In the previous example, notice that an increase in the yield by 10%

1

• In the previous example, notice that an increase
  in the yield by 10
• ? Decreases the price of the one-year bond by
  0.9
• ? Decreases the price of the two-year bond by
  1.71
• Thus, the difference between the two price falls
  is 0.81

2

• Suppose that the bond in the previous example had
  a three-year maturity
• The 10 increase in the yield lowers the bond
  price by 2.44
• Thus, the difference in the price fall between
  the two- and the three-year bond is 0.73
• Result The fall in the value of longer-term
  fixed-income asset or liability increases at a
  diminishing rate for any given increase in
  interest rates

3
MATURITY OF A PORTFOLIO OF ASSETS OR LIABILITIES

• The maturity of a portfolio of assets or
  liabilities is given by
• Mi wi1Mi1 wi2Mi2 winMin
• where
• Mi weighted average maturity of assets or
  liabilities i A, L
• Mij maturity of jth asset (liability)
• j 1 n

4

• wij importance of each asset (liability) in
  the asset (liability) portfolio as measured by
  the market value of that asset (liability)
  position relative to the market value of all the
  assets (liabilities)
• Result The net effect of rising or falling
  interest rates on a banks balance sheet depends
  on the extent and direction in which the bank
  mismatches its asset and liability portfolios

5

• Thus, the extent of interest rate risk depends on
  the maturity gap (MA ML)
• Recall that ?E ?A - ?L
• E.g. Suppose a banks balance sheet contains the
  following
• ? A three-year bond with a face value 100 and
  annual coupon of 10
• ? A one-year deposit of 90 with a promise to
  pay 10 rate

6

• ? Suppose that the initial yield is 10, but it
  increases to 11
• ? The banks own equity (capital) is 10
• Initially, the bank funds the purchase of the
  three-year bond with the 90 deposit and the 10
  equity
• After the rise in yield, the price of the bond is
  97.56 (calculate price)

7

• The value of the deposit is
• PD (909)/1.11 89.19
• Thus, the banks net worth is changed by
• ?E (-2.44) (-0.81) - 1.63
• In this example, the banks net worth declines by
  16.3 after a one percentage change in interest
  rates

8

• This decline is due to the maturity gap between
  assets and liabilities
• ? MA 3 years
• ? ML 1 year
• ? Maturity Gap 2 years
• Question Given the maturity gap, by how much
  should interest rates rise to make the bank
  insolvent?

9

• In our example, that would be a rise of 7
  percentage points
• Question How can the bank immunize itself from
  interest rate risk?
• Based on the maturity model, the bank must have
• MA ML
• Implication If maturity gap 0, bank does not
  take any risk to make a profit

10
MATURITY VS. DURATION

• A lower maturity gap reduces, but does not
  eliminate, interest rate risk
• The reason is that, even if the maturity gap is
  zero, the timing of cash flows may not be
  perfectly matched
• A bank must take into consideration the duration
  or average life of asset or liability cash flows
• Duration captures interest rate sensitivity
  better because it considers both the time of
  arrival of cash flows and maturity

11

• E.g. Consider the following bank that
• ? Borrows 100 through a one-year CD that
  promises to pay 15 interest
• ? Lends 100 for one year at an annual rate of
  15
• Note that MAML1 year ? supposedly no interest
  rate risk
• But, assume that bank requires that half of the
  loan is paid after 6 months

12

• Cash flows are as follows
• Cash Flow on one-year CD
• - Cash flow after 1 year
• Principal 100
• Interest 15
• Total 115

13

• Cash flow on one-year loan
• - Cash flow after 6 months
• Principal 50
• Interest 7.5
• Total 57.5
• - Cash flow after 1 year
• Principal 50
• Interest 3.75
• Reinvestment Income 4.3125
• Total 115.5625 (including flows at 6 months)

14

• In this case, bank makes a profit
• But, what if rates fall to 12 during second half
  of year?
• Then, cash flows on loan for second half of year
  are
• Principal 50
• Interest 3.75 (fixed-rate loan)
• Reinvestment Income 3.45
• Total 114.70

15

• Bank ends up with a loss of 0.3
• Definition Duration is the average life of an
  asset or liability
• Duration is given by the weighted-average time to
  maturity using the relative present values of the
  asset or liability cash flows as weights
• To illustrate, lets use the above example of the
  one-year loan

16
(No Transcript)
17

• To calculate duration
• ? weight the time at which cash flows are
  received
• ? use as weights the ratio of the present value
  of a cash flow to the sum of present values of
  all cash flows
• Duration ?t(CFt ? DFt ? t/?t(CFt ? DFt)
• ?t(PVt ? t)/?t PVt
• t 1 N

18

• where
• CFt cash flow at end of period t
• DFt discount factor (1/(1r)t)
• In the example of the one-year loan, the weights
  are
• ? X1/2 53.49/100 .5349
• ? X1 46.51/100 .4651
• Note By definition, sum of weights 1

19

• Thus, duration is
• D X1/2 (1/2) X1 (1)
• .5349(1/2) .4651 .7326 years
• Implication Even though maturity of loan is 1
  year, duration is .7326 years because .5349 of
  cash flows from loan are received after 6 months

20

• What is the duration of one-year CD?
• Cash flows are 115 after one year
• Present value of cash flows is 115/1.15 100
• Duration is
• D X1 ? 1 1 year

21

• Notice that
• ? Maturity Gap 0
• ? Duration Gap .7326 1 - 0.2674 years
• E.g. Find the duration of a two-year treasury
  bond that pays semiannual coupon of 8, with face
  value 1,000 and yield 12

22
(No Transcript)
23

• Duration of this bond is
• D 1,752.4/930.7 1.88 years
• E.g. What is the duration of a zero-coupon bond?
• For zero-coupon bonds Duration Maturity
• This is because there is only one payment at
  maturity

24

• E.g. What is the duration of a six-year Eurobond
  with annual coupon of 8, face value 1,000 and
  yield also 8?

25
(No Transcript)
26

• Duration of this bond is
• D 4,992.71/1,000 4.993 years