Ch6 slides

‘Introductory Econometrics for Finance’ © Chris Brooks 2013 1
Chapter 6
Univariate time series modelling and
forecasting

• Where we attempt to predict returns using only information contained in their
past values.
Some Notation and Concepts
• A Strictly Stationary Process
A strictly stationary process is one where
i.e. the probability measure for the sequence {yt} is the same as that for {yt+m}  m.
• A Weakly Stationary Process
If a series satisfies the next three equations, it is said to be weakly or covariance
stationary
1. E(yt) =  , t = 1,2,...,
2.
3.  t1 , t2
Univariate Time Series Models
P y b y b P y b y b
t t n t m t m n
n n
{ ,..., } { ,..., }
1 1
1 1
    
 
E y y
t t t t
( )( )
1 2 2 1
   
  
E y y
t t
( )( )
    
  2

• So if the process is covariance stationary, all the variances are the same and all
the covariances depend on the difference between t1 and t2. The moments
, s = 0,1,2, ...
are known as the covariance function.
• The covariances, s, are known as autocovariances.
• However, the value of the autocovariances depend on the units of measurement
of yt.
• It is thus more convenient to use the autocorrelations which are the
autocovariances normalised by dividing by the variance:
, s = 0,1,2, ...
• If we plot s against s=0,1,2,... then we obtain the autocorrelation function or
correlogram.
Univariate Time Series Models (cont’d)



s
s

0
E y E y y E y
t t t s t s s
( ( ))( ( ))
  
  

• A white noise process is one with (virtually) no discernible structure. A
definition of a white noise process is
• Thus the autocorrelation function will be zero apart from a single peak of 1
at s = 0. s  approximately N(0,1/T) where T = sample size
• We can use this to do significance tests for the autocorrelation coefficients
by constructing a confidence interval.
• For example, a 95% confidence interval would be given by . If
the sample autocorrelation coefficient, , falls outside this region for any
value of s, then we reject the null hypothesis that the true value of the
coefficient at lag s is zero.
A White Noise Process
E y
Var y
if t r
otherwise
t
t
t r
( )
( )












2
2
0

s
T
1
196
. 


• We can also test the joint hypothesis that all m of the k correlation coefficients
are simultaneously equal to zero using the Q-statistic developed by Box and
Pierce:
where T = sample size, m = maximum lag length
• The Q-statistic is asymptotically distributed as a .
• However, the Box Pierce test has poor small sample properties, so a variant
has been developed, called the Ljung-Box statistic:
• This statistic is very useful as a portmanteau (general) test of linear dependence
in time series.
Joint Hypothesis Tests
m
2



m
k
k
T
Q
1
2

  2
1
2
~
2 m
m
k
k
k
T
T
T
Q 








• Question:
Suppose that a researcher had estimated the first 5 autocorrelation coefficients
using a series of length 100 observations, and found them to be (from 1 to 5):
0.207, -0.013, 0.086, 0.005, -0.022.
Test each of the individual coefficient for significance, and use both the Box-
Pierce and Ljung-Box tests to establish whether they are jointly significant.
• Solution:
A coefficient would be significant if it lies outside (-0.196,+0.196) at the 5%
level, so only the first autocorrelation coefficient is significant.
Q=5.09 and Q*=5.26
Compared with a tabulated 2(5)=11.1 at the 5% level, so the 5 coefficients
are jointly insignificant.
An ACF Example

• Let ut (t=1,2,3,...) be a sequence of independently and identically
distributed (iid) random variables with E(ut)=0 and Var(ut)= , then
yt =  + ut + 1ut-1 + 2ut-2 + ... + qut-q
is a qth order moving average model MA(q).
• Its properties are
E(yt)=; Var(yt) = 0 = (1+ )2
Covariances
Moving Average Processes

