Predicting Interest Rates Statistical Models. Economic vs. Statistical Models Economic models are...

Predicting Interest RatesPredicting Interest Rates

Statistical ModelsStatistical Models

Economic vs. Statistical ModelsEconomic vs. Statistical Models

Economic models are designed to match Economic models are designed to match correlations between interest rates and correlations between interest rates and other economic aggregate variables other economic aggregate variables Pro: Economic (structural) models use all the Pro: Economic (structural) models use all the

latest information available to predict interest latest information available to predict interest rate movementsrate movements

Con: They require a lot of data, the equation Con: They require a lot of data, the equation can be quite complex, and over longer time can be quite complex, and over longer time periods are very inaccurateperiods are very inaccurate

Economic vs. Statistical ModelsEconomic vs. Statistical Models

Statistical models are designed to match Statistical models are designed to match the dynamics of interest rates and the the dynamics of interest rates and the yield curve using past behavior.yield curve using past behavior.Pro: Statistical Models require very little data Pro: Statistical Models require very little data

and are generally easy to calculateand are generally easy to calculateCon: Statistical models rely entirely on the Con: Statistical models rely entirely on the

past. They don’t incorporate new information. past. They don’t incorporate new information.

The Yield CurveThe Yield Curve

Recall that the yield curve is a collection of current spot Recall that the yield curve is a collection of current spot ratesrates

2.12 2.61 2.944.3

5.07

0

2

4

6

1 yr 2 yr 5yr 10 yr 20yrS(1) S(2) S(5) S(10) S(20)

Forward RatesForward Rates

Forward rates are interest rates for Forward rates are interest rates for contracts to be written in the future. (F)contracts to be written in the future. (F)

F(1,1) = Interest rate on 1 year loans contracted 1 F(1,1) = Interest rate on 1 year loans contracted 1 year from nowyear from now

F(1,2) = Interest rate on 2 yr loans contracted 2 F(1,2) = Interest rate on 2 yr loans contracted 2 years from now years from now

F(2,1) = interest rate on 1 year loans contracted 2 F(2,1) = interest rate on 1 year loans contracted 2 years from nowyears from now

S(1) = F(0,1)

Spot/Forward RatesSpot/Forward Rates

Now 1yr 2yrs 4yrs3yrs 5yrs

F(0,1) F(1,1) F(2,1)

S(1)

S(2)

S(3)

F(0,2)

F(1,2)

Spot Rates

Forward Rates

F(2,2)

F(1,3)

Calculating Forward RatesCalculating Forward Rates Forward rates are not observed, but are implied in the Forward rates are not observed, but are implied in the

yield curveyield curve Suppose the current annual yield on a 2 yr Treasury is Suppose the current annual yield on a 2 yr Treasury is

2.61% while a 1 yr Treasury pays an annual rate of 2.61% while a 1 yr Treasury pays an annual rate of 2.12%2.12%


F(1,1)

S(1)

S(2) 2.61%/yr

2.12%/yr

Calculating Forward RatesCalculating Forward Rates


F(1,1)

S(1)

S(2) 2.61%/yr

2.12%/yr

Strategy #1: Invest $1 in a two year Treasury

$1(1.0261)(1.0261) = 1.053 (5.3%)

Strategy #2: Invest $1 in a 1 year Treasury and then reinvest in 1 year

For these strategies, to pay the same return, the one year forward rate would need to be 3.1%

$1(1.0261)(1.0261) = $1(1.0261)(1+F(1,1)

$1(1.0212)(1 + F(1,1))

1+F(1,1) = $1(1.0261)(1.0261)

$1(1.0212) =1.031

Calculating Spot RatesCalculating Spot Rates We can also do this in reverse. If we knew the path for We can also do this in reverse. If we knew the path for

forward rates, we can calculate the spot rates:forward rates, we can calculate the spot rates:


3.3%

S(1)

S(2) ???

2%

2% 2.9%

S(3) ???

Calculating Spot RatesCalculating Spot Rates


3.3%

S(2) ???

Strategy #1: Invest $1 in a two year Treasury

$1(1+(S(2))(1+S(2))

Strategy #1: Invest $1 in a 1 year Treasury and then reinvest in 1 year

For these strategies, to pay the same return, the two year spot rate would need to be 2.6%

$1(1.02)(1.033) = $1(1+S(2))

$1(1.02)(1.033) = 1.054 (5.4%)

1+S(2) = ((1.02)(1.033)) =1.026

2%

2

1/2

Arithmetic vs. Geometric AveragesArithmetic vs. Geometric Averages


3.3%

S(2) 2.6%

In the previous example, we calculated the Geometric Average of expected forward rates to get the current spot rate

The Arithmetic Average is generally a good approximation

2%

1+S(2) = ((1.02)(1.033)) =1.026 (2.6%)1/2

S(2) = 2% + 3.3%

2= 2.65%


3.3%

S(1)

S(2) 2.65%

2%

2% 2.9%

S(3) 2.73%

S(2) = 2

= 2.65%2% + 3.3%

S(3) = 3

= 2.73%2% + 3.3% + 2.9%

Spot rates are equal to the averages of the corresponding forward rates (expectations hypothesis)

However, the expectations hypothesis assumes that investing in long term bonds is an equivalent strategy to investing in short term bonds


3.3%

S(2) 2.65%

2%This rate is flexible at time 0

This rate is “locked in” at time 0

Long term bondholders should be compensated for inflexibility of their portfolios by adding a “liquidity premium” to longer term rates (preferred habitat hypothesis)

Statistical Models Statistical Models


3.3%F(0,1) F(3,1)F(2,1)F(1,1) F(4,1)

First, write down a model to explain movements in the forward rates

Then, calculate the yield curve implied by the forward rates. Does it look like the actual yield curve?


