important of using examples to see the pattern of cash flows. Timeline
2: This timeline has the interest and principle withdrawn from timeline
1 Timeline 3: This is a standard trick for decreasing cashflow. (1) First
raise the first amount by the constant
an interest swap the t = i value is the expected one-year forward rate
fi1,i It is sometimes convenient to equivalently use the one-year
forward rate factor 1 + fi1,i It is instructive to work out why use of
the factor vs. rate is eqiuvalent 22.22 Unusual
=95% L 2 150% I 10% 95% L 150% x 10% 95% L 10% 95% L = 5% 95% L
(95%)2 L Exponent 2 = t=2 3 150% I 10%(95%)2 L 5% (95%)2 L (95%)3 L
Exponent 3= t=3 10 (95%)10 L Justified by Pattern of
examples 11 X 12 X 13 14-19 20 X Table 2: We break the problem
into tw
Solution reflects Dr Hendels approach to these problems This problem
requires 6 timelines and 5 equations of value TIMELINE 1 0-100
(t=1)-100(t=2) - An exchange is made at t = 5. So we need the the
outstanding balance (OB) at t=5. This outstanding balance
time different than 0. The recalculation may involve different i,n. It may
also involve conversions. This problem has 3 timelines and 5 equations.
So it affords us an opportunity to review how to approach solving
several equations in several variables Tim
equations and unknowns: (2 equations; 1 unknown) Subtract or
divide We can solve for i using the calculator TV Line N I PV PMT FV 20
CPT =4.2 -5341.12/400 1 0 i = 8.4%So i/2 = 4.2% QIT#55 Solutions
Sp 2015, Dr Hendel Solutions Reflect Dr Hendels Approach
problem It points out that definitionally a swap can refer to any
actuarially equivalent sequence of payments This is true But I feel it
misleading (I dont think the problem is good) The standard use of a
swap is to find an actuarially equivalent sequence
from now. So we need a discount factor, actually an accumulation
factor. 3) Adding (1) and (2) but subtracting (3), we get 9792.39 +
3525.26 - 2460.38 = 10857.27. The published SOA solutions noticed a
further pattern. They noticed that 198 = 36 x 5.5. So
pay the forward price S0e T e rT Using the concept of a synthetic
long forward we can neatly explain the cash and carry arbitrage In the
pen example, we shorted a pen high and longed a pen low and profited
on the difference Similarly in the cash and carry
swap, swap spread, deferred swap, accretizing swap, amortizing swap,
swaption You can get a good idea of their meaning by googling them
We are focusing on the numerical solutions and mathematical methods
in this chapter We also briefly mention LIBOR and
actuarially equivalent payment scheme Typically swaps refer to
actuarially equivalent level payment schemes Swaps are used to
hedge (avoid losses from) varying prices 259 260 CHAPTER 22. SWAPS
22.3 Swaps - Basic Timeline Approach Given: You are given a se
Key Phrases Underlined A 20-year loan of 1000 is repaid with payments
at the end of each year. Each of the first ten payments equals 150% of
the amount of interest due. Each of the last ten payments is X. The
lender charges interest at an annual effective
butterfly spread E. Sell a butterfly spread 19.18 D-SQ#8 - Solution
Volatile means you expect stock market values to go away from the
strike price Butterflys have graphical form HUDH; Written Butterflys
have graphical form HDUH. So if you are away from t
issues The last payment is the 8th payment The first payment of the
new loan is at the time of the 9th payment of the old loan The original
loan is in half years while the refinanced loan is in months The first
loan is at 6.09% per half year; the 2nd load
tables gives a 3% yield for all redemption values and prices. Hence we
use table IA: The maximum, indeed only price, is 1261.80. Why? By
Table IB, the buyer may get either 3%, 3.07% or 3.10% depending on
when the seller calls. So indeed the buyer gets at
you need to synthetically create a forward 21.17 More on cash and
carry Cash and Carry Arbitrage: by definition means an arbitrage that
buys the underlying asset and sells/shorts a forward You use a cash
and carry when Actual market forward price > Synthe
% % OLB8 I 1.0099 OLB8 I 1.00992 OLB8 I 1.00993 OLB8 I 1.00993
1.0075 OLB8 I 1.00993 1.00752 OLB8 I =L2 To understand what is going
on I made a timeline with 4 rows of labels. Notice how Chapter 2 is the
hard part of the problem not amortization as you mi
different for calls and puts (Since their profit regions are different) If
the current market price would motivate not exercising the option, the
option is OTM 19.10 Why Moneyness Remember in lecture 18 how
we used a put for insurance in the Happy Jalapen
approach advocated in these notes is powerful 224 CHAPTER 19.
DERIVATIVES - GRAPHING METHODS 19.20 D-SQ#15 The current price
of a non-dividend paying stock is 40 and the continuously compounded
risk-free rate of return is 8%. You enter into a short positi
EOV for Timeline #2 is: P V2 = Xv1 + Xv2 2 + Xv3 3 262 CHAPTER 22.
SWAPS Here vi , i = 1, 2, 3 is the present value or discount factor The
relationship between vi and ri is given by the following basic equation
v i i = 1 (1 + ri) i = Price of a 1 dollar
to solve OR we could solve as each equation comes up. That is a matter
of taste. But an important principle of solving many equations in many
unknowns is when each equation gives one more unknown. Summary:
Go back through the document. Make sure you under
is Y (13) Well X = 0.13 0.12 = 0.01 But slope = 10000 = X Y and
hence Y = 10000 0.01 = 100 So Y (0.13) = 100 + Y (0.12) = 289.91
18.23 DS-Q#3 Revisited We can now discuss DS-Q#3 more thoroughly
Suppose you wanted to make a Jalapeno deal You might make a
and that takes time which you have to spend TIP: Use variables not
numbers since it helps you see the pattern. T R I P OLB 0 1000 = L 1 10%
L 10% L 0 1000 2 10% L 10% L 0 1000 10 10% L 10% L 0 1000= L 11
15% L 10% L 5% L L -5%L=95% L 12 15% 95%L 10% 95%L
970.45 is the fair forward rate 21.20 Step 2: Test for Arbitrage Actual
market forward price: 965 Fair forward price: 970.45 Actual market
forward price is lower than the fair forward price Two different prices
for same entity arbitrage 21.21 Step 3: Arbi
#2 Fundamental technique: The first 12 payments of 2000 have a value
of at t=0. Similarly the next 12 payments of 2000(1.02) have a present
value at t=12 of We can now factor out of TIMELINE #2 the factor of We
have left TIMELINE #3 1 (t=0)-1.02 (t=1)-1.0
modified rate of i where 1/(1+i) i=0%.= 1.08/1.08=1 TIMELINE III
0-X(t=1)-X(t=2)-X(t=3).X(t=24) We now
have to find the outstanding balance of TIMELINE II at t=10. We also
need the payments at t=10 and t=11. For this we use the pattern
technique. Payment
with 8% annual coupons is bought at a premium to yield an annual
effective rate of 6%. Calculate the interest portion of the 7th coupon.
Figure 1: Text of QIT#10. Text is copyright SOA and reprinted with
permission English Algebra correspondence English A