2
  
1
2
2
2 2
  
... q











 


q
s
for
q
s
for
s
q
q
s
s
s
s
0
,...,
2
,
1
)
...
( 2
2
2
1
1 









1. Consider the following MA(2) process:
where ut is a zero mean white noise process with variance .
(i) Calculate the mean and variance of Xt
(ii) Derive the autocorrelation function for this process (i.e. express the
autocorrelations, 1, 2, ... as functions of the parameters 1 and
2).
(iii) If 1 = -0.5 and 2 = 0.25, sketch the acf of Xt.
Example of an MA Problem
2
2
1
1 
 

 t
t
t
t u
u
u
X 

2


(i) If E(ut)=0, then E(ut-i)=0  i.
So
E(Xt) = E(ut + 1ut-1+ 2ut-2)= E(ut)+ 1E(ut-1)+ 2E(ut-2)=0
Var(Xt) = E[Xt-E(Xt)][Xt-E(Xt)]
but E(Xt) = 0, so
Var(Xt) = E[(Xt)(Xt)]
= E[(ut + 1ut-1+ 2ut-2)(ut + 1ut-1+ 2ut-2)]
= E[ +cross-products]
But E[cross-products]=0 since Cov(ut,ut-s)=0 for s0.
Solution
2
2
2
2
2
1
2
1
2

 
 t
t
t u
u
u 


So Var(Xt) = 0= E [ ]
=
=
(ii) The acf of Xt.
1 = E[Xt-E(Xt)][Xt-1-E(Xt-1)]
= E[Xt][Xt-1]
= E[(ut +1ut-1+ 2ut-2)(ut-1 + 1ut-2+ 2ut-3)]
= E[( )]
=
=
Solution (cont’d)
2
2
2
2
2
1
2
1
2

 
 t
t
t u
u
u 

2
2
2
2
2
1
2




 

2
2
2
2
1 )
1
( 

 

2
2
2
1
2
1
1 
  t
t u
u 


2
2
1
2
1 



 
2
2
1
1 )
( 


 

2 = E[Xt-E(Xt)][Xt-2-E(Xt-2)]
= E[Xt][Xt-2]
= E[(ut +1ut-1+2ut-2)(ut-2 +1ut-3+2ut-4)]
= E[( )]
=
3 = E[Xt-E(Xt)][Xt-3-E(Xt-3)]
= E[Xt][Xt-3]
= E[(ut +1ut-1+2ut-2)(ut-3 +1ut-4+2ut-5)]
= 0
So s = 0 for s > 2.
Solution (cont’d)
2
2
2 
t
u

2
2


Solution (cont’d)
We have the autocovariances, now calculate the autocorrelations:
(iii) For 1 = -0.5 and 2 = 0.25, substituting these into the formulae above
gives 1 = -0.476, 2 = 0.190.



0
0
0
1
 



3
3
0
0
 



s
s
s
   
0
0 2
)
1
(
)
(
)
1
(
)
(
2
2
2
1
2
1
1
2
2
2
2
1
2
2
1
1
0
1
1
























)
1
(
)
1
(
)
(
2
2
2
1
2
2
2
2
2
1
2
2
0
2
2



















Thus the acf plot will appear as follows:
ACF Plot
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6
s
acf

• An autoregressive model of order p, an AR(p) can be expressed as
• Or using the lag operator notation:
Lyt = yt-1 Liyt = yt-i
• or
or where .
Autoregressive Processes
   
( ) ( ... )
L L L L
p
p
   
1 1 2
2
t
p
t
p
t
t
t u
y
y
y
y 




 

 


 ...
2
2
1
1


 


p
i
t
i
t
i
t u
y
y
1







p
i
t
t
i
i
t u
y
L
y
1


t
t u
y
L 
 
 )
(

• The condition for stationarity of a general AR(p) model is that the
roots of all lie outside the unit circle.
• A stationary AR(p) model is required for it to have an MA()
representation.
• Example 1: Is yt = yt-1 + ut stationary?
The characteristic root is 1, so it is a unit root process (so non-
stationary)
• Example 2: Is yt = 3yt-1 – 2.75yt-2 + 0.75yt-3 +ut stationary?
The characteristic roots are 1, 2/3, and 2. Since only one of these lies
outside the unit circle, the process is non-stationary.
The Stationary Condition for an AR Model
1 0
1 2
2
    
  
z z z
p
p
...

