v. the
mortgages in the GSE-sponsored MBSs?
• What are the four components of prepayments in PIMCO’s agency prepayment model?
• How do GNMA prepayments compare to FNMA prepayments?
• How do the durations of GNMAs and FNMAs in Figure 8 vary as a function of the
coupon rate?
1
4
Practice problems (6 points)
Provide Excel templates for the solution of the following problems (2 points each).
1. Assume cm = 0.10/12. Calculate the mortgage balance after 5 and 10 years, for a levelpayment, fixed-rate mortgage with maturity n = 360, and principal M B = 100, 000.
Solution:
To calculate the mortgage balance at a future date, we first need to calculate the monthly
payment. To do this, we solve the mortgage amortization formula for the mortgage payment
MP :

1
M P = M B n × cm × 1 −
(1 + cm )n
−1
Plugging in n = 360, M B360 = 100, 000, and cm = 0.10
, we get M P = 877.57. We can now
12
calculate the mortgage balance at any time as the present value of the remaining n mortgage
payments:


MP
1
M Bn =
1−
cm
(1 + cm )n
, and n = 300, we get M B300 = 96, 574.32. And plugging
Plugging in M P = 877.57, cm = 0.10
12
in n = 240 we get M B240 = 90, 938.02.
For the following problem assume the interest-rate tree:
y1 = 4%
y1u = 5.5%
y1d = 3.5%
2. Consider now a mortgage with maturity n = 2, initial balance M B2 = 100, 000, and cm =
4.29222%.
Calculate the mortgage balance at time 1.
2
Calculate the price of the prepayment-free mortgage at each node of the tree.
Calculate the market value of the prepayment option, C, assuming optimal prepayment.
(Remember that that the mortgage cannot be prepaid at time 0.)
Calculate the current market price of the mortgage.
Also, calculate the prices of a PO and an IO strip backed by the mortgage and verify that
the sum of the prices of the two strips equals the price of the mortgage.
Solution:
Pricing the mortgage
We first calculate the mortgage payment:

M P = M B n × cm × 1 −
1
(1 + cm )n

= 100, 000 × 4.29222% × 1 −
−1
1
(1.0429222)2
−1
= 53, 241.71
We then get the mortgage balance at time 1 by calculating the present value of future cash
flows as of that point, discounted at cm . As of time one, there is only one payment remaining:
M B10 =
MP
= 51, 050.51
1.0429222
We calculate the value of the prepayment-free mortgage as we do any stream of cash flows:
PNuuP = 0
MP
PNu P = 1+y
u = 50, 466.08
1
PN P =
0
E[PN
P ]+M P
1+y1
PNudP = 0
= 100, 187.87
MP
PNd P = 1+y
d = 51, 441.27
1
PNddP = 0
3
Given these values, we can calculate the value of the prepayment option C:
C u = max{0, PNu P − M B10 } = 0.00
C = W = 187.86
C d = max{0, PNd P − M B10 } = 390.76
Note that, at origination, the mortgage cannot be prepaid and C = W .
Finally, the market price of the mortgage is the price of the prepayment-free mortgage minus
the value of the call option:
Pm = P N P − C
= 100, 000.00
Pricing the P O and IO strips
We now calculate the prices of the a PO and IO strip at each node in the tree. At time 1,
the formula for a PO strip is
PP0 O = I(N P 0 ) ×
E[PP00O ] + M P − cm M B10
+ [1 − I(N P 0 )] × M B10 ,
0
1 + y1
where M B10 is the remaining mortgage balance at the particular node in the tree and I(N P 0 )
is an indicator function that equals one, if there is no prepayment, and equals zero otherwise.
4
We use this formula in our pricing tree as follows:
E u [PP00O ] + M P − cm M B10
+ [1 − I(N P u )] × M B10
1 + y1u
0 + 53, 241.71 − 4.29222% × 51, 050.51
=
1.055
PPuO = I(N P u ) ×
= 48, 389.11
PPd O = I(N P d ) ×
E d [PP00O ] + M P − cm M B10
+ [1 − I(N P d )] × M B10
1 + y1d
= 51, 050.51
E[PP0 O ] + M P − cm M B2
1 + y1
1
(48, 389.11 + 51, 050.51) + 53, 241.71 − 4.29222% × 100, 000
= 2
1.04
PP O =
= 94, 874.33
Note that, at origination, we are assuming that the mortgage cannot be prepaid, and the
price of the PO simply equals the discounted expected price next period plus principal
payment.
We now price the interest-only mortgage strip using the formula given in the notes:
PIO = I(N P ) ×
5
0
E[PIO
] + cm M B 2
1 + y1
We use this formula in our pricing tree as follows:
00
E u [PIO
] + cm M B10
1 + y1u
0 + 4.29222% × 51, 050.51
=1×
1.055
u
PIO
= I(N P u ) ×
= 2, 076.97
d
PIO
= I(N P d ) ×
00
E d [PIO
] + cm M B10
1 + y1d
=0
0
E[PIO
] + cm M B 2
1 + y1
1
(2, 076.97 + 0) + 4.29222% × 100, 000
= 2
1.04
PIO =
= 5, 125.68
Again, at origination, the mortgage cannot be prepaid, and the price of the PO equals the
expected price next period plus interest payment. We can also verify that in each node, the
mortgage price P equals the sum of the prices of the P O and IO strips:
u
PPuO + PIO
= 48, 389.11 + 2, 076.97 = 50, 466.08 = P u
d
PPd O + PIO
= 51, 050.51 + 0 = 51, 050.51 = P d
PP O + PIO = 94, 874.33 + 5, 125.68 = 100, 000 = P.
3. Calculate the price of the mortgage and of the mortgage strips of the previous problem
assuming behavioral prepayment, and λ0 = 99.90% (λ = 100%). Again, verify that the sum
of the prices of the two strips equal the price of the mortgage.
Solution:
Pricing the mortgage
Now, we calculate the price of the mortgage using behavioral repayment. The recursive
6
formula is:
Pm0 = λ0 ×


