Capitulo 5
Modelagem e Previsão de Series
Temporais Univariadas
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
1
Modelos de Series Temporais
• Onde tentamos prever retornos usando somente informação contida nos seus
valores passados.
Alguma Notação e Conceitos
• Um Processo Estritamente Estacionario
Um processo estritamente estacionario é um onde
P{yt1  b1,..., ytn  bn}  P{yt1  m  b1,..., ytn  m  bn}
i.e.Uma medida de probabilidade para a sequencia {yt} é a mesma para {yt+m}  m.
• Um Processo Fracamente Estacionario
Se uma serie satisfaz as três equações abaixo, se diz fracamente ou covariancia
estationaria
1. E(yt) =  ,
t = 1,2,...,
2
2. E ( yt   )( yt   )    
3. E ( yt1   )( yt 2   )   t 2  t1  t1 , t2
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2
Modelos de Series Temporais (cont’d)
• Então se o processo é covariancia estationario, as variancias são as mesmas e as
covariancias dependem da differença entre t1 and t2. Os momentos
E ( yt  E ( yt ))( yt  s  E ( yt  s ))   s , s = 0,1,2, ...
são conhecidos como a função covariancia.
• As covariancias, s, são conhecidas como autocovariancias.
• Todavia, o valor das autocovariancias dependem das unidades de medida de yt.
• É então mais conveniente usar as autocorrelações que são as autocovariancias
normalizadas dividindo-se pela variancia:
, s = 0,1,2, ...

s  s
• Se fizermos o grafico de  0s against s=0,1,2,..então se obtém a função de
autocorrelação ou o correlograma.
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
3
Um Processo Ruido Branco
• Um processo ruido branco é um com (virtualmente) estrutura discernivel.
Uma definição de processo ruido branco
é
E( y )  
t
Var ( yt )   2
 2 if t  r
 t r  
otherwise
0
• Então, a função de autocorrelação será zero exceto para um unico peak de
1 at s = 0. s  aproximadamente N(0,1/T) onde T = tamanho da amostra
• Pode-se usar isto para realizar testes de significancia para os coeficientes
de autocorrelação através da construção de intervalo de confiança.
 .196 
por
1
T
• Por exemplo, um intervalo de confiança a 95% seria dado
.
Se o coeficiente de autocorrelação, s
fica fora dessa região para
qualquer valor de s, então rejeitamos a hipotese nula de que o valor
verdadeiro do coeficiente no lag s é zero.
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
4
Testes de Hipótese Conjunta
• Podemos tambem testar a hipotese conjunta de que todos os m dos coeficientes
de correlação k são simultaneamente iguais a zero com a estatistica Q
m
desenvolvida por Box and
Pierce:
Q  T   k2
k 1
onde T = tamanho da amostra, m = extensão maxima da defasagem
• A estatistica Q é assintoticamente distribuida
. m2
• Todavia, o teste de Box Pierce tem propriedades pobres em pequenas amostras,
tendo sido desenvolvida uma variante, chamada a estatistica Ljung-Box :
m
Q  T T  2

k 1
 k2
T k
~  m2
• Esta estatistica é muito util como um teste de dependencia linear em time series.
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
5
An ACF Example
• 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 BoxPierce 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.
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
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Processos de Médias Móveis
• Seja ut (t=1,2,3,...) uma sequencia de variaveis aleatórias distribuidas
identicamente e indepedentemente com E(ut)=0 e Var(ut)= ,2, então
yt =  + ut + 1ut-1 + 2ut-2 + ... + qut-q
é um modelo de media movel de qth ordem, MA(q).
• Suas propriedades são
E(yt)=; Var(yt) = 0 = (1+12   22 ... q2 )2
Covariancias
( s   s 1 1   s  2 2  ...   q q  s ) 2
s 
0 for s  q
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for
s  1,2,..., q
7
Example of an MA Problem
1. Consider the following MA(2) process:
X t  u t   1u t 1   2 u t  2
where t is a zero mean white noise process with variance  2 .
(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.
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
8
Solution
(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)
but E(Xt)
Var(Xt)
= E[Xt-E(Xt)][Xt-E(Xt)]
= 0, so
= E[(Xt)(Xt)]
= E[(ut + 1ut-1+ 2ut-2)(ut + 1ut-1+ 2ut-2)]
= E[ u t2   12 u t21   22 u t2 2 +cross-products]
But E[cross-products]=0 since Cov(ut,ut-s)=0 for s0.
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
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Solution (cont’d)
So Var(Xt) = 0= E [ u t   1 u t 1   2 u t  2 ]
2
2 2
2 2
=  1   2 
2
2
2
= (1   1   2 )
2
2
2
2
2
(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)]
2
2
= E[(  1u t 1   1 2 u t  2 )]
=  1  2   1 2  2
= ( 1   1 2 ) 2
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
10
Solution (cont’d)
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[(  2 u t2 2 )]
2
=  2
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.
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
11
Solution (cont’d)
We have the autocovariances, now calculate the autocorrelations:

