GARCH 1

GARCH Models

Introduction

ARMA models assume a constant volatility In finance, correct specification of volatility isessential

ARMA models are used to model the conditionalexpectation

They write Yt as a linear function of the past plus awhite noise term

GARCH 2

1970 1975 1980 1985 1990 1995

102

101

100

year

|chan

ge| +

1/2 %

Absolute changes in weekly AAA rate

GARCH 3

1994 1996 1998 2000 2002 2004 200640

30

20

10

0

10

20

30

year

perc

ent n

et re

turn

Cree Daily Returns

GARCH 4

1994 1996 1998 2000 2002 2004 20060

5

10

15

20

25

30

35

year

perc

ent n

et re

turn

Cree Daily Returns

GARCH 5

GARCH models of nonconstant volatility ARCH = AutoRegressive ConditionalHeteroscedasticity

heteroscedasticity = non-constant variance homoscedasticity = constant variance

GARCH 6

ARMA unconditionally homoscedastic

conditionally homoscedastic

GARCH unconditionally homoscedastic, but

conditionally heteroscedastic

Unconditional or marginal distribution of Rt meansthe distribution when none of the other returns are

known.

GARCH 7

Modeling conditional means and variances

Idea: If is N(0, 1), and Y = a+ b, then E(Y ) = aand Var(Y ) = b2.

general form for the regression of Yt on X1.t, . . . , Xp,tis

Yt = f(X1,t, . . . , Xp,t) + t (1)

Frequently, f is linear so thatf(X1,t, . . . , Xp,t) = 0 + 1X1,t + + pXp,t.

Principle: To model the conditional mean of Ytgiven X1.t, . . . , Xp,t, write Yt as the conditional mean

plus white noise.

GARCH 8

Let 2(X1,t, . . . , Xp,t) be the conditional variance of Ytgiven X1,t, . . . , Xp,t. Then the model

Yt = f(X1,t, . . . , Xp,t) + (X1,t, . . . , Xp,t)t (2)

gives the correct conditional mean and variance.

Principle: To allow a nonconstant conditionalvariance in the model, multiply the white noise term

by the conditional standard deviation. This product

is added to the conditional mean as in the previous

principle.

(X1,t, . . . , Xp,t) must be non-negative since it is astandard deviation

GARCH 9

ARCH(1) processes

Let 1, 2, . . . be Gaussian white noise with unitvariance, that is, let this process be independent

N(0,1).

ThenE(t|t1, . . .) = 0,

and

Var(t|t1, . . .) = 1. (3) Property (3) is called conditionalhomoscedasticity.

GARCH 10

at = t

0 + 1a2t1. (4)

It is required that 0 0 and 1 0 It is also required that 1 < 1 in order for at to bestationary with a finite variance.

If 1 = 1 then at is stationary, but its variance is Define

2t = Var(at|at1, . . .)

GARCH 11

From previous slide:

at = t

0 + 1a2t1.

Therefore

a2t = 2t{0 + 1a2t1}.

Since t is independent of at1 and Var(t) = 1E(at|at1, . . .) = 0, (5)

and

2t = 0 + 1a2t1. (6)

GARCH 12

From previous slide:

2t = 0 + 1a2t1.

If at1 has an unusually large deviation then the conditional variance of at is larger than

usual

at is also expected to have an unusually large

deviation

volatility will propagate since at having a large

deviation makes 2t+1 large so that at+1 will tend to

be large.

GARCH 13

The conditional variance tends to revert to theunconditional variance provided that 1 < 1 so that

the process is stationary with a finite variance.

The unconditional, i.e., marginal, variance of atdenoted by a(0)

GARCH 14

The basic ARCH(1) equation is2t = 0 + 1a

2t1. (7)

This gives us

a(0) = 0 + 1a(0).

This equation has a positive solution if 1 < 1:a(0) = 0/(1 1).

GARCH 15

If 1 = 1 then a(0) is infinite. It turns out that at is stationary nonetheless.