E(Pm00 ) + M P
+ (1 − λ0 ) × M B10
1 + y1
Applying this formula to our interest rate tree, we have
Pmu = λ0 ×
h u
i
+ (1 − λ0 ) × M B10 = 50, 466.66
Pmd = λ0 ×
h d
i
+ (1 − λ0 ) × M B10 = 51, 440.88
00 )+M P
E (Pm
1+y1u
0
m )+M P
= 100, 187.96
Pm = E(P1+y
1
00 )+M P
E (Pm
1+y1d
where we used the fact that at origination λ = 100%. Notice that when calculating Pmu and
Pmd , E u (Pm00 ) = E d (Pm00 ) = 0, since the mortgage will have ended by the end of the period.
And when calculating Pm , E(Pm0 ) = 12 (50, 466.66 + 51, 440.88).
Pricing the P O and IO strips
We now calculate the PO and IO strips. We have the following formula for PP0 O :
PP0 O = λ0 ×
E[PP00O ] + M P − cm M B10
+ (1 − λ0 ) × M B10
1 + y10
Notice that the formula is the same as in the case of optimal prepayment, but we’ve replaced
7
I(N P 0 ) with λ0 . We apply this formula to the pricing tree as follows:
E u [PP00O ] + M P − cm M B10
+ (1 − λ0 ) × M B10
1 + y1u
0 + 53, 241.71 − 4.29222% × 51, 050.51
+ 0.1% × 51, 050.51
= 99.9% ×
1.055
PPuO = λ0 ×
= 48, 391.77
E d [PP00O ] + M P − cm M B10
+ (1 − λ0 ) × M B10
1 + y1d
0 + 53, 241.71 − 4.29222% × 51, 050.51
= 99.9% ×
+ 0.1% × 51, 050.51
1.035
PPd O = λ0 ×
= 49, 325.89
E[PP0 O ] + M P − cm M B2
1 + y1
1
(48, 391.77 + 49, 325.89) + 53, 241.71 − 4.29222% × 100, 000
= 2
1.04
PP O =
= 94, 046.46
where we used the fact that, at origination, the mortgage cannot be prepaid.
We now price the IO mortgage strip using the formula
00
] + cm M B10
E[PIO
0
0
PIO = λ ×
1 + y10
8
Our price tree is as follows:
00
E u [PIO
] + cm M B10
1 + y1u
0 + 4.29222% × 51, 051.51
= 99.9% ×
1.055
u
PIO
= λ0 ×
= 2, 074.89
00
E d [PIO
] + cm M B10
1 + y1d
0 + 4.29222% × 51, 051.51
= 99.9% ×
1.035
d
PIO
= λ0 ×
= 2, 114.98
0
E[PIO
] + cm M B
1 + y1
1
(2, 074.89 + 2, 114.98) + 4.29222% × 100, 000
= 2
1.04
PIO =
= 6, 141.50
where we used the fact that, at origination, the mortgage cannot be prepaid. We can verify
that Pm = PP O + PIO at each node:
u
PPuO + PIO
= 48, 397.77 + 2, 047.89 = 50, 466.66 = Pmu
d
PPd O + PIO
= 49, 325.89 + 2, 114.98 = 51, 440.88 = Pmd
PP O + PIO = 94, 046.46 + 6, 141.50 = 100, 187.96 = Pm .
9
MODULE 11 ASSIGNMENT, PROBLEM 1
c_m=
MB_360=
n=
0.83%
$ 100,000.00
360
MP=
?
MB_300=
MB_240=
?
?
MODULE 11 ASSIGNMENT, PROBLEM 2
INTEREST RATE TREE
t=0
y_1=
m= 0.50%
t=1
y_1^u=
5.5%
y_1^d=
3.5%
MP=
?
sigma= 1%
4.0%
c_m=
n=
4.29222%
2
MB_0=
MB_1’=
$100,000.00
?
NON-PREPAYABLE MORTAGE, PRICE TREE
t=0
P_NP^u=
P_NP=
?
P_NP^d=
t=1
?
?