0   0  1
0
( 1   1 2 ) 2
1
( 1   1 2 )
1 


 0 (1   12   22 ) 2 (1   12   22 )
( 2 ) 2
2
2
2 


 0 (1   12   22 ) 2 (1   12   22 )

3   3  0
0

s   s  0s  2
0
(iii) For 1 = -0.5 and 2 = 0.25, substituting these into the formulae above
gives 1 = -0.476, 2 = 0.190.
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
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Grafico da ACF
Thus the acf plot will appear as follows:
1.2
1
0.8
0.6
acf
0.4
0.2
0
0
1
2
3
4
5
6
-0.2
-0.4
-0.6
s
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13
Processos Autoregressivos
• Um modelo autoregressivo de ordem p, um AR(p) pode ser expresso
como
y t    1 y t 1   2 y t  2  ...   p y t  p  u t
• Ou usando a notação do operador de defasagem:
Lyt = yt-1
Liyt = yt-i
p
y t      i y t i  u t
i 1
p
y t      i Li y t  u t
• or
i 1
 ( L) y t    u t
or
 ( L)  1  (1 L  2 L2 ... p Lp )
where
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.
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A Condição de Estacionaridade para um Modelo AR
• The condition for stationarity of a general AR(p) model is that the
roots of 1  1z  2 z2 ... p z p  0 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 nonstationary)
• Example 2: Is yt = 3yt-1 - 0.25yt-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.
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
15
Wold’s Decomposition Theorem
• 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,  ( L) yt  ut , ignoring the intercept, the Wold
decomposition is
yt   ( L)ut
where,
 ( L)  (1  1 L  2 L2 ... p Lp ) 1
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
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The Moments of an Autoregressive Process
• The moments of an autoregressive process are as follows. The mean is
given by
0
E ( yt ) 
1  1  2  ...   p
• The autocovariances and autocorrelation functions can be obtained by
solving what are known as the Yule-Walker equations:
1  1  12  ...   p 1 p
 2  11  2  ...   p  2 p



 p   p 11   p  22  ...   p
• If the AR model is stationary, the autocorrelation function will decay
exponentially to zero.
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
17
Sample AR Problem
• Consider the following simple AR(1) model
yt    1 yt 1  ut
(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.
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Solution
(i) Unconditional mean:
E(yt) = E(+1yt-1)
= +1E(yt-1)
But also
So E(yt)=  +1 ( +1E(yt-2))
=  +1  +12 E(yt-2))
E(yt) =  +1  +12 E(yt-2))
=  +1  +12 ( +1E(yt-3))
=  +1  +12  +13 E(yt-3)
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Solution (cont’d)
An infinite number of such substitutions would give
E(yt) =  (1+1+12 +...) + 1y0
So long as the model is stationary, i.e. , then 1 = 0.
So E(yt) =  (1+1+12 +...) =