GARCH 16

For an ARCH(1) process with 1 < 1:

Var(at+k|at, at1, . . .) = (0)+k1{a2t(0)} (0) as k .

In contrast, for any ARMA process:

Var(at+k|at, at1, . . .) = (0).

GARCH 17

0 5 10 15 200.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

k

Cond

itiona

l var

ianc

e

ARCH(1) case 1

ARCH(1) case 2

AR(1)

Var(at+k|at, . . .) for AR(1) and ARCH(1). (0) = 1in both cases. For ARCH(1), 1 = .9. Case 1:

a2t = 1.5. Case 2: a2t = .5.

GARCH 18

independence implies zero correlation but not viceversa

GARCH processes are good examples

dependence of the conditional variance on the

past is the reason the process is not independent

independence of the conditional mean on the past

is the reason that the process is uncorrelated

GARCH 19

Example:

0 = 1, 1 = .95, = .1, and = .8

0 20 40 603

2

1

0

1

2

3White noise

0 20 40 601

2

3

4

5

6Conditional std dev

0 20 40 606

4

2

0

2

4ARCH(1)

0 20 40 6015

10

5

0

5AR(1)/ARCH(1))

GARCH 20

0 5004

2

0

2

4White noise

0 5000

20

40

60Conditional std dev

0 50050

0

50

100ARCH(1)

0 50040

20

0

20

40

60AR(1)/ARCH(1))

40 20 0 20 400.0010.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99

0.9970.999

Data

Prob

abilit

y

normal plot of ARCH(1)

Parameters: 0 = 1, 1 = .95, = .1, and = .8.

GARCH 21

Comparison of AR(1) and ARCH(1)

AR(1)

Yt = (Yt1 ) + t.ARCH(1)

at = tt.

t =0 + 1a2t1.

GARCH 22

Comparison of AR(1) and ARCH(1)

AR(1)

E(Yt) = .

Et(Yt) = + (Yt1 ).ARCH(1)

E(at) = 0.

Et(at) = 0.

GARCH 23

Comparison of AR(1) and ARCH(1)

AR(1)

2t = 2.

ARCH(1)

2 =0

1 1 .

2t = 0 + 1a2t1.

Recall: 2t = Var(at|at1, . . .).

GARCH 24

The AR(1)/ARCH(1) model

Let at be an ARCH(1) process Suppose that

ut = (ut1 ) + at.

ut looks like an AR(1) process, except that the noiseterm is not independent white noise but rather an

ARCH(1) process.

GARCH 25

at is not independent white noise but is uncorrelated Therefore, ut has the same ACF as an AR(1)

process:

u(h) = |h| h.

a2t has the ARCH(1) ACF:

a2(h) = |h|1 h.

need to assume that both || < 1 and 1 < 1 inorder for u to be stationary with a finite variance

GARCH 26

ARCH(q) models

let t be Gaussian white noise with unit variance at is an ARCH(q) process if

at = tt

and

t =

0 + qi=1

ia2ti

GARCH 27

GARCH(p, q) models

the GARCH(p, q) model isat = tt

where

t =

0 + qi=1

ia2ti +pi=1

i2ti.

very general time series model: at is GARCH(pG, qG) and

at is the noise term in an ARIMA(pA, d, qA) model

GARCH 28

Heavy-tailed distributions

stock returns have heavy-tailed oroutlier-prone distributions

reason for the outliers may be that the conditionalvariance is not constant

GARCH processes exhibit heavy-tails Example 90% N(0, 1) and 10% N(0, 25) variance of this distribution is (.9)(1) + (.1)(25) = 3.4 standard deviation is 1.844

distribution is MUCH different that a N(0, 3.4)distribution

GARCH 29

5 0 5 100

0.1

0.2

0.3

0.4Densities

normalnormal mix

4 6 8 10 120

0.005

0.01

0.015

0.02

0.025Densities detail

normalnormal mix

2 1 0 1 20.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Data

Prob

abilit

y

Normal plot normal

10 0 100.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Data

Prob

abilit

y

Normal plot normal mix

Comparison on normal and heavy-tailed distributions.