S(1)

S(2)

S(3)

Lattice Methods (Discrete)Lattice Methods (Discrete)

Lattice models assume that the interest Lattice models assume that the interest rate makes discrete jumps between time rate makes discrete jumps between time periods (usually calibrated monthly)periods (usually calibrated monthly)

Binomial: Two Possibilities each PeriodBinomial: Two Possibilities each PeriodTrinomial: Three Possibilities each PeriodTrinomial: Three Possibilities each Period

An ExampleAn Example

At time zero, the interest rate 5%: F(0,1) = S(1)At time zero, the interest rate 5%: F(0,1) = S(1)


In the first year, the interest rate has a 50% In the first year, the interest rate has a 50% chance of rising to 5.7% or falling to 4.8%: F(1,1)chance of rising to 5.7% or falling to 4.8%: F(1,1)


In the second year, there is also a 50% chance of rising In the second year, there is also a 50% chance of rising or falling conditional on what happened the previous or falling conditional on what happened the previous year: F(2,1)year: F(2,1)

Calculating the Yield CurveCalculating the Yield Curve

5.7%

5%

4.8%

Path 1: (1.05)(1.057) = 1.10985 (10.985%)Path 1: (1.05)(1.057) = 1.10985 (10.985%)

Path 2: (1.05)(1.048) = 1.10040 (10.04%)Path 2: (1.05)(1.048) = 1.10040 (10.04%)

(.5)(1.10985) + (.5)(1.10040) = 1.105125 (10.5125%)Expected two year cumulative return =

Annualized Return = (1.105125)1/2

= 1.0512 (5.12%) = S(2)

S(1)

5.7%

5%

4.8%

Path 1: (1.05)(1.057)(1.064) = 1.181 (18.1%)Path 1: (1.05)(1.057)(1.064) = 1.181 (18.1%)

Path 2: (1.05)(1.057)(1.052) = 1.168 (16.8%)Path 2: (1.05)(1.057)(1.052) = 1.168 (16.8%)

(.25)(1.181) + (.25)(1.168) + (.25)(1.157) +(.25)(1.151) = 1.164Expected three year cumulative return =

Annualized Return = (1.164)1/3

= 1.0519 (5.19%) = S(3)

4.6%

5.2%

6.4%

Path 3: (1.05)(1.048)(1.052) = 1.157 (15.7%)Path 3: (1.05)(1.048)(1.052) = 1.157 (15.7%)

Path 4: (1.05)(1.048)(1.046) = 1.151 (15.1%)Path 4: (1.05)(1.048)(1.046) = 1.151 (15.1%)

Future Yield CurvesFuture Yield Curves

5.7%

5%

4.8%

4.6%

5.2%

6.4%

Suppose that next months interest rate turns out to be 4.8% = S(1)’

Path 1: (1.048)(1.052) = 1.1025 (10.25%)Path 1: (1.048)(1.052) = 1.1025 (10.25%)

Path 2: (1.048)(1.046) = 1.096 (9.6%)Path 2: (1.048)(1.046) = 1.096 (9.6%)

(.5)(1.1025) + (.5)(1.096) = 1.0993(9.3%)

S(2)’ = (1.099)1/2

= 1.049 (4.9%)

Volatility & Term StructureVolatility & Term Structure

A common form for a binomial tree is as follows: A common form for a binomial tree is as follows:

.5y probabilit with

)1(

.5y probabilit with 1

1

t

t

t i

i

i

Sigma is measuring volatility

7.85

7.9

7.95

8

8.05

8.1

8.15

8.2

8.25

8.3

1 2 3 4 5 6 7 8 9 10 11 12

High Sigma

Low Sigma

Higher volatility raises the probability of very large or very small future interest rates. This will be reflected in a steeper yield curve

Continuous Time ModelsContinuous Time Models

dztidttiadi ttt ,,

Change in the interest rate at time ‘t’

Deterministic (Non-Random) component Random component

Random Error term with N(0,1) distribution

VasicekVasicek

The Vasicek model is a particularly simple form:The Vasicek model is a particularly simple form:

dzdtidi tt

Controls Persistence

Controls Mean

Controls Variance

Using the Vasicek ModelUsing the Vasicek Model Choose parameter valuesChoose parameter values Choose a starting valueChoose a starting value Generate a set of random numbers with mean 0 and Generate a set of random numbers with mean 0 and

variance 1variance 1

%6

262.