PREPAYMENT OPTION PRICE TREE
t=0
C^u=
C=
?
C^d=
t=1
?
?
MORTGAGE PRICE TREE, BEHAVIORAL PREPAYMENT
t=0
t=1
P_m^u=
?
P_m=
?
P_m^d=
?
NO-PREPAYMENT INDICATOR FUNCTION, I(y_1)
t=2
I(y_1)^u=
?
PO PRICE TREE, OPTIMAL PREPAYMENT
t=0
P_PO^u=
P_PO=
?
P_PO^d=
IO PRICE TREE, OPTIMAL PREPAYMENT
t=0
t=1
P_IO^u=
?
P_IO=
?
P_IO^d=
?
VERIFY THAT P_m=P_PO+P_IO
t=0
P_PO^u+P_IO^u=
P_PO+P_IO=
?
P_PO^d+P_IO^d=
t=1
?
?
t=1
?
?
I(y_1)^d=
?
MODULE 11 ASSIGNMENT, PROBLEM 3
INTEREST RATE TREE
t=0
y_1=
m= 0.50%
t=1
y_1^u=
5.5%
y_1^d=
3.5%
4.0%
c_m=
n=
4.29222%
2
MB_0=
MB_1=
$100,000.00
?
MORTGAGE PAYMENT, MP=
MORTGAGE PRICE TREE, BEHAVIORAL PREPAYMENT
t=0
t=1
lambda=
100%
99.90%
P_m=
P_m^u=
?
P_m^d=
?
?
PO PRICE TREE, BEHAVIORAL PREPAYMENT
P_PO=
P_PO^u=
?
P_PO^d=
?
?
IO PRICE TREE, BEHAVIORAL PREPAYMENT
P_IO=
P_IO^u=
?
P_IO^d=
?
P_PO^u+P_IO^u=
?
P_PO^d+P_IO^d=
?
?
VERIFY THAT P_m=P_PO+P_IO
P_PO+P_IO=
sigma= 1%
?
?
MODULE 11 ASSIGNMENT, PROBLEM 1
c_m=
MB_360=
n=
0.83%
$ 100,000.00
360
MP=
$877.57
MB_300=
MB_240=
$96,574.32
$90,938.02
MODULE 11 ASSIGNMENT, PROBLEM 2
INTEREST RATE TREE
t=0
y_1=
4.0%
c_m=
n=
4.29222%
2
MB_0=
MB_1’=
$100,000.00
$51,050.51
m= 0.50%
t=1
y_1^u=
5.5%
y_1^d=
3.5%
MP=
$53,241.71
NON-PREPAYABLE MORTAGE, PRICE TREE
t=0
P_NP^u=
$100,187.87
P_NP=
P_NP^d=
t=1
$50,466.08
$51,441.27
sigma= 1%
PREPAYMENT OPTION PRICE TREE
t=0
t=1
C^u=
$0.00
C=
$187.86
C^d=
$390.76
MORTGAGE PRICE TREE, BEHAVIORAL PREPAYMENT
t=0
t=1
P_m^u=
$ 50,466.08
P_m=
$ 100,000.00
P_m^d=
$ 51,050.51
NO-PREPAYMENT INDICATOR FUNCTION, I(y_1)
t=2
TRUE
I(y_1)^u=
PO PRICE TREE, OPTIMAL PREPAYMENT
t=0
P_PO^u=
P_PO=
$ 94,874.33
P_PO^d=
IO PRICE TREE, OPTIMAL PREPAYMENT
t=0
t=1
P_IO^u= $ 2,076.97
P_IO=
$ 5,125.68
P_IO^d= $
–
VERIFY THAT P_m=P_PO+P_IO
t=0
P_PO^u+P_IO^u=
P_PO+P_IO= $ 100,000.00
P_PO^d+P_IO^d=
t=1
$ 48,389.11
$ 51,050.51
t=1
$ 50,466.08
$ 51,050.51
I(y_1)^d=
FALSE
MODULE 11 ASSIGNMENT, PROBLEM 3
INTEREST RATE TREE
t=0
y_1=
4.0%
c_m=
n=
4.29222%
2
MB_0=
MB_1=
$100,000.00
$51,050.51
m= 0.50%
t=1
y_1^u=
5.5%
y_1^d=
3.5%
MORTGAGE PAYMENT, MP= $
MORTGAGE PRICE TREE, BEHAVIORAL PREPAYMENT
t=0
t=1
lambda=
100%
99.90%
P_m=
$100,187.96
P_m^u=
$50,466.66
P_m^d=
$51,440.88
PO PRICE TREE, BEHAVIORAL PREPAYMENT
P_PO=
$94,046.46
P_PO^u=
$48,391.77
P_PO^d=
$49,325.89
IO PRICE TREE, BEHAVIORAL PREPAYMENT
P_IO=
$6,141.50
P_IO^u=
$2,074.89
P_IO^d=
$2,114.98
P_PO^u+P_IO^u=
$50,466.66
P_PO^d+P_IO^d=
$51,440.88
VERIFY THAT P_m=P_PO+P_IO
P_PO+P_IO=
$100,187.96
sigma= 1%
53,241.71