Time Series AnalysisIntroduction

• Variables over a long time period usually exhibit a pattern

• Regressing one Time series variable on another often gives high t-vaues even though theoretically there may not be any logical relationship between the two.

• Eg: GDP and height of a child • This is the problem of spurious or nonsense

regression• Challenge in time series analysis is to identify long

run equilibrium variables to enable forecasting • Conditions for valid forecasting

• A random/stochastic process is a collection of random variable ordered in time.

• A stationary stochastic process has constant mean and variance over time and value of covariance between the two time periods depends only on the lag between time periods and not actual time at which covariance is computed.

• If these conditions are not met, variable is non-stationary time series

• A special type of stationary stochastic process is “Purely random” or “White noise”: with zero mean, constant variance and no serial correlation.

• This is the assumed nature of error term in CLRM• Most time series are non-stationary. Means,

variance and autocovariance change over time • OLS requires condition of stationarity and

therefore using OLS on non-stationary variables can lead to spurious results.

• We first need to establish variables to be stationary or convert non-stationary variables into stationary variables before undertaking OLS for time series

Non-stationary stochastic process• A common form of non-stationarity is

“random walk model” (RWM)• Pure random walk, Yt=Yt-1+ut

• Random walk with drift, Yt=b1+Yt-1+ut

• random walk with drift and deterministic trend, Yt=b1+b2t+Yt-1+ut

• All three processes are non-stationary but can be converted to statonarity

Pure random walk• In RWM without drift, value of Y at time 't' is equal to its

value in previous time period (t-1) plus a random shock (ut) which is white noise. Yt=Yt-1+ut

• Stock prices, exchange rates etc are thot to follow this process. Therefore impossible to predict

• Y1=Y0+u1

• Y2=Y1+u2=Y0+u1+u2

• Y3=Y2+u3=Y0+u1+u2+u3

• RWM without drift is therefore non-stationary

• First difference operator converts RWM without drift to stationary series

2

00

0

)va r(

)()(

tY

Yu+YE=YE

u+Y=Y

t

tt

tt

uY=YY tttt 1

• Non-stationarity in RWM with drift can also be eliminated by taking first difference. Such a series is integrated of order 1 Y~I(I). If variable has to be differenced twice to make it stationary its integrated of order 2 Y~I(2)...

• A stationary series is integrated of order 0.

Non Stationary variables and CLRMAssume Yt = C + bXt + Et• If Xt is non-stationary due to increasing variance

because of absence of asymptotic distribution theory where Var(Xt) is given by

• It shows if n increases then numerator would increase with higher ratio and standard error of would decrease and t-statistic would increase as:

and

Spurious Regression• If there are two variables and relationship

between them develops by chance then such regression is spurious.

• If dependent and independent variables are trend determined then CLRM would produce highly significant coefficients and very high value of R2 even if variables are not related to each other. So in that case relationship will be determined by chance.

Earlier Solution to non-Stationary Problem

• Earlier solution to spurious relation was to take first difference of all the variables and then estimate the relationship.

• Then people started to take log and difference which almost becomes growth rate as:

log(Yt/Yt-1) = log (Yt) – log (Yt-1)And growth rate is given by

(Yt – Yt-1) / Yt-1 = (Yt / Yt-1 )-1Where LHS is equal to RHS except 1 which is

ingnorable

Yt = B1 + B2Xt + Et

Yt-1 = B1 + B2Xt-1 + Et-1

∆Yt = B2∆Xt + Vt Where Vt = Et – Et-1

In difference form constant is equal to zero. If it is actually equal to zero then its fine but if it is not equal to zero then why to force it to be zero?

Problems with Differencing 1

Problems with Differencing 2Yt = B1 + B2Xt + Et

Yt-1 = B1 + B2Xt-1 + Et-1

∆Yt = B2∆Xt + Vt Where Vt = Et – Et-1

1. Vt is autocorrelated

Vt = Et – Et-1 , Vt-1 = Et-1 – Et-2

Et-1 = Vt-1 + Et-2 , Vt = Et – Vt-1 –Et-2

2. Vt is autocorrelated

Problems with Differencing 3Assume Yt = B0 + B1 XtThen one of the following situations may happen

a)Yt-1 = B0 + B1 Xt-1

b)Yt-1 > B0 + B1 Xt-1

c)Yt-1 < B0 + B1 Xt-1

• If (a) holds then ∆Yt = B1∆Xt (No Problem but it ignores disequilibrium information as it moves from equilibrium to equilibrium)

• If (b) holds then ∆Yt < B1∆Xt (We are estimating a change more than actual change)

• If (c) holds then ∆Yt > B1∆Xt (Actual change is greater than estimated change it means we are loosing some information as Yt depends not only on Xt rather previous values of X as well)

Problems with Differencing 4Assume ∆Yt = B2∆Xt

If Xt is constant(∆Xt=0) then sooner or later Yt will also become constant. In such case we will achieve the equilibrium (i.e. static equilibrium).

