Econ 7933: Financial Time Series: Introduction

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Econ 7933: Financial Time Series: Introduction

Sheng-Kai Chang (NTU)

Fall, 2015

Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 1 / 39

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Examples of Fnancial time series

Daily log returns of Apple stock: 2005 to 2014.

U.S. monthly unemployment rates

U.S. daily interesrate rate of 10-year Treasury Notes The VIX index

Exchange rate between US Dollar vs Euro

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setwd("D:/R/FTS/Data") library(quantmod) library(downloader)

getSymbols("AAPL",from="2005-01-01",to="2014-12-31")

## [1] "AAPL"

chartSeries(AAPL,theme="white")

100 200 300 400 500 600 700

AAPL [2005−01−03/2014−12−31]

Last 110.379997

Volume (millions):

41,403,400

0 200 400 600 800

... 03 2005 ... 02 2007 ... 04 2010 ... 02 2012 ... 31 2014

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chartSeries(AAPL$AAPL.Adjusted,theme="white")

0 20 40 60 80 100 120

$ AAPL AAPL.Adjusted

[2005−01−03/2014−12−31]

Last 108.99538

... 03 2005 ... 02 2007 ... 04 2010 ... 02 2012 ... 31 2014

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AAPL.rtn=diff(log(AAPL$AAPL.Adjusted)) hist(AAPL.rtn)

Histogram of AAPL.rtn

AAPL.rtn

Frequency

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02004006008001000

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chartSeries(AAPL.rtn, theme="white")

−0.20

−0.15

−0.10

−0.05 0.00 0.05 0.10

AAPL.rtn [2005−01−04/2014−12−31]

Last −0.0192020203673824

... 04 2005 ... 02 2007 ... 04 2010 ... 02 2012 ... 31 2014

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getSymbols("UNRATE",src="FRED")

## [1] "UNRATE"

chartSeries(UNRATE,theme="white")

4 6 8 10 UNRATE [1948−01−01/2015−08−01]

Last 5.1

... 1948 ... 1960 ... 1975 ... 1990 ... 2005

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U.S. Government Bonds: Treasury Bills, Treasury Notes, Treasury Bonds Treasury bills (T-Bills) mature in one year or less. They do not pay interest prior to maturity and are sold at a discount of the face value (par value).

The commonly used maturities are 28 days (1 months), 91 days (3 months), 182 days (6 months), and 364 days (1 year).

Treasury Notes (T-Notes) matures in 1-10 years. They have a coupon payment every 6 months and face value of $1, 000

The 10-year T-Notes has become the security most frequently quoted when discussing U.S. government bond market.

Treasury Bonds (T-Bonds) have longer maturities ranging from 20 to 30 years. Thay have coupon payment every 6 months.

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Bond Yields and Prices

Current Yield= Annual interest paid in dollars

Market price of the bond × 100%

For zero-coupon bonds, Current Yield= (Purchase price^{Face value} )^1/k− 1, where k is time to maturity in years.

EX: if an investor paid $90 for a bond with face value of $100 and the bond paid a coupon rate 5% per annum, then the current yield of the bond is (0.05 ∗ 100)/90 ∗ 100% = 5.56%.

The current yield does not consider the time value of money since it does not consider the present value of the coupon payments the investor will receive in the future.

A more commonly used measurement of bond investment is the Yield to Maturity (YTM).

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Yield to Maturity (YTM)

YTM is the yield obtained by equating the bond price to the present value of all future payments.

Suppose that the bond holder will receive k payments between purchase and maturity. Let y and P be the YTM and price of the bond respectively. Then

P = C1

1 + y + C2

(1 + y )² + ... + C_k+ F (1 + y )^k

where F denotes the face value and Ci us the i th cash flow of coupon payment.

Suppose that the coupon rate is α per annum, the number of payments is m per year, and the time to maturity is n year. Then cash flow of coupon payment is F α/m and the number of payments is k = mn.