• States that any stationary series can be decomposed into the sum of two
unrelated processes, a purely deterministic part and a purely stochastic
part, which will be an MA().
• For the AR(p) model, , ignoring the intercept, the Wold
decomposition is
where,
Wold’s Decomposition Theorem
   
( ) ( ... )
L L L L
p
p
     
1 1 2
2 1
t
t u
y
L 
)
(

t
t u
L
y )
(



• The moments of an autoregressive process are as follows. The mean is
given by
• The autocovariances and autocorrelation functions can be obtained by
solving what are known as the Yule-Walker equations:
• If the AR model is stationary, the autocorrelation function will decay
exponentially to zero.
The Moments of an Autoregressive Process
p
t
y
E









...
1
)
(
2
1
0
p
p
p
p
p
p
p
p


































...
...
...
2
2
1
1
2
2
1
1
2
1
2
1
1
1




• Consider the following simple AR(1) model
(i) Calculate the (unconditional) mean of yt.
For the remainder of the question, set =0 for simplicity.
(ii) Calculate the (unconditional) variance of yt.
(iii) Derive the autocorrelation function for yt.
Sample AR Problem
t
t
t u
y
y 

 1
1



(i) Unconditional mean:
E(yt) = E(+1yt-1)
= +1E(yt-1)
But also
So E(yt)=  +1 ( +1E(yt-2))
=  +1  +1
2 E(yt-2))
E(yt) =  +1  +1
2 E(yt-2))
=  +1  +1
2 ( +1E(yt-3))
=  +1  +1
2  +1
3 E(yt-3)
Solution

An infinite number of such substitutions would give
E(yt) =  (1+1+1
2 +...) + 1
y0
So long as the model is stationary, i.e. 1 < 1, then 1
 = 0.
So E(yt) =  (1+1+1
2 +...) =
(ii) Calculating the variance of yt:
From Wold’s decomposition theorem:
Solution (cont’d)
1
1 


t
t
t u
y
y 
 1
1

t
t u
L
y 
 )
1
( 1

t
t u
L
y 1
1 )
1
( 

 
t
t u
L
L
y ...)
1
( 2
2
1
1 


 


So long as , this will converge.
Var(yt) = E[yt-E(yt)][yt-E(yt)]
but E(yt) = 0, since we are setting  = 0.
Var(yt) = E[(yt)(yt)]
= E[ ]
= E[
= E[
=
=
=
Solution (cont’d)
1
1 

...
2
2
1
1
1 


 
 t
t
t
t u
u
u
y 

  
..
.. 2
2
1
1
1
2
2
1
1
1 




 


 t
t
t
t
t
t u
u
u
u
u
u 



)]
...
(
2
2
4
1
2
1
2
1
2
products
cross
u
u
u t
t
t 



 
 

...)]
(
2
2
4
1
2
1
2
1
2


 
 t
t
t u
u
u 

...
2
4
1
2
2
1
2


 u
u
u 




...)
1
( 4
1
2
1
2


 

u
)
1
( 2
1
2



u

(iii) Turning now to calculating the acf, first calculate the autocovariances:
1 = Cov(yt, yt-1) = E[yt-E(yt)][yt-1-E(yt-1)]
Since a0 has been set to zero, E(yt) = 0 and E(yt-1) = 0, so
1 = E[ytyt-1]
1 = E[ ]
= E[
=
=
Solution (cont’d)
...)
( 2
2
1
1
1 

 
 t
t
t u
u
u 
 ...)
( 3
2
1
2
1
1 

 

 t
t
t u
u
u 

]
...
2
2
3
1
2
1
1 products
cross
u
u t
t 


 
 

...
2
5
1
2
3
1
2
1 

 





)
1
( 2
1
2
1





Solution (cont’d)
For the second autocorrelation coefficient,
2 = Cov(yt, yt-2) = E[yt-E(yt)][yt-2-E(yt-2)]
Using the same rules as applied above for the lag 1 covariance
2 = E[ytyt-2]
= E[ ]
= E[
=
=
=
...)
( 2
2
1
1
1 

 
 t
t
t u
u
u 
 ...)
( 4
2
1
3
1
2 

 

 t
t
t u
u
u 

]
...
2
3
4
1
2
2
2
1 products
cross
u
u t
t 


 
 

