predicting interest rates statistical models. economic vs. statistical models economic models are...
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Predicting Interest RatesPredicting Interest Rates
Statistical ModelsStatistical Models
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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
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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.
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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)
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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)
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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)
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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%
Now 1yr 2yrs 4yrs3yrs 5yrs
F(1,1)
S(1)
S(2) 2.61%/yr
2.12%/yr
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Calculating Forward RatesCalculating Forward Rates
Now 1yr 2yrs 4yrs3yrs 5yrs
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
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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:
Now 1yr 2yrs 4yrs3yrs 5yrs
3.3%
S(1)
S(2) ???
2%
2% 2.9%
S(3) ???
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Calculating Spot RatesCalculating Spot Rates
Now 1yr 2yrs 4yrs3yrs 5yrs
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
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Arithmetic vs. Geometric AveragesArithmetic vs. Geometric Averages
Now 1yr 2yrs 4yrs3yrs 5yrs
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%
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Now 1yr 2yrs 4yrs3yrs 5yrs
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)
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However, the expectations hypothesis assumes that investing in long term bonds is an equivalent strategy to investing in short term bonds
Now 1yr 2yrs 4yrs3yrs 5yrs
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)
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Statistical Models Statistical Models
Now 1yr 2yrs 4yrs3yrs 5yrs
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?
Now 1yr 2yrs 4yrs3yrs 5yrs
S(1)
S(2)
S(3)
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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
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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)
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An ExampleAn Example
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)
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An ExampleAn Example
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)
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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)
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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%)
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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%)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Table 2Table 2Summary Statistics for Vasicek ModelSummary Statistics for Vasicek Model
ShapeNormal Inverted Humped Other41.6% 54.8% 3.6% 0.0%
Yield Statistics1 yr. 3 yr. 5 yr. 10 yr.
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
Percentiles1 yr. 3 yr. 5 yr. 10 yr.
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
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Table 3Table 3Summary Statistics for CIR ModelSummary Statistics for CIR Model
Shape
Normal Inverted Humped Other
47.7% 47.6% 4.7% 0.0%
Yield Statistics1 yr. 3 yr. 5 yr. 10 yr.
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
Percentiles1 yr. 3 yr. 5 yr. 10 yr.
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
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Table 4Table 4Summary Statistics for HJM ModelSummary Statistics for HJM Model
Yield Statistics1 yr. 3 yr. 5 yr. 10 yr.
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
Percentiles1 yr. 3 yr. 5 yr. 10 yr.
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