GARCH 30

For a N(0, 2) random variable X,P{|X| > x} = 2(1 (x/)).

Therefore, for the normal distribution with variance3.4,

P{|X| > 6} = 2(1 (6/3.4)) = .0011.

GARCH 31

For the normal mixture population which hasvariance 1 with probability .9 and variance 25 with

probability .1 we have that

P{|X| > 6} = 2{.9(1 (6)) + .1(1 (6/5))}= (.9)(0) + (.1)(.23) = .023.

Since .023/.001 21, the normal mixture distributionis 21 times more likely to be in this outlier range than

the normal distribution.

GARCH 32

Property Gaussian ARMA GARCH ARMA/

WN GARCH

Cond. mean constant non-const 0 non-const

Cond. var constant constant non-const non-const

Cond. distn normal normal normal normal

Marg. mean & var. constant constant constant constant

Marg. distn normal normal heavy-tailed heavy-tailed

GARCH 33

All of the processes are stationary marginal meansand variances are constant

Gaussian white noise is the baseline process. conditional distribution = marginal distribution

conditional means and variances are constant

conditional and marginal distributions are normal

Gaussian white noise is the source of randomess forthe other processes

therefore, they all have normal conditional

distributions

GARCH 34

Fitting GARCH models

Fit to 300 observation from a simulated AR(1)/ARCH(1)

Listing of the SAS program for the simulated data

options linesize = 65 ;

data arch ;

infile c:\courses\or473\sas\garch02.dat ;

input y ;

run ;

title Simulated ARCH(1)/AR(1) data ;

proc autoreg ;

model y =/nlag = 1 archtest garch=(q=1);

run ;

GARCH 35

SAS output

Q and LM Tests for ARCH Disturbances

Order Q Pr > Q LM Pr > LM

1 119.7578

GARCH 36

Standard Approx

Variable DF Estimate Error t Value Pr > |t|

Intercept 1 0.4810 0.3910 1.23 0.2187

AR1 1 -0.8226 0.0266 -30.92

GARCH 37

standard errors of the ARCH parameters are ratherlarge

approximate 95% confidence interval for 1 is.70 (2)(0.117) = (.446, .934)

GARCH 38

Example: S&P 500 returns

60 65 70 75 80 85 90 950.2

0.15

0.1

0.05

0

0.05

0.1

0.15

year

retu

rn re

sidu

al

Residuals when the S&P 500 returns are

regressed against the change in the 3-month

T-bill rates and the rate of inflation.

GARCH 39

This analysis uses RETURNSP = the return on the S&P 500

DR3 = change in the 3-month T-bill rate

GPW = the rate of wholesale price inflation

RETURNSP is regressed on DR3 and GPW (factormodel)

GARCH 40

Model

RETURNSP = 0 + 1DR3 + 2GPW+ ut (8)

ut is an AR(1)/GARCH(1,1) process Therefore,

ut = 1ut1 + at,

at is a GARCH(1,1) process:at = tt

wheret =

0 + 1a2t1 + 1

2t1.

GARCH 41

SAS Program

Key command:

proc autoreg ;

model returnsp = DR3 gpw/nlag = 1 archtest garch=(p=1,q=1);

returnsp = DR3 gpw specifies the regression model nlag = 1 specifies the AR(1) structure. garch=(p=1,q=1) specifies the GARCH(1,1)structure.

archtest specifies that tests of conditionalheteroscedasticity be performed

GARCH 42

SAS output

The p-values of the Q and LM tests are all very small,less than .0001. Therefore, the errors in the regression

model exhibit conditional heteroscedasticity.