P = αF

m [ 1

1 + y + 1

(1 + y )² + ... + 1

(1 + y )^k] + F (1 + y )^k

= αF

my[1 − 1

(1 + y )^k] + F (1 + y )^k

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The table below shows some results between bond price and YTM assuming that F = $100 and coupon rate is 5% per annum payable semiannually, and time to maturity is 3 years.

Yield to Maturity (%) Semiannual Rat (% ) Bond price ($)

6 3.0 97.29

7 3.5 94.67

8 4.0 92.14

9 4.5 89.68

10 5.0 87.31

YTM is inversely proportional to the bond price.

In practice, we observe bond price, so that YTM can be calculated.

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getSymbols("^TNX",from="2007-01-01",to="2015-6-30")

## [1] "TNX"

chartSeries(TNX,theme="white")

2 3 4 5

TNX [2007−01−03/2015−06−30]

Last 2.335

Volume (100s):

0

−1.0

−0.5 0.0 0.5 1.0

... 03 2007 ... 02 2009 ... 03 2011 ... 02 2013 ... 02 2015

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TNX.rtn=diff(log(TNX$TNX.Adjusted)) chartSeries(TNX.rtn,theme="white")

−0.15

−0.10

−0.05 0.00 0.05 0.10

TNX.rtn [2007−01−04/2015−06−30]

Last 0.00171453106724051

... 04 2007 ... 02 2009 ... 03 2011 ... 02 2013 ... 02 2015

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getSymbols("^VIX",from="2007-01-01",to="2015-6-30")

## [1] "VIX"

chartSeries(VIX,theme="white")

20 40 60 80

VIX [2007−01−03/2015−06−30]

Last 18.23

Volume (100,000s):

0

0 5 10 15 20

... 03 2007 ... 02 2009 ... 03 2011 ... 02 2013 ... 02 2015

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getSymbols("DEXUSEU",src="FRED")

## [1] "DEXUSEU"

chartSeries(DEXUSEU,theme="white")

0.8 1.0 1.2 1.4 1.6 DEXUSEU [1999−01−04/2015−09−11]

Last 1.1338

... 04 1999 ... 02 2003 ... 02 2007 ... 03 2011 ... 02 2015

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Asset Returns:Let P_t be the price of an asset at time t, and assume no dividend.

One-period simple return Gross return: 1 + Rt= _P^P^t

t−1

Simple return: Rt = _P^P^t

t−1 − 1.

Multiperiod simple return Gross return: 1 + Rt(k) = _P^P^t

t−k = _P^P^t

t−1 ×^P_P^t−1

t−2 × ... ×^P_P^t−k+1

t−k

The k-period Simple net return: R_t(k) = _P^P^t

t−k − 1.

Time interval is important, default is one year.

Annualized (average) return:

Annualized [Rt(k)] = [Qk−1

j =0(1 + Rt−j)]^1/k− 1 Annualized [R_t(k)] ≈ _k¹P_k−1

j =0 R_t−j

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Continuously compounding: Let C be the initial capital, r be the interest annum, and n be the number of years.

A = C × exp[r × n]

Present value: C = A × exp[−r × n]

Continuously compounding (log) return:

rt= ln(1 + Rt) = ln(_P^P^t

t−1) = pt− p_t−1 Multiperiod log return

rt(k) = ln(1 + Rt(k)) = ln[_P^P^t

t−1 ×^P_P^t−1

t−2 × · · · ×^P_P^t−k+1

t−k ]

= r_t+ r_t−1+ · · · + r_t−k+1

Portfolio return with N assets and weights w : R_p,t =PN

i =1w_iR_it

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Dividend (Dt): Rt = ^P_P^t^+D^t

t−1 − 1,r_t = ln(Pt+ Dt) − ln(Pt−1)

Excess return: Zt = Rt− R_0t, zt = rt− r_0t, where r0t denoted the log return of a risk-free interest rate.

rt= ln(1 + Rt), Rt= e^r^t − 1.

If the returns are in percentage, then rt = 100 × ln(1 +₁₀₀^R^t ), Rt= [exp(₁₀₀^r^t ) − 1] × 100.

