Lesson 8: Testing for IID Hypothesis with the correlogram · n should fall between the bounds 1p:96...

Lesson 8: Testing for IID Hypothesis withthe correlogram

Umberto Triacca

Dipartimento di Ingegneria e Scienze dell’Informazione e MatematicaUniversita dell’Aquila,

[email protected]

Umberto Triacca Lesson 8: Testing for IID Hypothesis with the correlogram

Testing for i.i.d. Hypothesis

Given a time series {x1, x2, ..., xT}, we want establish if it can beconsidered a realization of an i.i.d. process

xt ∼ i .i .d .(0, σ2)

An i.i.d. process is a sequence of independent and identicallydistributed (i.i.d.) random variables with zero mean and varianceσ2



We want to test the null hypothesis

H0 : ρk = 0



The decision rule could be:

Reject H0 if |ρk | > c

where c is a constant.

How do we can choose the constant c?


We can choose c such that

P(|ρk | > c |H0) = 0.05

Now, we have

P(|ρk | > c |H0) = 1− P(|ρk | ≤ c|H0) = 0.05

This implies that

P(|ρk | ≤ c |H0) = P(−c√T ≤

√T ρk ≤ c

√T)

= 0.95


Testing for i.i.d. Hypothesis with the correlogram

Ifxt ∼ i .i .d .(0, σ2)

then

√T ρk → N(0, 1)

This means that the standard normal distribution provides a goodapproximation to the true distribution of

√T ρk for large T .



It follows that

P(−c√T ≤

√T ρk ≤ c

√T)

= 0.95

if and only ifc√T = 1.96

and hence

c =1.96√T



Reject H0 if

|ρk | >1.96√T

that is if

ρk /∈[−1.96√

T,

1.96√T

]



If the data {x1, ..., xT} were really generated by an i.i.d. process,then about 95% of the sample autocorrelations ρ1, ρ2, ...ρn shouldfall between the bounds ±1.96√

T.

In other terms, if the considered process is i.i.d., we would expect5% of sample autocorrelations to lie outside the blue dashed lines.



For example if we calculate the first 40 values of ρk , then oneexpects only two values which fall outside these limits.



Consider the following time series



The correlogram for the data of this example is

We see that 2 of the first 40 values of ρk lie just outside thebounds ±1.96/

√T . As these occur not at relevant time lags, we

conclude that there is no evidence to reject the hypothesis that theobservations are independently distributed.



If we compute the sample autocorrelations up to lag 40 and findthat more than two or three values fall outside the bounds, or thatone value falls far outside the bounds, we reject the i.i.d.hypothesis.






We reject the i.i.d. hypothesis.






In this case, only one value of ρk lies outside the bounds±1.96/

√T . However, this occurs at lag 1, a relevant time lag.

Thus, we reject the i.i.d. hypothesis. As we will see, in this case,an MA(1) model could be appropriate.


Test of the random walk hypothesis for financial data

Are the prices of financial assets random walk?



The process {xt ; t ∈ Z} is a random walk if

xt = xt−1 + ut

where ut ∼ i .i .d .(0, σ2).The increments, or first differences of x , are independently andidentically distributed (i.i.d.). Thus the increments areunpredictable.



Usually, to investigate whether the data are RW, the firstdifference data

∆t = xt − xt−1

are used.The difference data should be i.i.d. (0, σ2) if the system is a RW.



We examine the logarithm of the daily close prices of IBM stockfrom 3 January 2000 to 1 October 2002.



The graph of the differenced (the returns) series is



The corellogram is given by

We accept the random walk hypothesis



Here, we consider monthly returns on Bank of New York stockfrom 1990.01 through 1998.12.




We accept the random walk hypothesis


Squared Returns

Consider the series of the squared returns




We conclude that the the squared returns are not i.i.d.



Whereas the sample autocorrelations of the returns are close tozero, the correlogram of the squared returns shows quite a differentpicture: the squared return seems significantly correlated.


Portmanteau testing for i.i.d. processes

In addition to assess the individual significance of sampleautocorrelogram, at a specific lag, the researchers are ofteninterested to the joint significance of a set of sampleautocorrelations.

H0 : ρk = 0 for k = 1, ...,K



If xt is an i.i.d. sequence with mean zero and finite variance, thenfor T large and K < T , the random variable

QK = TK∑

k=1

ρ2k

is approximately chi-square with K degrees of freedom.



Thus, the joint statistical significance of ρ1, . . . , ρK may be testedusing the Box-Pierce Portmanteau statistic

QK = TK∑

k=1

ρ2k

.

We reject the i.i.d. hypothesis

H0 : ρ1 = ρ2 = . . . = ρK = 0

at level α if QK > χ21−α,K , where χ2

1−α,K is the 1− α quantile ofthe chi-squared distribution with K degrees of freedom.The value K is chosen, somewhat arbitrarily, equal to 20.



A refinement of this test, formulated by Ljung and Box (1978), isobtained replacing QK with

QLBK = T (T + 2)

K∑k=1

ρ2k/(T − k)

whose distribution is better approximated by the chi-squareddistribution with K degrees of freedom. Large values of QLB

K leadto a rejection of the null hipothesis.


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Lesson 8: Testing for IID Hypothesis with the correlogram · n should fall between the bounds 1p:96...

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