...
2
4
1
2
2
1 
 



...)
1
( 4
1
2
1
2
2
1 

 



)
1
( 2
1
2
2
1





Solution (cont’d)
• If these steps were repeated for 3, the following expression would be
obtained
3 =
and for any lag s, the autocovariance would be given by
s =
The acf can now be obtained by dividing the covariances by the
variance:
)
1
( 2
1
2
3
1




)
1
( 2
1
2
1




s

Solution (cont’d)
0 =
1 = 2 =
3 =
…
s =
1
0
0



1
2
1
2
2
1
2
1
0
1
)
1
(
)
1
(































 2
1
2
1
2
2
1
2
2
1
0
2
)
1
(
)
1
(
































3
1

s
1


• Measures the correlation between an observation k periods ago and the
current observation, after controlling for observations at intermediate lags
(i.e. all lags < k).
• So kk measures the correlation between yt and yt-k after removing the effects
of yt-k+1 , yt-k+2 , …, yt-1 .
• At lag 1, the acf = pacf always
• At lag 2, 22 = (2-1
2) / (1-1
2)
• For lags 3+, the formulae are more complex.
The Partial Autocorrelation Function (denoted kk)

• The pacf is useful for telling the difference between an AR process and an
ARMA process.
• In the case of an AR(p), there are direct connections between yt and yt-s only
for s p.
• So for an AR(p), the theoretical pacf will be zero after lag p.
• In the case of an MA(q), this can be written as an AR(), so there are direct
connections between yt and all its previous values.
• For an MA(q), the theoretical pacf will be geometrically declining.
The Partial Autocorrelation Function (denoted kk)
(cont’d)

• By combining the AR(p) and MA(q) models, we can obtain an ARMA(p,q)
model:
where
and
or
with
ARMA Processes
   
( ) ...
L L L L
p
p
    
1 1 2
2
q
qL
L
L
L 


 



 ...
1
)
( 2
2
1
t
t u
L
y
L )
(
)
( 

 

t
q
t
q
t
t
p
t
p
t
t
t u
u
u
u
y
y
y
y 








 




 





 ...
... 2
2
1
1
2
2
1
1
s
t
u
u
E
u
E
u
E s
t
t
t 


 ,
0
)
(
;
)
(
;
0
)
( 2
2


• Similar to the stationarity condition, we typically require the MA(q) part of
the model to have roots of (z)=0 greater than one in absolute value.
• The mean of an ARMA series is given by
• The autocorrelation function for an ARMA process will display
combinations of behaviour derived from the AR and MA parts, but for lags
beyond q, the acf will simply be identical to the individual AR(p) model.
The Invertibility Condition
E yt
p
( )
...

   

  
1 1 2

An autoregressive process has
• a geometrically decaying acf
• number of spikes of pacf = AR order
A moving average process has
• Number of spikes of acf = MA order
• a geometrically decaying pacf
Summary of the Behaviour of the acf for
AR and MA Processes

The acf and pacf are not produced analytically from the relevant formulae for a model of that
type, but rather are estimated using 100,000 simulated observations with disturbances drawn
from a normal distribution.
ACF and PACF for an MA(1) Model: yt = – 0.5ut-1 + ut
Some sample acf and pacf plots
for standard processes
-0.45
-0.4
-0.35
-0.3
-0.25
-0.2
-0.15
-0.1
-0.05
0
0.05
1 2 3 4 5 6 7 8 9 10
Lag
acf
and
pacf
acf
pacf

ACF and PACF for an MA(2) Model:
yt = 0.5ut-1 - 0.25ut-2 + ut
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
1 2 3 4 5 6 7 8 9 10
Lags
acf
and
pacf
acf
pacf

-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 2 3 4 5 6 7 8 9 10
Lags
acf
and
pacf
acf
pacf
ACF and PACF for a slowly decaying AR(1) Model:
yt = 0.9yt-1 + ut