1  1
(ii) Calculating the variance of yt: yt  1 yt 1  ut
From Wold’s decomposition theorem:
yt (1  1 L)  ut
yt  (1  1 L) 1 u t
y t  (1  1 L  1 L2  ...)u t
2
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Solution (cont’d)
So long as 1  1 , this will converge.
yt  ut  1ut 1  1 ut 2  ...
2
Var(yt) = E[yt-E(yt)][yt-E(yt)]
but E(yt) = 0, since we are setting  = 0.
Var(yt) = E[(yt)(yt)]
= E[ ut  1ut 1  12ut 2  .. ut  1ut 1  12ut 2  .. ]
= E[(ut 2  12ut 12  14ut 2 2  ...  cross  products)]
2
2
2
4
2
= E[(ut  1 ut 1  1 ut 2  ...)]
=  u2  12 u2  14 u2  ...
2
2
4
=  u (1  1  1  ...)
2

u
=
(1  12 )


‘Introductory Econometrics for Finance’ © Chris Brooks 2002

21
Solution (cont’d)
(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]
2
2
1 = E[(ut  1ut 1  1 ut 2  ...)(u t 1  1u t 2  1 u t 3  ...) ]
2
3
2

u


u
= E[ 1 t 1
1
t  2  ...  cross  products]
2
3 2
5 2
= 1  1   1   ...
=
1 2
(1  12 )
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
22
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]
2
2
= E[(ut  1ut 1  1 ut  2  ...)(ut 2  1ut 3  1 ut 4  ...) ]
= E[ 1 2 u t  2 2  1 4 u t 3 2  ...  cross  products]
= 12 2  14 2  ...
2 2
2
4
= 1  (1  1  1  ...)
=
12 2
(1  12 )
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23
Solution (cont’d)
• If these steps were repeated for 3, the following expression would be
obtained
3 =
13 2
(1  12 )
and for any lag s, the autocovariance would be given by
s =
1s  2
(1  12 )
The acf can now be obtained by dividing the covariances by the
variance:
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24
Solution (cont’d)
0
1
0 =
0


 1 2 

2 
(
1



1 )
1 

 1
1 =  


0
 2 

2 
(
1



1 )


2 =
2
0
 2 2 
 1  

2 
(
1



1 )



 12


 2 

2 
(
1



1 )


3
3 =  1
…
s = 1s
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A Função de Autocorrelação Parcial ( kk)
• Mensura a correlação entre uma observação k periodos atras e a observação
atual, após controlar as observações de defasagens intermediárias (i.e. all
lags < k).
• Portanto, kk mede a correlação entre yt and yt-k após eliminar os efeitos de ytk+1 , yt-k+2 , …, yt-1 .
• No lag 1, a acf = pacf sempre
• No lag 2, 22 = (2-12) / (1-12)
• Para lags 3+, as formulas são mais complexas.
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26
A Função de Autocorrelação Parcial ( kk)
(cont’d)
• A pacf é util para dizer a diferença entre um processo AR e um
processo ARMA .
• No caso de um AR(p), há conexões diretas entre yt e yt-s somente
para s p.
• Então para um AR(p), a pacf teórica será zero após o lag p.
• No caso de um MA(q), isto pode ser escrito como um AR(), daí haver
conexões diretas entre yt e todos os seus valores anteriores.
• Para um MA(q), o pacf teórico será geometricamente declinante.
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27
Processos ARMA
• Combinando os modelos AR(p) e MA(q) , podemos obter um modelo
ARMA(p,q) :
 ( L) yt     ( L)ut
where  ( L)  1  1 L  2 L2 ... p Lp
and  ( L)  1  1L   2 L2  ...   q Lq
or
yt    1 yt 1   2 yt 2  ...   p yt  p   1u t 1   2 u t 2  ...   q u t q  u t
2
2
with E(ut )  0; E(ut )   ; E(ut u s )  0, t  s
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28
The Invertibility Condition
• 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
E ( yt ) 