Aggregation of the returns

1 + R_t(k) = (1 + R_t)(1 + R_t−1) · · · (1 + R_t−k+1) r_t(k) = r_t+ r_t−1+ · · · + r_t−k+1

These two relations are important in practice, for example, obtain annual returns from monthly returns.

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Distributional properties of returns: What is the distribution of r_it? Moment of a random variable X with density f (x ): r -th moment E (X^r) =R∞

−∞x^rf (x )dx . First moment: mean of X .

r -th central moment: E ((X − µ_X)^r) =R∞

−∞(x − µ_X)^rf (x )dx Second central moment: variance of X .

Skewness (symmetry): S (x ) = E [^{(X −µ}_σ3^x⁾³

x ].

Kurtosis (fat-tails) K (x ) = E [^{(X −µ}_σ4^x⁾⁴

x ], K (x ) − 3 is excess kurtosis.

The mean and variance are related to long-term retrun and risk, respectively. Symmetry has important implications in holding short or long financial positions and in risk management. Moreover, high kurtosis implies heavy (or long) tails in distribution.

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Estimation: Data {x1, ..., xT}.

Sample mean: ˆµx = _T¹ PT t=1xt. Sample variance: ˆσ²_x = _{T −1}¹ PT

t=1(xt− ˆµx)². Sample skewness: ˆS (x ) =_{(T −1)ˆ}¹ _σ3

x

PT

t=1(xt− ˆµx)³. Sample kurtosis: ˆK (x ) = _{(T −1)ˆ}¹ _σ4

x

PT

t=1(xt− ˆµx)⁴.

Assume that x_t is normally distributed, ˆS (x ) ∼ N(0,_T⁶) and K (x ) − 3 ∼ N(0,ˆ ²⁴_T).

Test for symmetry: S^∗ = √^{S(x )}^ˆ

6/T ∼ N(0, 1) under normality. Decision rule: Reject H0 of a symmetric distribution if |S^∗| > Z_α/2 or p-value is less than α.

Test for tail thickness: K^∗= ^{K (x )−3}√^ˆ

24/T ∼ N(0, 1) under normality.

Decision rule: Reject H₀ of a symmetric distribution if |K^∗| > Z_α/2 or p-value is less than α.

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A joint test (Jarque-Bera test): JB = (S^∗)²+ (K^∗)² ∼ χ²₂ under normality. Decision rule: Reject H₀ of normality if JB > χ²₂(α) or p-value is less than α.

Empirical distribution of asset returns tend to be skewed to the left with heavy tails and has a higher peak than normal distribution.

Normal and lognormal: Y is lognormal if X = ln(Y ) is normal.

If X ∼ N(µ, σ²), then Y = exp(X ) is lognormal with E (Y ) = exp(µ +^σ₂²), V (Y ) = exp(2µ + σ²)[exp(σ²) − 1].

If Y is lognormal with mean µ_y and variance σ²_y, then X = ln(Y ) is normal with E (X ) = ln[s^µ^y

1+^σ2^y

µ2y

], V (Y ) = ln[1 + ^σ

2y

µ²_y].

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library(fBasics)

#da=read.table("d-ibm3dx7008.txt",header=T) # Load data with header into R da=read.table("m-ibm6708.txt",header=T) # Load data with header into R da[1,]

## date ibm sprtn

## 1 19670331 0.048837 0.03941

dim(da) # Check dimension of the data (row = sample size, col = variables)

## [1] 502 3

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ibm=da[,2] # Select the simple returns of IBM stock stored in Column 2.

plot(ibm,type='l')# Plot the simple returns.

0 100 200 300 400 500

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Index

ibm

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basicStats(ibm) # Compute the descriptive statistics of simple returns.

## ibm

## nobs 502.000000

## NAs 0.000000

## Minimum -0.261905

## Maximum 0.353799

## 1. Quartile -0.036942

## 3. Quartile 0.049636

## Mean 0.008865

## Median 0.005274

## Sum 4.450050

## SE Mean 0.003262

## LCL Mean 0.002456

## UCL Mean 0.015273

## Variance 0.005341

## Stdev 0.073085

## Skewness 0.263914

## Kurtosis 1.822178

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libm=log(ibm+1) # Compute the IBM log returns

basicStats(libm) # Compute descriptive statistics of log returns.