Assume ∆Yt = B1 + B2∆Xt

If ∆Xt=0 but B1 is not equal to zero then

∆Yt = B1 => Yt – Yt-1= B1 => Yt = B1 + Yt-1 Then there will be no static equilibrium and in

economics and finance most of the times we are concerned with static equilibrium. Thus simple differencing does not provide satisfactory solution to the problem of non-stationary variables

Error Correction Model (ECM)Assume

yt = C0 + C1Yt-1 + B0Xt + B1Xt-1 + et

Subtract yt-1 from both sides:

Yt- Yt-1 = C0 + C1Yt-1- Yt-1 + B0Xt + B1Xt-1+et

Then add B0Xt-1 and -B0Xt-1 to the right hand side:

∆Yt= B0- (1-C1)Yt-1 + B0Xt –B0Xt-1+ B0Xt-1 +B1Xt-1 + et

Rearrange terms

∆Yt= B0- (1-C1)Yt-1 + B0 ∆ Xt + (B0+B1)Xt-1 + et

Assume 1-C1 = ƛ, (B0/ ƛ) = b0, ((B0+B1)/ ƛ) = b1

∆Yt= ƛ b0- ƛ Yt-1 + B0 ∆ Xt + ƛ b1Xt-1 + et

∆Yt= B0 ∆ Xt + ƛ b0- ƛ Yt-1 + ƛ b1Xt-1 + et

∆Yt= B0 ∆ Xt - ƛ (Yt-1 -b0- b1Xt-1 )+ et

ECM with Two Variables Assume

yt = C0 + C1Yt-1 + B0Xt + B1Xt-1 + G0Zt + G1Zt-1 + et

Subtract yt-1 from both sides:

Yt- Yt-1 = C0 + C1Yt-1- Yt-1 + B0Xt + B1Xt-1 + G0Zt + G1Zt-1 +et

Then add and subtract B0Xt-1 and G0Zt-1 to the right hand side:

∆Yt= B0- (1-C1)Yt-1 + B0Xt –B0Xt-1+ B0Xt-1 +B1Xt-1 + G0Zt -G0Zt-1 + G0Zt-1 +G1Zt-1 + et

Rearrange terms

∆Yt= B0- (1-C1)Yt-1 +B0∆Xt +(B0+B1)Xt-1 +G0 ∆ Zt + (G0 +G1) Zt-1 + et

Assume 1-C1 = ƛ, (B0/ ƛ) = b0, ((B0+B1)/ ƛ) = b1 and ((G0+G1)/ ƛ) = g1

∆Yt= ƛ b0- ƛ Yt-1 + B0 ∆ Xt + ƛ b1Xt-1 +G0∆ Zt + ƛ g1Zt-1 + et

∆Yt= B0 ∆ Xt +G0∆ Zt - ƛ Yt-1 + ƛ b0+ ƛ b1Xt-1 + ƛ g1Zt-1 + et

∆Yt= B0 ∆ Xt - ƛ (Yt-1 -b0- b1Xt-1 - g1Zt-1)+ et

ECM versus First Differencing• In first differencing error term is auto

correlated but in ECM it is not.• In first differencing disequilibrium effect is

ignored but in ECM it is captured.• In first differencing constant term is

eliminated but in ECM it is included.

Advantages of ECM• It avoids problems of first differencing.• The ECM is formulated in the terms of stationary variables and the

disequilibrium terms. The problem of spurious regression is resolved as the trend is taken away by stationary values.

• ECM clearly differentiates between short run and long run effects so ECM can be used to test the economic theories as economic theory is very definite about long run but not about short run. Adding or dropping terms in difference form may change only the path to equilibrium but not the equilibrium in itself.

• ECM solves the problem of multicollinearity. In disequilibrium model the independent variables such as Xt andXt-1 are highly correlated but in ECM they are transformed to ∆Xt and Xt-1 which are not highly correlated.– There may be two reasons of low t-values (due to irrelevance of independent

variables and due to multicollinearity). If we know that variables are not collinear then we can use t values to decide whether the variables should be included or excluded from the model.

How to Estimate ECM? 1

Two Step ApproachEstimate the equation:

Yt = b0 + b1Xt + Ut

Then get Ut and lag it as Ut-1and then substitute in

∆Yt= B0 ∆ Xt - ƛ (Yt-1 -b0- b1Xt-1 )+ et As

∆Yt= B0 ∆ Xt - ƛ (Ut-1)+ etAnd estimate the above

How to Estimate ECM? 2

One Step ApproachEstimate the equation:

∆Yt= ƛ b0 + B0 ∆ Xt + ƛ b1Xt-1 - ƛ Yt-1+ et

Then divide ƛ b0 and ƛ b1 by ƛ to get b0 and b1

One step procedure is better than two step procedure.

Chow test for Parameter Stability• Suppose we have following equation

Yt = B0 + B1Xt + B2Zt + Et (Eq1: Restricted)

Suppose there is a breakpoint in this relation after n1 period i.e. one or more parameters, B0, B1, B2 change after n1 period but there is no further change.

To divide the sample into two periods we create a dummy variable D as D=0 for n1 and D=1 for n2 and estimate the following equation:

Yt = B0 + b0D +B1Xt + b1(XtD)+ B2Zt +b2(ZtD)+ Et (Eq2: Pooled)

In Eq2 For n1 Eq1 holds but for n2 Eq2 becomes as:

Yt =(b0+B0)+(b1+B1)Xt+(b2 +B2)Zt+Et (Eq3: Unrestricted)

If bs are zero then we get eq1 and parameters are stable but if bs are not equal to zero then parameters are not stable.