ACF and PACF for a more rapidly decaying AR(1)
Model: yt = 0.5yt-1 + ut
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
1 2 3 4 5 6 7 8 9 10
Lags
acf
and
pacf
acf
pacf

ACF and PACF for a more rapidly decaying AR(1)
Model with Negative Coefficient: yt = -0.5yt-1 + ut
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
1 2 3 4 5 6 7 8 9 10
Lags
acf
and
pacf
acf
pacf

ACF and PACF for a Non-stationary Model
(i.e. a unit coefficient): yt = yt-1 + ut
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 2 3 4 5 6 7 8 9 10
Lags
acf
and
pacf
acf
pacf

ACF and PACF for an ARMA(1,1):
yt = 0.5yt-1 + 0.5ut-1 + ut
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1 2 3 4 5 6 7 8 9 10
Lags
acf
and
pacf
acf
pacf

• Box and Jenkins (1970) were the first to approach the task of estimating an
ARMA model in a systematic manner. There are 3 steps to their approach:
1. Identification
2. Estimation
3. Model diagnostic checking
Step 1:
- Involves determining the order of the model.
- Use of graphical procedures
- A better procedure is now available
Building ARMA Models
- The Box Jenkins Approach

Step 2:
- Estimation of the parameters
- Can be done using least squares or maximum likelihood depending
on the
model.
Step 3:
- Model checking
Box and Jenkins suggest 2 methods:
- deliberate overfitting
- residual diagnostics
Building ARMA Models
- The Box Jenkins Approach (cont’d)

• Identification would typically not be done using acf’s.
• We want to form a parsimonious model.
• Reasons:
- variance of estimators is inversely proportional to the number of degrees of
freedom.
- models which are profligate might be inclined to fit to data specific features
• This gives motivation for using information criteria, which embody 2 factors
- a term which is a function of the RSS
- some penalty for adding extra parameters
• The object is to choose the number of parameters which minimises the
information criterion.
Some More Recent Developments in
ARMA Modelling

• The information criteria vary according to how stiff the penalty term is.
• The three most popular criteria are Akaike’s (1974) information criterion
(AIC), Schwarz’s (1978) Bayesian information criterion (SBIC), and the
Hannan-Quinn criterion (HQIC).
where k = p + q + 1, T = sample size. So we min. IC s.t.
SBIC embodies a stiffer penalty term than AIC.
• Which IC should be preferred if they suggest different model orders?
– SBIC is strongly consistent but (inefficient).
– AIC is not consistent, and will typically pick “bigger” models.
Information Criteria for Model Selection
AIC k T
 
ln(  ) /
 2
2
p p q q
 
,
T
T
k
SBIC ln
)
ˆ
ln( 2

 
))
ln(ln(
2
)
ˆ
ln( 2
T
T
k
HQIC 
 

• As distinct from ARMA models. The I stands for integrated.
• An integrated autoregressive process is one with a characteristic root
on the unit circle.
• Typically researchers difference the variable as necessary and then
build an ARMA model on those differenced variables.
• An ARMA(p,q) model in the variable differenced d times is equivalent
to an ARIMA(p,d,q) model on the original data.
ARIMA Models

• Another modelling and forecasting technique
• How much weight do we attach to previous observations?
• Expect recent observations to have the most power in helping to forecast
future values of a series.
• The equation for the model
St =  yt + (1-)St-1 (1)
where
 is the smoothing constant, with 01
yt is the current realised value
St is the current smoothed value
Exponential Smoothing

• Lagging (1) by one period we can write
St-1 =  yt-1 + (1-)St-2 (2)
• and lagging again
St-2 =  yt-2 + (1-)St-3 (3)
• Substituting into (1) for St-1 from (2)
St =  yt + (1-)( yt-1 + (1-)St-2)
=  yt + (1-) yt-1 + (1-)2 St-2 (4)
• Substituting into (4) for St-2 from (3)
St =  yt + (1-) yt-1 + (1-)2 St-2
=  yt + (1-) yt-1 + (1-)2( yt-2 + (1-)St-3)
=  yt + (1-) yt-1 + (1-)2 yt-2 + (1-)3 St-3
Exponential Smoothing (cont’d)