Ordinary least squares estimates of the regressionparameters are:

Standard Approx

Variable DF Estimate Error t Value Pr > |t|

Intercept 1 0.0120 0.001755 6.86

GARCH 43

Using residuals from the OLS estimates, theestimated residual autocorrelations are:

Estimates of Autocorrelations

Lag Covariance Correlation

0 0.00108 1.000000

1 0.000253 0.234934

GARCH 44

Also, using OLS residuals, the estimate ARparameter is:

Estimates of Autoregressive Parameters

Standard

Lag Coefficient Error t Value

1 -0.234934 0.046929 -5.01

GARCH 45

Assuming AR(1)/GARCH(1,1) errors, the estimatedparameters of the regression are:

Standard Approx

Variable DF Estimate Error t Value Pr > |t|

Intercept 1 0.0125 0.001875 6.66

GARCH 46

The estimated GARCH parameters are:AR1 1 -0.2016 0.0603 -3.34 0.0008

ARCH0 1 0.000147 0.0000688 2.14 0.0320

ARCH1 1 0.1337 0.0404 3.31 0.0009

GARCH1 1 0.7254 0.0918 7.91 > ARCH1 (0.1337) reasonably long persistence of volatility.

GARCH 47

I-GARCH models

I-GARCH or integrated GARCH processes designedto model persistent changes in volatility

A GARCH(p, q) process is stationary with a finitevariance if

qi=1

i +

pi=1

i < 1.

GARCH 48

A GARCH(p, q) process is called an I-GARCH process if

qi=1

i +

pi=1

i = 1.

I-GARCH processes are either non-stationary or havean infinite variance.

Here are some simulations of ARCH(1) processes:

GARCH 49

0 0.5 1 1.5 2 2.5 3 3.5 4x 104

150

100

50

0

501 = .9

0 0.5 1 1.5 2 2.5 3 3.5 4x 104

100

0

100

2001 = 1

0 0.5 1 1.5 2 2.5 3 3.5 4x 104

5

0

5 x 104 1 = 1.8

Simulated ARCH(1) processes with 1 = .9, 1, and 1.8.

GARCH 50

100 80 60 40 20 0 20 40

0.0010.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99

0.9970.999

Prob

abilit

y

1 = .9

50 0 50 100 150

0.0010.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99

0.9970.999

Prob

abilit

y

1 = 1

4 3 2 1 0 1 2 3 4x 104

0.0010.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99

0.9970.999

Prob

abilit

y

1 = 1.8

Normal plots of ARCH(1) processes with 1 = .9, 1, and 1.8.

GARCH 51

Comments on the figures

all three processes do revert to their mean, 0 larger the value of 1 the more the volatility comes insharp bursts

processes with 1 = .9 and 1 = 1 looks similar none of the processes in the figure show muchpersistence of higher volatility

to model persistence of higher volatility, one needs anI-GARCH(p, q) process with q 1

Next figure shows simulations from I-GARCH(1,1)processes

GARCH 52

0 1000 20000

5

10

15

20

25

301 = 0.95, Cond. std dev

0 1000 200020

10

0

10

20

301 = 0.95, GARCH(1,1)

0 1000 20000

5

10

15

20

25

301 = 0.4, Cond. std dev

0 1000 200040

20

0

20

401 = 0.4, GARCH(1,1)

0 1000 20000

50

100

150

2001 = 0.2, Cond. std dev

0 1000 2000300

200

100

0

100

200

3001 = 0.2, GARCH(1,1)

0 1000 20005

10

15

20

25

30

35

401 = 0.05, Cond. std dev

0 1000 2000100

50

0

50

1001 = 0.05, GARCH(1,1)

Simulations of I-GARCH(1,1) processes. 1 + 1 = 1

GARCH 53

To fit I-GARCH in SAS:

proc autoreg ;

model returnsp =/nlag = 1 garch=(p=1,q=1,type=integrated);

run ;

The default value of type is nonneg which onlyconstrains the GARCH coefficients to be

non-negative.

type=integrated in addition imposes thesum-to-one constraint of the I-GARCH model

GARCH 54

What does infinite variance mean?

let X be a random variable with density fX the expectation of X is

xfX(x)dx

provided that this integral is defined.