0

i

dzdtidi tt

t=0t=0 t=1t=1 t=2t=2 t=3t=3 t=4t=4

i 6% 6.8% 6.84% 4.202% 5.5616%

.2(6-i).2(6-i) 00 -.16-.16 -.168-.168 .3596.3596

dzdz .4.4 .2.2 -1.1-1.1 .5.5

didi .8.8 .04.04 -2.368-2.368 1.35961.3596 -.9-.9

Vasicek (sigma = 2, kappa = .17)Vasicek (sigma = 2, kappa = .17)

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

Path1

Path2

Path 3

Path 4

Path 5

Average

Vasicek (sigma = 4, kappa = .17 )Vasicek (sigma = 4, kappa = .17 )

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

Path1

Path2

Path 3

Path 4

Path 5

Average

Vasicek (sigma = 2, kappa = .4)Vasicek (sigma = 2, kappa = .4)

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

Path1

Path2

Path 3

Path 4

Path 5

Average

Cox, Ingersoll, Ross (CIR)Cox, Ingersoll, Ross (CIR)

The CIR framework allows for volatility that The CIR framework allows for volatility that depends on the current level of the interest depends on the current level of the interest rate (higher volatilities are associated with rate (higher volatilities are associated with higher rates)higher rates)

dzrdtrdrt

dzidtidi tt

Heath,Jarow,Morton (HJM)Heath,Jarow,Morton (HJM) Vasicek and CIR assume a process for a single forward Vasicek and CIR assume a process for a single forward

rate and then use that to construct the yield curverate and then use that to construct the yield curve In this framework, the correlation between different In this framework, the correlation between different

interest rates of different maturities in automatically one interest rates of different maturities in automatically one (as is the case with any one factor model)(as is the case with any one factor model)

HJM actually model the evolution of the entire array of HJM actually model the evolution of the entire array of forward ratesforward rates

dzTtfTtdtTtfTtaTtdf ),(,,),(,,,

Change it the forward rate of maturity T ant time t

Table 1Table 1Summary Statistics for Historical RatesSummary Statistics for Historical Rates

ShapeNormal Inverted Humped Other68.8% 11.6% 13.4% 6.3%

Yield Statistics1 yr. 3 yr. 5 yr. 10 yr.

Mean 6.08 6.47 6.64 6.81S.D. 3.01 2.88 2.84 2.81Skewness 0.97 0.84 0.77 0.68Exc. Kurtosis 1.10 0.69 0.48 0.16

Percentiles1 yr. 3 yr. 5 yr. 10 yr.

1% 1.07 1.59 1.94 2.385% 2.05 2.52 2.72 2.9050% 5.61 6.20 6.44 6.6895% 12.08 12.48 12.59 12.5699% 15.17 14.69 14.59 14.29

Corr (1 yr,10 yr) = 0.944

Tables 1-4 from Ahlgrim, D’Arcy, and Gorvett, CAS 1999 DFA Call Paper Program

Table 2Table 2Summary Statistics for Vasicek ModelSummary Statistics for Vasicek Model

ShapeNormal Inverted Humped Other41.6% 54.8% 3.6% 0.0%


Mean 8.81 8.75 8.68 8.52S.D. 3.83 3.24 2.77 1.95Skewness -0.16 -0.16 -0.16 -0.16Exc. Kurtosis -0.19 -0.19 -0.19 -0.19


1% -0.38 0.97 2.04 3.845% 2.33 3.27 4.00 5.2250% 8.94 8.86 8.77 8.5995% 14.69 13.73 12.94 11.5399% 17.22 15.87 14.76 12.82

Corr (1 yr,10 yr) = 1.000

Notes: Number of simulations = 10,000, = 0.1779, = 0.0866, = 0.0200

Table 3Table 3Summary Statistics for CIR ModelSummary Statistics for CIR Model

Shape

Normal Inverted Humped Other

47.7% 47.6% 4.7% 0.0%


Mean 8.08 8.04 7.98 7.86S.D. 2.89 2.31 1.88 1.20Skewness 0.92 0.92 0.92 0.92Exc. Kurtosis 1.49 1.49 1.49 1.49


1% 2.92 3.90 4.62 5.715% 3.95 4.73 5.29 6.1450% 7.71 7.73 7.73 7.7095% 13.42 12.31 11.45 10.0999% 17.19 15.33 13.90 11.66

Corr (1 yr,10 yr) = 1.000

Notes: Number of simulations = 10,000, = 0.2339, = 0.0808, = 0.0854

Table 4Table 4Summary Statistics for HJM ModelSummary Statistics for HJM Model


Mean 7.39 7.51 7.60 7.80S.D. 2.26 2.27 2.31 2.44Skewness 0.51 0.53 0.54 0.54Exc. Kurtosis -0.88 -0.85 -0.85 -0.86


1% 4.45 4.48 4.52 4.595% 4.79 4.85 4.90 4.9950% 7.48 7.58 7.65 7.8395% 11.57 11.74 11.92 12.3899% 12.09 12.26 12.44 12.89

Corr (1 yr,10 yr) = 0.999

Notes: Number of simulations = 100

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