1  1  2 ...p
• 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.
‘Introductory Econometrics for Finance’ © Chris Brooks 2002
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Sumário do Comportamento da acf para os
Processos AR e MA
Um processo autoregressivo tem
• Uma acf que cai geometricamente
• Numero de spikes da pacf = ordem de AR
Um processo de media movel tem
• Numero de spikes da acf = ordem de MA
• Uma pacf que cai geometricamente
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30
Alguns gráficos de acf e pacf para processos padrão
acf e a pacf não são produzidas analiticamente de formulas relevantes para um modelo
daquele tipo, mas, ao invés, são estimadas usando 100,000 observações simuladas com
disturbancias extraidas de uma distribuição normal.
ACF and PACF for an MA(1) Model: yt = – 0.5ut-1 + ut
A
0.05
0
1
2
3
4
5
6
7
8
9
10
-0.05
acf and pacf
-0.1
-0.15
-0.2
-0.25
-0.3
acf
-0.35
pacf
-0.4
-0.45
Lag
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31
ACF and PACF for an MA(2) Model:
yt = 0.5ut-1 - 0.25ut-2 + ut
0.4
acf
0.3
pacf
0.2
acf and pacf
0.1
0
1
2
3
4
5
6
7
8
9
10
-0.1
-0.2
-0.3
-0.4
Lags
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32
ACF e PACF para um Modelo AR(1), que caem
lentamente:
yt = 0.9yt-1 + ut
1
0.9
acf
pacf
0.8
0.7
acf and pacf
0.6
0.5
0.4
0.3
0.2
0.1
0
1
2
3
4
5
6
7
8
9
10
-0.1
Lags
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ACF e PACF para um Modelo AR(1) que cai mais
rapidamente AR(1) Model: yt = 0.5yt-1 + ut
0.6
0.5
acf
pacf
acf and pacf
0.4
0.3
0.2
0.1
0
1
2
3
4
5
6
7
8
9
10
-0.1
Lags
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ACF e PACF para um Modelo AR(1), que cai mais
rapidamente, com Coeficiente Negativo: yt = -0.5yt-1 +
ut
0.3
0.2
0.1
acf and pacf
0
1
2
3
4
5
6
7
8
9
10
-0.1
-0.2
-0.3
-0.4
acf
pacf
-0.5
-0.6
Lags
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ACF e PACF para um Modelo Não-estacionario
(i.e. a unit coefficient): yt = yt-1 + ut
1
0.9
acf
pacf
0.8
acf and pacf
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
1
2
3
4
6
5
7
8
9
10
Lags
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ACF e PACF para um ARMA(1,1):
yt = 0.5yt-1 + 0.5ut-1 + ut
0.8
0.6
acf
pacf
acf and pacf
0.4
0.2
0
1
2
3
4
5
6
7
8
9
10
-0.2
-0.4
Lags
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A Construção de Modelos ARMA
- A Abordagem de Box Jenkins
• Box e Jenkins (1970) foram os primeiros a abordarem a tarefa de estimar
um modelo ARMA de uma maneira sistematica. Há 3 etapas na sua
abordagem:
1. Identificação
2. Estimação
3. Diagnóstico do Modelo
Etapa 1:
- Envolve a determinação da ordem do modelo.
- Uso de procedimentos de graficos
- Um procedimento melhor está disponivel agora
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A Construção de Modelos ARMA
- A Abordagem de Box Jenkins
Etapa 2:
- Estimação dos parametros
- Pode ser feito usando MQO ou MV a depender do modelo..
Etapa 3:
- Checagem do Modelo
Box and Jenkins sugerem 2 metodos:
- deliberado sobreajustamento
- diagnostico
- diagnostico dos resíduos
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Alguns Desenvolvimentos Mais Recentes na
Modelagem de ARMA
• A identificação não seria tipicamente feita usando acf”s.
• Queremos formar um modelo parcimonioso.
• Razões:
- a variancia dos estimadores é inversamente proporcional ao numero de
graus de liberdade freedom.
- modelos que são profligate podem ser inclinados a dados de carateristicas
especificas
• Isto motiva o uso de criterios de informação, que contém 2 fatores
- um termo que é uma função de RSS
- alguma penalidade por acréscimo de parâmetros extra
• O objetivo é escolher o numero de parametros que minimiza o criterio de
informação.
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Critérios de Informação para Escolha de Modelo
• Os criterios de informação variam de acordo com o termo da penalidade.
• Os três critérios mais populares são o critério de informação de Akaike
(1974) (AIC), o critério Bayesiano de informação de Schwarz (1978)
(SBIC), e o critério de Hannan-Quinn (HQIC).
AIC  ln( 2 )  2k / T
k
SBIC  ln( ˆ 2 )  ln T
T
2k
HQIC  ln( ˆ 2 ) 
ln(ln( T ))
T
where k = p + q + 1, T = sample size. So we min. IC s.t. p  p, q  q
SBIC incorpora uma penalidade maior que AIC.
• Que C I deve ser preferido se eles sugerem ordens diferentes de modelo?
– SBIC é fortemente consistente mas (ineficiente).
– AIC não é consistente, e pegará tipicamente modelos maiores
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Modelos ARIMA
• Diferentes dos modelos ARMA. O I significa integrado.
• Um processo autoregressivo integrado é um com uma raiz
caracteristica no circulo unitário.
• Os pesquisadores diferenciam a variavel como necessaria e então
constroem um modelo ARMA com aquelas variaveis diferenciadas.
• Um modelo ARMA(p,q) na
variavel diferenciada d vezes é
equivalente a um modelo ARIMA(p,d,q) nos dados originais.
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Exponential Smoothing
• 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
where