• T successive substitutions of this kind would lead to
since 0, the effect of each observation declines exponentially as we
move another observation forward in time.
• Forecasts are generated by
ft+s = St
for all steps into the future s = 1, 2, ...
• This technique is called single (or simple) exponential smoothing.
    0
0
1
1 S
y
S
T
T
i
i
t
i
t 

 








 



• It doesn’t work well for financial data because
– there is little structure to smooth
– it cannot allow for seasonality
– it is an ARIMA(0,1,1) with MA coefficient (1-) - (See Granger & Newbold, p174)
– forecasts do not converge on long term mean as s
• Can modify single exponential smoothing
– to allow for trends (Holt’s method)
– or to allow for seasonality (Winter’s method).
• Advantages of Exponential Smoothing
– Very simple to use
– Easy to update the model if a new realisation becomes available.

• Forecasting = prediction.
• An important test of the adequacy of a model.
e.g.
- Forecasting tomorrow’s return on a particular share
- Forecasting the price of a house given its characteristics
- Forecasting the riskiness of a portfolio over the next year
- Forecasting the volatility of bond returns
• We can distinguish two approaches:
- Econometric (structural) forecasting
- Time series forecasting
• The distinction between the two types is somewhat blurred (e.g, VARs).
Forecasting in Econometrics

• Expect the “forecast” of the model to be good in-sample.
• Say we have some data - e.g. monthly FTSE returns for 120 months:
1990M1 – 1999M12. We could use all of it to build the model, or keep
some observations back:
• A good test of the model since we have not used the information from
1999M1 onwards when we estimated the model parameters.
In-Sample Versus Out-of-Sample

How to produce forecasts
• Multi-step ahead versus single-step ahead forecasts
• Recursive versus rolling windows
• To understand how to construct forecasts, we need the idea of conditional
expectations:
E(yt+1  t )
• We cannot forecast a white noise process: E(ut+s  t ) = 0  s > 0.
• The two simplest forecasting “methods”
1. Assume no change : f(yt+s) = yt
2. Forecasts are the long term average f(yt+s) = y

Models for Forecasting
• Structural models
e.g. y = X + u
To forecast y, we require the conditional expectation of its future
value:
=
But what are etc.? We could use , so
= !!
t
kt
k
t
t u
x
x
y 



 

 
2
2
1
   
t
kt
k
t
t
t u
x
x
E
y
E 




  

 
2
2
1
1
   
kt
k
t x
E
x
E 

 

 
2
2
1
)
( 2t
x
 2
x
  k
k
t x
x
y
E 

 


 
2
2
1
y

Models for Forecasting (cont’d)
• Time Series Models
The current value of a series, yt, is modelled as a function only of its previous
values and the current value of an error term (and possibly previous values of
the error term).
• Models include:
• simple unweighted averages
• exponentially weighted averages
• ARIMA models
• Non-linear models – e.g. threshold models, GARCH, bilinear models, etc.

The forecasting model typically used is of the form:
where ft,s = yt+s , s 0; ut+s = 0, s > 0
= ut+s , s  0
Forecasting with ARMA Models

 



 


q
j
j
s
t
j
p
i
i
s
t
i
s
t u
f
f
1
1
,
, 



• An MA(q) only has memory of q.
e.g. say we have estimated an MA(3) model:
yt =  + 1ut-1 +  2ut-2 +  3ut-3 + ut
yt+1 =  +  1ut +  2ut-1 +  3ut-2 + ut+1
yt+2 =  +  1ut+1 +  2ut +  3ut-1 + ut+2
yt+3 =  +  1ut+2 +  2ut+1 +  3ut + ut+3
• We are at time t and we want to forecast 1,2,..., s steps ahead.
• We know yt , yt-1, ..., and ut , ut-1
Forecasting with MA Models

ft, 1 = E(yt+1  t ) = E( +  1ut +  2ut-1 +  3ut-2 + ut+1)
=  +  1ut +  2ut-1 +  3ut-2
ft, 2 = E(yt+2  t ) = E( +  1ut+1 +  2ut +  3ut-1 + ut+2)
=  +  2ut +  3ut-1
ft, 3 = E(yt+3  t ) = E( +  1ut+2 +  2ut+1 +  3ut + ut+3)
=  +  3ut
ft, 4 = E(yt+4  t ) = 
ft, s = E(yt+s  t ) =   s  4
Forecasting with MA Models (cont’d)

