Estimating Volatilities and
Correlations
Chapter 21
1

Outline
 Standard Approach to Estimating Volatility
 Weighting Scheme
 ARCH(m) Model
 EWMA Model
 GARCH model
 Maximum Likelihood Methods
2

Standard Approach to Estimating
Volatility
 Define s n as the volatility per day between day n 1 and day n , as estimated at end of day n 1
 Define S i as the value of market variable at end of day i
 Define u i
= ln ( S i
/S i1
) s n
2  m
1

1 i m 

1
( u n
 i
 u )
2 u

1 m i m 

1 u n
 i
3

Simplifications Usually Made



Define u i as ( S i
−S i1
) /S i1
Assume that the mean value of u i
Replace m− 1 by m is zero
This gives s n
2 
1 m
 i m

1 u
2
4

Weighting Scheme
Instead of assigning equal weights to the observations we can set s n
2 
 i m

1
 u
2 i n i where i m 

1
 i

1
5

ARCH(m) Model
In an ARCH(m) model we also assign some weight to the long-run variance rate, V
L
: s
2 n
 
V
L
  i m

1
 i u
2 n
 i where
  i m 

1
 i
Define

1
  
V
L s n
2     i m

1
 i u
2 n
 i
6

EWMA Model (1/3)


In an exponentially weighted moving average model, the weights assigned to the u 2 decline exponentially as we move back through time, and the decline rate is

This leads to s 2 n
 s 2 n

1

( 1
 
) u
2 n

1
7

EWMA Model (2/3) replace s n

1
2 s n
2  
[
s n
2

2

( 1
 
)(
 n
2

1 substitute s n
2

2

( 1
 
)

2 n

2
]

( 1
 
)
 n
2

1
 
2 n

2
)
 
2 s
2 n

2 s n
2 
( 1
 
)(

2 n

1
  n
2

2
 
2
 n
2

3
)
 
3 s n
2

3 continuing the way gives s n
2 
( 1
 
) i m 

1
 i

1

2 n
 i
  m s n
2
 m
8

EWMA Model (3/3)
For the large m,
 m s
2 n
 m is sufficient ly small the equation is same as the quation of weight scheme where
 i

( 1
 
)
 i

1 or
 i

1
  i
9

Attractions of EWMA
 Relatively little data needs to be stored
 We need only remember the current estimate of the variance rate and the most recent observation on the market variable


Tracks volatility changes
RiskMetrics uses

= 0.94 for daily volatility forecasting
10

GARCH (1,1)
In GARCH (1,1) we assign some weight to the long-run average variance rate s
2 n
 
V
L
  u
2 n

1
 bs
2 n

1
Since weights must sum to 1
    b 1
11

GARCH (1,1)
Setting
  
V the GARCH (1,1) model is s n
2
    u
2 n

1
 bs
2 n

1 and
V
L


1
   b
12

GARCH (p,q) s 2 n
   i p 

1
 i u
2 n
 i
 j q 

1 b j s 2 n
 j
13

The weight substitute for s n
2
1 in GARCH(1,1) s
2 n

1
 
 


 b
2 n

1

 b
(


2 n

1
 
2 n

2
 b
2 n

2
 bs
2 n

2
)
 b
2 s
2 n

2 substitute for s
2 n 2 s
2 n

2
   b  b
2
  
2 n

1
 b
2 n

2
 b
2

2 n

3
 b
3 s
2 n

3
The weights decline exponentially at rate β
The GARCH(1,1) is similar to EWMA except that assign some weight to the long-run volatility
14

Mean Reversion
 The GARCH(1,1) model recognizes that over time the variance tends to get pulled back to a long-run avarage level
L
 The GARCH(1,1) is equivalent to a model where the variance V follows the stochastic process dV
 a ( V
L

V ) dt
 
Vdz where time measured in days, a

1 -

b
,
  
2
15

Prove(1/2) s n
2
   so that s n
2
 u
2 n

1
 s

2 n

1 bs
 
E (

2 n

1
)

Var (

2 n

1
)
Var (
 n

1
)

E (

4 n

1
)
2 n

1


  u
2 n

1
[ E (
 n

1
[ E (

2 n

1

( b
)]
2
)]
2

1 ) s
2 n

1
 s
2 n

1

2 s
4 n

1
moment function
16

Prove(2/2) assume
 i
2 are generated by Wiener process dz, we can therefore write

2 n 1
 s where

2 n

1 dt

2 s
2 n

1
 dt is a random sample
 s
2 n

1

2 s
2 n

1
 from standard normal substitute this equation for s
2 n
 s
2 n

1 s
2 n
 s
2 n

1 let dV

 s n
2
 
(
 
 s
2 n

1
, V b 
1 ) s
2 n

1
 s
2 n

1
, a
 
2 s
2 n

1


1
   b
, a V
L and
  
2 so that
  dV
 a ( V
L dV
 a ( V
L


V )

V ) dt


V


Vdz a ( V
L

V ) dt
 
V dt
17

Example
 Suppose s n
2 
.

.
u
2 n

1
 s n
2

1
 The long-run variance rate is 0.0002 so that the long-run volatility per day is 1.4%
18

Example continued
 Suppose that the current estimate of the volatility is 1.6% per day and the most recent percentage change in the market variable is 1%.
 The new variance rate is

.

.

.

.

.
The new volatility is 1.53% per day
19

Maximum Likelihood Methods
 In maximum likelihood methods we choose parameters that maximize the likelihood of the observations occurring
20

Example 1
 We sample 10 stocks at random on a certain day and find that the price of one of them declined and the price of others remaind same or increased what is the best estimate of the probability of a price decline?


The probability of the event happening on one particular trial and not on the others is
9 p ( 1
 p )
We maximize this to obtain a maximum likelihood estimate. Result: p = 0.1
21

Example 2(1/2)
Estimate the variance of observations from a normal distribution with mean zero
Normal PDF for X when X
 u i
Maximize
1
2
 v exp


  u
2 v i
2



: i m 

1



1
2
 v exp


  u
2 v i
2






22

Example 2(2/2)
Taking logarithms this is equivalent to maximizing
1
2 i m 

1


 ln( 2
 v )
 u i
2

 v remove constant f ( x )
 i m 

1


 ln( v )
 u i
2 v

 f
'
( x )
 
1
 m
 v i m 

1 u i
2 v
2
Result

0
: v

1 m i m 

1 u i
2
:
23

24

Z ~ N ( 0 , 1 )
M z
( t )
 e
0
 t

1
2

1
2  t
2
M
' z
( t )
 te
1
2 t
2
 e
1
2 t
2
M
' ' z
( t )
 e
1
2 t
2
 t
M z
' ' '
( t )

( 2 t ) e
1
2 t
2
2 e
1
2 t
2

( 1


( 1
 t
2
) e
1
2 t
2 t
2
) te
1
2 t
2

( 3 t
 t
3
) e
1
2 t
2
M ( 4 ) z
( t )

( 3

3 t 2 ) e
1
2 t
2
E
E
( Z
( Z
E [(
4
4
)
)
 n

1



M
E (
( z
4 ) x
( t
 s
)

| t
)

0
4
0 )
4
]

3 s


4 n

1
E [

4 n

1
]

3 s
4 n

1
3
3

( 3 t
 t 3 ) te
1
2 t
2
back
25