is the smoothing constant, with 01
yt
is the current realised value
St
is the current smoothed value
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43
Exponential Smoothing (cont’d)
• Lagging (1) by one period we can write
St-1 =  yt-1 + (1-)St-2
• and lagging again
St-2 =  yt-2 + (1-)St-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
(2)
(3)
(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
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Exponential Smoothing (cont’d)
• T successive substitutions of this kind would lead to
 T

i
T
St    1    yt i   1    S0
 i 0

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.
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Exponential Smoothing (cont’d)
•
–
–
–
–
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.
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Previsão em Econometria
• Forecasting = prediction.
• Um teste importante da adequacão de um modelo.
e.g.
- Prever o retorno de uma determinada ação amanhã
- Prever o preço de uma casa dadas as suas caracteristicas
- Prever o risco de um portfolio ao longo do próximo ano
- Prever a volatilidade de retornos de bonds
• Podemos distinguir duas abordagens:
- Previsão econometrica (estructural)
- Previsão de series temporais
• The distinction between the two types is somewhat blurred (e.g, VARs).
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In-Sample Versus Out-of-Sample
• 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.
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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
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Models for Forecasting
• Structural models
e.g.
y = X + u
yt  1   2 x2t     k xkt  ut
To forecast y, we require the conditional expectation of its future
value:
Eyt t 1   E1   2 x2t     k xkt  ut 
= 1   2 E x2t      k E xkt 
But what are ( x 2t ) etc.? We could use x 2 , so
E  yt   1   2 x2     k xk
= y !!
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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.
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Forecasting with ARMA Models
The forecasting model typically used is of the form:
p
q
i 1
j 1
f t , s      i f t , s  i    j ut  s  j
where ft,s = yt+s , s 0; ut+s = 0, s > 0
= ut+s , s  0
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Forecasting with MA Models
• 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
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Forecasting with MA Models (cont’d)
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
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Forecasting with AR Models
• 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
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Forecasting with AR Models (cont’d)
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.
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How can we test whether a forecast is accurate or not?
•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:
1
MSE 
N
1
MAE is given by MAE 
N
N

t 1
N

t 1
( yt  s  f t , s ) 2
yt  s  f t , s
1 N yt s  ft ,s
Mean absolute percentage error: MAPE  100 
N t 1
yt s
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How can we test whether a forecast is accurate or not?
(cont’d)
• 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:
1 N
 zt  s
% correct sign predictions =
N t 1
where
zt+s = 1 if (xt+s . ft,s ) > 0
zt+s = 0 otherwise
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Forecast Evaluation Example
• Given the following forecast and actual values, calculate the MSE, MAE and
percentage of correct sign predictions:
Steps Ahead
1
2
3
4
5
Forecast
0.20
0.15
0.10
0.06
0.04
Actual
-0.40
0.20
0.10
-0.10
-0.05
• MSE = 0.079, MAE = 0.180, % of correct sign predictions = 40
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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
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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
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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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