## libm

## nobs 502.000000

## NAs 0.000000

## Minimum -0.303683

## Maximum 0.302915

## 1. Quartile -0.037641

## 3. Quartile 0.048443

## Mean 0.006208

## Median 0.005260

## Sum 3.116457

## SE Mean 0.003237

## LCL Mean -0.000151

## UCL Mean 0.012567

## Variance 0.005259

## Stdev 0.072517

## Skewness -0.135343

## Kurtosis 1.693092

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t.test(ibm) # Perform t-test for mean being zero.

##

## One Sample t-test

##

## data: ibm

## t = 2.7176, df = 501, p-value = 0.006804

## alternative hypothesis: true mean is not equal to 0

## 95 percent confidence interval:

## 0.002455871 0.015273412

## sample estimates:

## mean of x

## 0.008864641

t.test(ibm,alternative=c("greater"))# Perform one-sided test.

##

## One Sample t-test

##

## data: ibm

## t = 2.7176, df = 501, p-value = 0.003402

## alternative hypothesis: true mean is greater than 0

## 95 percent confidence interval:

## 0.003489285 Inf

## sample estimates:

## mean of x

## 0.008864641

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hist(ibm,nclass=40) # Obtain histogram of IBM simple returns.

Histogram of ibm

ibm

Frequency

−0.3 −0.2 −0.1 0.0 0.1 0.2 0.3

010203040506070

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d1=density(libm) # Compute density function of ibm log returns plot(d1$x,d1$y,type='l') # Plot the sample density of log returns

mu=mean(libm); s1=sd(libm)# compute the sample mean and standard deviation of IBM log returns.

x=seq(-0.4,0.4,0.01)# create a sequence of real numbers from -0.4 to 0.4 with increment 0.01.

y=dnorm(x,mean=mu,sd=s1)# obtain normal density with mean mu and standard deviation s1.

lines(x,y,lty=2)# impose a dashed line on the density plot for comparison with normal density.

lines(x,y,col="red")#will plot a red curve.

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0123456

d1$x

d1$y

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normalTest(libm,method='jb') # Perform normality test.

##

## Title:

## Jarque - Bera Normalality Test

##

## Test Results:

## STATISTIC:

## X-squared: 62.8363

## P VALUE:

## Asymptotic p Value: 2.265e-14

##

## Description:

## Thu Sep 17 21:08:52 2015 by user: Sheng-Kai

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Dependence: Consider two random variables X and Y , correlation coefficient: ρ = Cov (X ,Y )

std (X )std (Y ).

Kendall’s tau: Let ( ˜X , ˜Y ) be an random copy of (X , Y ).

ρ_τ = P[(X − ˜X )(Y − ˜Y ) > 0] − P[(X − ˜X )(Y − ˜Y ) < 0] = E [sign[(X − ˜X )(Y − ˜Y )]].

Kendall’s tau quantifies the probability of concordant over discordant.

Here concordant means (X − ˜X )(Y − ˜Y ) > 0.

Spearman’s rho: rank correlation. Let Fx(x ) and Fy(y ) be the cumulative distribution of X and Y . ρ_S = ρ(F_x(X ), F_y(Y )).

Spearman’s rho is the correlation of probability-transformed variables.

It is the correlation coefficient of the ranks of the data.

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x= rnorm(1000) # Generate 1000 N(0,1) random numbers cor(x,x)

## [1] 1 cor(x,exp(x))

## [1] 0.7510654

cor(x,exp(x),method="kendall")

## [1] 1

cor(x,exp(x),method="spearman")

## [1] 1

ibm=da[,2]

sp=da[,3]

cor(ibm,sp)

## [1] 0.5949041

cor(ibm,sp,method="kendall")

## [1] 0.4223351

cor(ibm,sp,method="spearman")

## [1] 0.5897587

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Basic concepts of Time Series Models Stationarity:

- Strict: distributions are time-invariant.