• Say we have estimated an AR(2)
yt =  + 1yt-1 +  2yt-2 + ut
yt+1 =  +  1yt +  2yt-1 + ut+1
yt+2 =  +  1yt+1 +  2yt + ut+2
yt+3 =  +  1yt+2 +  2yt+1 + ut+3
ft, 1 = E(yt+1  t )= E( +  1yt +  2yt-1 + ut+1)
=  +  1E(yt) +  2E(yt-1)
=  +  1yt +  2yt-1
ft, 2 = E(yt+2  t )= E( +  1yt+1 +  2yt + ut+2)
=  +  1E(yt+1) +  2E(yt)
=  +  1 ft, 1 +  2yt
Forecasting with AR Models

ft, 3 = E(yt+3  t ) = E( +  1yt+2 +  2yt+1 + ut+3)
=  +  1E(yt+2) +  2E(yt+1)
=  +  1 ft, 2 +  2 ft, 1
• We can see immediately that
ft, 4 =  +  1 ft, 3 +  2 ft, 2 etc., so
ft, s =  +  1 ft, s-1 +  2 ft, s-2
• Can easily generate ARMA(p,q) forecasts in the same way.
Forecasting with AR Models (cont’d)

•For example, say we predict that tomorrow’s return on the FTSE will be 0.2, but
the outcome is actually -0.4. Is this accurate? Define ft,s as the forecast made at
time t for s steps ahead (i.e. the forecast made for time t+s), and yt+s as the
realised value of y at time t+s.
• Some of the most popular criteria for assessing the accuracy of time series
forecasting techniques are:
MAE is given by
Mean absolute percentage error:
How can we test whether a forecast is accurate or not?
2
,
1
)
(
1
s
t
s
t
N
t
f
y
N
MSE 
 


s
t
s
t
N
t
f
y
N
MAE ,
1
1

 


s
t
s
t
s
t
N
t y
f
y
N
MAPE





  ,
1
1
100

• It has, however, also recently been shown (Gerlow et al., 1993) that the
accuracy of forecasts according to traditional statistical criteria are not
related to trading profitability.
• A measure more closely correlated with profitability:
% correct sign predictions =
where zt+s = 1 if (yt+s . ft,s ) > 0
zt+s = 0 otherwise
How can we test whether a forecast is accurate or not?
(cont’d)



N
t
s
t
z
N 1
1

• Given the following forecast and actual values, calculate the MSE, MAE and
percentage of correct sign predictions:
• MSE = 0.079, MAE = 0.180, % of correct sign predictions = 40
Forecast Evaluation Example
Steps Ahead Forecast Actual
1 0.20 -0.40
2 0.15 0.20
3 0.10 0.10
4 0.06 -0.10
5 0.04 -0.05

What factors are likely to lead to a
good forecasting model?
• “signal” versus “noise”
• “data mining” issues
• simple versus complex models
• financial or economic theory

Statistical Versus Economic or
Financial loss functions
• Statistical evaluation metrics may not be appropriate.
• How well does the forecast perform in doing the job we wanted it for?
Limits of forecasting: What can and cannot be forecast?
• All statistical forecasting models are essentially extrapolative
• Forecasting models are prone to break down around turning points
• Series subject to structural changes or regime shifts cannot be forecast
• Predictive accuracy usually declines with forecasting horizon
• Forecasting is not a substitute for judgement

Back to the original question: why forecast?
• Why not use “experts” to make judgemental forecasts?
• Judgemental forecasts bring a different set of problems:
e.g., psychologists have found that expert judgements are prone to the
following biases:
– over-confidence
– inconsistency
– recency
– anchoring
– illusory patterns
– “group-think”.
• The Usually Optimal Approach
To use a statistical forecasting model built on solid theoretical
foundations supplemented by expert judgements and interpretation.

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