Computational Finance II: Time Series

K.Ensor

What is a time series? Anything observed

sequentially (by time?)

Returns, volatility, interest rates, exchange rates, bond yields, …

Hourly temperature, variations of the thickness of a wire as a function of length

Time in quarters

earn

ing

05

1015

Jan 60 Jan 64 Jan 68 Jan 72 Jan 76 Jan 80

Quarterly Earning per shar Johnson and Johnson

What is different? The observations

are not independent.

There is correlation from observation to observation.

Consider the log of the J&J series.

Is there correlation in the observations over time?

Time in quarters

log

earn

ing

01

2

Jan 60 Jan 64 Jan 68 Jan 72 Jan 76 Jan 80

Log Quarterly Earning per share Johnson and Johnson

What are our objectives? Understanding / Modeling Estimating summary measures (e.g.

mean) Prediction / Forecasting

If correlation is present between the observations then our typical approaches are not correct (assume iid samples).

Autocorrelation?

How would you determine or show correlation over time?

lagged 1S

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Lagged Scatterplots : xLagged Scatterplots : x

Autocorrelation Function In theory…

How to estimate this quantity?

Sample ACF and PACF Sample ACF – sample estimate of the

autocorrelation function. Substitute sample estimates of the covariance

between X(t) and X(t+h). Note: We do not have “n” pairs but “n-h” pairs.

Subsitute sample estimate of variance. Sample PACF – correlation between

observations X(t) and X(t+h) after removing the linear relationship of all observations in that fall between X(t) and X(t+h).

Summary Plots

Time in quarters

01

2

Jan 60 Jan 64 Jan 68 Jan 72 Jan 76 Jan 80

Log Quarterly Earnings for J&J

-1 0 1 2 3

05

1015

Histogram

Log Quarterly Earnings for J&J Lag

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0 5 10 15

-1.0

-0.5

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Lag

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Detrending by taking first difference.

Time in quarters

first

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eren

ce o

f log

ear

ning

-0.6

-0.4

-0.2

0.0

0.2

0.4

Apr 60 Apr 64 Apr 68 Apr 72 Apr 76 Apr 80

First Difference Log Quarterly Earning per share J&JY(t)=X(t) – X(t-1)

What happens to the trend?

Suppose X(t)=a+bt+Z(t)

Z(t) is a random variable.

Detrended J&J series: Autocorrelation?

lagged 1

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ies

1

-0.6 -0.4 -0.2 0.0 0.2 0.4

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Lagged Scatterplots : xLagged Scatterplots : x

Sumary Plots of Detrended J&J log earnings per share.

Time in quarters

-0.6

0.0

0.4

Apr 60 Apr 64 Apr 68 Apr 72 Apr 76 Apr 80

Detrended Log Quarterly Earnings for J&J

-0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6

010

2030

Histogram

Detrended Log Quarterly Earnings for J&J Lag

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F

0 5 10 15

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Lag

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F

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PACF

Removing Seasonal Trend – one way to proceed. Suppose Y(t)=g(t)+W(t) where

g(t)=g(t-s) where s is our “season” for all t. W(t) is again a new random variable

Form a new series U(t) by taking the “s” difference

U(t)=Y(t)-Y(t-s) =g(t)-g(t-s) + W(t)-W(t-s) =W(t)-W(t-s) again a random variable

Summary of Transformed J&J Series

Time in quarters

-0.2

-0.1

0.0

0.1

0.2

Apr 61 Jan 65 Oct 68 Jul 72 Apr 76 Jan 80

Log J&J After Removing Linear and Seasonal Trend

-0.2 -0.1 0.0 0.1 0.2 0.3

05

1015

Histogram

Log J&J After Removing Linear and Seasonal Trend Lag

AC

F

0 5 10 15

-1.0

-0.5

0.0

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Lag

AC

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0 5 10 15

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Summary of Transformations: X(t) = log (Q(t)) Y(t)=X(t)-X(t-1) = (1-B)X(t) U(t)= (1-B4)Y(t)

U(t)=(1-B4) (1-B)X(t)

What is the next step? U(t) is a time series process called a

moving average of order 1 (or possibly a MA(1) plus a seasonal MA(1)) U(t)=(t-1) + (t)

Proceed to estimate and then we can estimate summary information about the earnings per share as well as predict.

Time

5 10 15 20

1015

2025

Two Year Forecast and 95% Bounds for Johnson and Johnson Quarterly Earnings Per Share

Quarter

Ear

ning

s

Forecast of J&J series

Why does the autocorrelation matter when making inferences? Consider estimation of the mean of a

stationary series E[X(t)]= for all t

If X(1),…,X(n) are iid what is the sampling distribution of the estimator for , namely the sample mean?

Why does the autocorrelation matter? What if X(t) has the following

structure (autoregressive model of order 1 AR(1) ) X(t)- = (X(t-1)- ) + (t)

Then Corr(X(t),X(t+h))= |h| for all h And Var(X)= (1+ Var(X)/n

Comparing the samples size? Let m denote the

number of iid obs. Let n denote the

number of correlated obs.

Setting the variances equal and solving for m as a function of n yields m=n(1-

Let n=100, then m=5 iid obs.

If n=100 and then the equivalent number of iid observations is 1900.

For positive and negative (correlation of lag 1) the equivalent sample sizes are 33 and 300.

Why? Why does the autocorrelation make

such a big difference in our ability to estimate the mean?

The same arguments for other mean functions of the process or other functions of the process we want to estimate.

Summary Times series is correlated data,

sequentially observed. The autocorrelation is a measure of the

this correlation over the time lag. This dependence structure along with

proper assumptions allows us to forecast the future of the process.

Correct inference requires incorporating knowledge of the dependence structure.