GARCH 55

If 0

xfX(x)dx = (9)

and 0

xfX(x)dx = (10)then the expectation is, formally, + not defined if both integrals are finite, then the expectation is thesum of these two integrals

GARCH 56

Exercise: fX(x) = 1/4 if |x| < 1 andfX(x) = 1/(4x

2) if |x| 1 fX is a density since

fX(x)dx = 1

Then expectation does not exist since 0

xfX(x)dx =

and 0

xfX(x)dx =

GARCH 57

What are the implications of having no

expectation?

assume sample of iid sample from fX law of large numbers sample mean will converge tothe expectation

law of large numbers doesnt apply if expectation isnot defined

there is no point to which the sample mean canconverge

it will just wander without converging

GARCH 58

0 0.5 1 1.5 2 2.5 3 3.5 4x 104

1

0.5

0

0.51 = .9

0 0.5 1 1.5 2 2.5 3 3.5 4x 104

6

4

2

0

2

41 = 1

0 0.5 1 1.5 2 2.5 3 3.5 4x 104

5

0

5

10

151 = 1.8

Sample means of ARCH(1) processes with 1 = .9, 1, and 1.8.

GARCH 59

What are the implications of having infinite

variance

now suppose that the expectation of X exists andequals X

the variance

(x X)2fX(x)dx

if this integral is + then the variance is infinite law of large numbers sample variance will convergeto the variance

variance of X is infinity the sample variance willconverge to infinity

GARCH 60

0 0.5 1 1.5 2 2.5 3 3.5 4x 104

0

2

4

6

81 = .9

0 0.5 1 1.5 2 2.5 3 3.5 4x 104

0

50

100

1501 = 1

0 0.5 1 1.5 2 2.5 3 3.5 4x 104

0

1

2

3

4 x 105 1 = 1.8

Sample variances: ARCH(1) with 1 = .9, 1, and 1.8.

GARCH 61

GARCH-M processes

if we fit a regression model with GARCH errors could use the conditional standard deviation (t)

as one of the regression variables

when the dependent variable is a return the market demands a higher risk premium for

higher risk

so higher conditional variability could cause higher

returns

GARCH 62

GARCH-M models in SAS add keyword mean,e.g.,

proc autoreg ;

model returnsp =/nlag = 1 garch=(p=1,q=1,mean);

run ;

or for I-GARCH-Mproc autoreg ;

model returnsp =/nlag = 1 garch=(p=1,q=1,mean,type=integrated);

run ;

GARCH 63

GARCH-M example: S&P 500

GARCH(1,1)-M was fit in SAS is the regression coefficient for t = .5150 standard error = .3695

t-value = 1.39 p-value = .1633

GARCH 64

since p-value = .1633 could accept the null hypothesis that = 0

no strong evidence that there are higher returns

during times of higher volatility.

volatility of S&P 500 is market risk so this issomewhat surprising (think of CAPM)

may be that the effect is small but not 0 ( ispositive, after all)

AIC criterion does select the GARCH-M model

GARCH 65

E-GARCH

E-GARCH models are used to model the leverageeffect

prices become more volatile as prices decrease

E-GARCH, model is

log(t) = 0 +

qi=1

1g(ti) +pi=1

i log(ti),

where

g(t) = t + {|t| E(|t|)} log(t) can be negative no constraints onparameters

GARCH 66

From the previous page:g(t) = t + {|t| E(|t|)}

To understand g note that

g(t) = E(|t|) + ( + )|t| if t > 0,and

g(t) = E(|t|) + ( )|t| if t < 0, typically, 1 < < 0 so that 0 < + < = .7 in the S&P 500 example E(|t|) =

2/pi = .7979 (good calculus exercise)

GARCH 67

4 2 0 2 41

01

2

34

56

t

g( t

)

= 0.7

4 2 0 2 41

01

2

34

56

t

g( t

)

= 0

4 2 0 2 41

01

2

34

56

t

g( t

)

= 0.7

4 2 0 2 41

01

2

34

56

t

g( t

)

= 1

The g function for the S&P 500 data (top left

panel) and several other values of .