- Weak: first 2 moments (mean, variance and covariance) are time-invariant.

Sample mean and sample variance are used to estimate the mean and variance.

Lag-k autocovariance: let rt be log return, γ_k = Cov (rt, t_t−k) = E [(rt− µ)(r_t−k− µ)].

Autocorrelations: ρ_l = ^{cov (r}_{var (r}^t^,r^t−l⁾

t) . Note that ρ₀ = 1 and ρ_k = ρ−k for k 6= 0.

Existence of serial correlations implies that the return is predictable and indicating market inefficiency.

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Basic concepts of Time Series Models

Sample autocorrelation function (ACF): ˆρ_l =

PT −l

t=1(rt−¯r )(r_t+l−¯r ) PT −l

t=1(rt−¯r )² , where

¯

r is the sample mean and T is the sample size.

Test zero serial corellation (market efficiency) - Individual test: H0: ρ1 = 0 v.s. H1 : ρ1 6= 0 t = √^ρ^ˆ¹

1/T =√

T ˆρ₁, which is asymptotically N(0, 1).

Decision rule: Reject H0 if |t| > Z_α/2 or p-value is less than α.

- Joint test (Ljung-Box statistics) :H₀: ρ₁ = · · · = ρ_m = 0 v.s.

H₁: ρ_i 6= 0

Q(m) = T (T + 2)Pm l =1

ˆ ρ²_l

T −l, which is asymptotically χ²_m.

Decision rule: Reject H0 if Q(m) > χ²_m(α) or p-value is less than α.

Significant sample ACF does not necessarily imply market inefficiency.

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library(timeSeries)

par(mfcol=c(2,1)) # Put two plots on a sinlge page. c(2,1) means two rows and one column.

ts.plot(ibm) ts.plot(sp)

Time

ibm

0 100 200 300 400 500

−0.20.00.2

Time

sp

0 100 200 300 400 500

−0.20.00.1

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tdx=(2+c(1:502))/12+1967# create a calendar time index. The data start at March 1967 so that the index is from 3:505 par(mfcol=c(1,1))# return to one plot per page

plot(tdx,ibm,type='l',xlab='year',ylab='ibm',main="Monthly IBM returns")

1970 1980 1990 2000 2010

−0.2−0.10.00.10.20.3

Monthly IBM returns

year

ibm

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acf(ibm,lag=20) # specify the number of ACF to compute

0 5 10 15 20

0.00.20.40.60.81.0

Lag

ACF

Series ibm

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pacf(ibm,lag=20)

5 10 15 20

−0.050.000.050.10

Lag

Partial ACF

Series ibm

Box.test(ibm,lag=10,type="Ljung")# Compute Ljung-Box Q(m) statistics

##

## Box-Ljung test

##

## data: ibm

## X-squared = 6.9444, df = 10, p-value = 0.7307

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Lag operator: Lrt= rt−1, L²rt = rt−2.

Available data at time t, Ft−1≡ {r₁, r2, · · · , rt−1}.

The return r_t is decomposed into two parts given information F_t−1: rt= µt+ at= E (rt|F_t−1) + σtt.

µt is conditional mean of rt. It is the function of elements of Ft−1, and it is predictable part of rt.

at is shock or innovation at time t, it is unpredictable part of rt.

_t is an i .i .d . sequence with mean zero and variance 1.

σ_t is conditional standard deviation. (volatility)

Model for µ_t: mean equation; Model for σ_t²: volatility equation.

Two purposes for time series models:

- A model for µ_t.

- Understanding models for σ_t². (properties, forcasting, etc.)

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Linear time series

r_t is linear if the predictable part is a linear function of F_t−1and a_ts are i .i .d . distributed.

That is, r_t can be written as r_t = µ +P∞

i =0ψ_ia_t−i, where µ is a constant, ψ0 = 1 and at is an i .i .d . sequence with mean zero and finite variance.

at is shock or innovation at time t and ψi are the impulse responses of r_t.