GARCH 68

SAS fits the E-GARCH model fixed as 1

estimated

E-GARCH model is specified by using type=exp asin:

proc autoreg ;

model returnsp =/nlag = 1 garch=(p=1,q=1,mean,type=exp);

run ;

GARCH 69

Back to the S&P 500 example

SAS can fit six different AR(1)/GARCH(1,1) models type = integrated, exp, or nonneg

GARCH-in-mean effect can be included or not

following table contains the AIC statistics models ordered by AIC (best fitting to worse)

GARCH 70

Model AIC AIC

E-GARCH-M 1783.9 0E-GARCH 1783.1 0.8GARCH-M 1764.6 19.3GARCH 1764.1 19.8I-GARCH-M 1758.0 25.9I-GARCH 1756.4 27.5

AIC statistics for six AR(1)/GARCH(1,1) models

fit to the S&P 500 returns data. AIC is change

in AIC between a given model and E-GARCH-M.

GARCH 71

AR(2) and E-GARCH(1,2)-M, E-GARCH(2,1)-M,and E-GARCH(2,2)-M models were tried

none of these lowered AIC

none had all parameters significant at p = .1

GARCH 72

Listing of SAS output for the E-GARCH-Mmodel:

The AUTOREG Procedure

Estimates of Autoregressive Parameters

Standard

Lag Coefficient Error t Value

1 -0.234934 0.046929 -5.01

Algorithm converged.

Exponential GARCH Estimates

SSE 0.44211939 Observations 433

MSE 0.00102 Uncond Var .

Log Likelihood 900.962569 Total R-Square 0.1050

SBC -1747.2885 AIC -1783.9251

Normality Test 24.9607 Pr > ChiSq

GARCH 73

Standard Approx

Variable DF Estimate Error t Value Pr > |t|

Intercept 1 -0.003791 0.0102 -0.37 0.7095

DR3 1 -1.2062 0.3044 -3.96

GARCH 74

The GARCH zoo

Heres a sample of other GARCH models mentioned in

Bollerslev, Engle, and Nelson (1994):

QARCH = quadratic ARCH TARCH = threshold ARCH STARCH = structural ARCH SWARCH = switching ARCH QTARCH = quantitative threshold ARCH vector ARCH diagonal ARCH factor ARCH

GARCH 75

GARCH Models in Finance

Remember the problem of implied volatility it

depended on K and T !

Black-Scholes assumes a constant variance But GARCH effects are common so the Black-Scholesmodel is not adequate

Having a volatility function (smile) is a quick fix but not logical

GARCH 76

3540

4550

55

0

50

100

1500.015

0.02

0.025

0.03

0.035

0.04

exercise pricematurity

impl

ied

vola

tility

GARCH 77

Options can be priced assuming the log-returns are a

GARCH process (rather than a random walk)

Multinomial (not binomial) tree to have differentlevels of volatility

Need to keep track of price and conditional variance

GARCH 78

Ritchken and Trevor use an NGARCH (nonlinearasymmetric GARCH) model:

log(St+1/St) = r + ht ht/2 +

htt+1

ht+1 = 0 + 1ht + 2ht(t+1 c)2where t is WhiteNoise(0, 1).

c = 0 is an ordinary GARCH model

is a risk premium

Under the risk-neutral (martingale) measure:log(St+1/St) = (r ht/2) +

htt+1,

ht+1 = 0 + 1ht + 2ht(t+1 c)2,where c = c+ and t is WhiteNoise(0, 1).

GARCH 79

Now there are five unknown parameters:

h0 1, 2, and 3 c

These parameters are estimated by nonlinear

least-squares

Implied GARCH parameters

GARCH 80

Volatility smile is explained as due to GARCH

effects:

nonconstant variance nonnormal marginal distribution

GARCH 81

From Chapter 7, Bank of Volatility, of When

Genius Failed by Roger Lowenstein:

Early in 1998, Long-Term began to short large amounts

of equity volatility.

This simple trade, second nature to Rosenfeld and David

Modest, would be indecipherable to 999 out of 1,000

Americans.

Equity vol comes straight from the Black-Scholes model.

GARCH 82

The stock market, for instance, typically varies

by about 15 percent to 20 percent a year.

And when the model told them that the markets were

mispricing equity vol, they were willing to bet the firm

on it.

The options market was anticipating volatility in

the stock market of roughly 20 percent. Long-term

viewed this as incorrect ... Thus, it figured that options

prices would sooner or later fall.

GARCH 83

Long-Term began to short options on the S&P 500 are

similar European options. They were selling

volatility.

In fact, it sold insurance (options) both

waysagainst a sharp downturn and against a sharp

rise.

GARCH 84

1994 1996 1998 2000 20020

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08S&

P 50

0 ab

solu

te lo

g re

turn

GARCH 85

1997 1998 1999 2000 20010.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

S&P5

00, c

ondi

tiona

l SD,

ann

ualiz

ed

AR(1)/EGARCH(1,1) model

Annualized SD =253 daily variance

GARCH 86

data capm ;

set Sasuser.capm ;

logR_sp500 = dif(log(close_sp500)) ;

logR_msft = dif(log(close_msft)) ;

run ;

proc gplot ;

plot logR_sp500*date ;

run ;

proc autoreg ;

model logR_sp500 = /nlag=1 garch=(p=1,q=1,type=exp) ;

output out=outdata cev=cev ;

run ;

proc gplot ;

plot cev*date ;

run ;

GARCH 87

nlag p q E-GARCH M-GARCH AIC

1 2 2 no no 14,3661 1 2 no no 15,0021 1 2 yes no 15,1051 1 2 yes yes 15,1031 2 1 yes no 15,1051 1 1 yes no 15,1070 1 1 yes no 14,3662 2 2 yes no 15,1121 2 2 yes no 15,1042 2 3 yes no 15,1002 3 2 yes no 15,1001 3 2 yes no 15,101

GARCH 88

In fact, it sold insurance (options) both

waysagainst a sharp downturn and against a sharp

rise.

0 0.2 0.4 0.6 0.8 190

100

110st

ock

price

0 0.2 0.4 0.6 0.8 10

5

10

call

price

0 0.2 0.4 0.6 0.8 10

2

4

time

put p

rice

GARCH 89

The Greeks

C(S, T, t,K, , r) = price of an option

=

SC(S, T, t,K, , r) Delta

=

tC(S, T, t,K, , r) Theta

R = rC(S, T, t,K, , r) Rho

V =

C(S, T, t,K, , r) Vega

GARCH 90

Put-call parity

Put and call prices with same K and T are related:

P (S, T, t,K, , r) = C(S, T, t,K, , r) + er(Tt)K S.Therefore, the call and put have the same vegas

P (S, T, t,K, , r) =

C(S, T, t,K, , r)

but difference deltas

SP (S, T, t,K, , r) =

SC(S, T, t,K, , r) 1

0 < (call) = (d1) < 1

1 < (put) = (d1) 1 < 0

GARCH 91

From previous slide:

1 < (put) = (d1) 1 < 0

Suppose we buy N1 call options and N2 put options.

Delta of the portfolio is

N1(d1) +N2((d1) 1)which is zero if

N1N2

=1 (d1)(d1)

Hedging is possible with a put and call with different K

and T each then has its own value of d1

GARCH 92

Selling equity vol was very clever, but as Lowenstein

remarks:

This wasso unlike the partners credorank

speculation.