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
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 2 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 3 / 39
chartSeries(AAPL$AAPL.Adjusted, theme="white")
0
20
40
60
80
100
120
$AAPLAAPL.Adjusted
[2005−01−03/2014−12−31]
Last 108.99538
...... 03 2005 ...... 02 2007 ...... 04 2010 ...... 02 2012 ......... 31 2014
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 4 / 39
AAPL.rtn=diff(log(AAPL$AAPL.Adjusted))
hist(AAPL.rtn)
Histogram of AAPL.rtn
AAPL.rtn
Fre
quen
cy
−0.20 −0.15 −0.10 −0.05 0.00 0.05 0.10 0.15
020
040
060
080
010
00
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 5 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 6 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 7 / 39
U.S. Government Bonds: Treasury Bills, Treasury Notes, Treasury Bonds
Treasury bills (T-Bills) mature in one year or less. They do not payinterest 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 (3months), 182 days (6 months), and 364 days (1 year).
Treasury Notes (T-Notes) matures in 1-10 years. They have a couponpayment every 6 months and face value of $1, 000
The 10-year T-Notes has become the security most frequently quotedwhen discussing U.S. government bond market.
Treasury Bonds (T-Bonds) have longer maturities ranging from 20 to30 years. Thay have coupon payment every 6 months.
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 8 / 39
Bond Yields and Prices
Current Yield= Annual interest paid in dollarsMarket price of the bond × 100%
For zero-coupon bonds, Current Yield= ( Face valuePurchase price )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 thebond paid a coupon rate 5% per annum, then the current yield of thebond is (0.05 ∗ 100)/90 ∗ 100% = 5.56%.
The current yield does not consider the time value of money since itdoes not consider the present value of the coupon payments theinvestor will receive in the future.
A more commonly used measurement of bond investment is the Yieldto Maturity (YTM).
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 9 / 39
Yield to Maturity (YTM)
YTM is the yield obtained by equating the bond price to the presentvalue of all future payments.Suppose that the bond holder will receive k payments betweenpurchase and maturity. Let y and P be the YTM and price of thebond respectively. Then
P =C1
1 + y+
C2
(1 + y)2+ ...+
Ck + F
(1 + y)k
where F denotes the face value and Ci us the ith cash flow of couponpayment.Suppose that the coupon rate is α per annum, the number ofpayments is m per year, and the time to maturity is n year. Thencash flow of coupon payment is Fα/m and the number of paymentsis 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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 10 / 39
The table below shows some results between bond price and YTMassuming that F = $100 and coupon rate is 5% per annum payablesemiannually, and time to maturity is 3 years.
Yield to Maturity (%) Semiannual Rat (% ) Bond price ($)
6 3.0 97.297 3.5 94.678 4.0 92.149 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.
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 11 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 12 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 13 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 14 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 15 / 39
Asset Returns:Let Pt be the price of an asset at time t, and assume nodividend.
One-period simple returnGross return: 1 + Rt = Pt
Pt−1
Simple return: Rt = PtPt−1− 1.
Multiperiod simple returnGross return: 1 + Rt(k) = Pt
Pt−k= Pt
Pt−1× Pt−1
Pt−2× ...× Pt−k+1
Pt−k
The k-period Simple net return: Rt(k) = PtPt−k− 1.
Time interval is important, default is one year.
Annualized (average) return:Annualized [Rt(k)] = [
∏k−1j=0 (1 + Rt−j)]1/k − 1
Annualized [Rt(k)] ≈ 1k
∑k−1j=0 Rt−j
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 16 / 39
Continuously compounding: Let C be the initial capital, r be the interestannum, 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( Pt
Pt−1) = pt − pt−1
Multiperiod log returnrt(k) = ln(1 + Rt(k)) = ln[ Pt
Pt−1× Pt−1
Pt−2× · · · × Pt−k+1
Pt−k]
= rt + rt−1 + · · ·+ rt−k+1
Portfolio return with N assets and weights w :Rp,t =
∑Ni=1 wiRit
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 17 / 39
Dividend (Dt): Rt = Pt+DtPt−1
− 1,rt = ln(Pt + Dt)− ln(Pt−1)
Excess return: Zt = Rt − R0t , zt = rt − r0t , where r0t denoted the logreturn of a risk-free interest rate.
rt = ln(1 + Rt), Rt = ert − 1.
If the returns are in percentage, then rt = 100× ln(1 + Rt100 ),
Rt = [exp( rt100 )− 1]× 100.
Aggregation of the returns1 + Rt(k) = (1 + Rt)(1 + Rt−1) · · · (1 + Rt−k+1)rt(k) = rt + rt−1 + · · ·+ rt−k+1
These two relations are important in practice, for example, obtainannual returns from monthly returns.
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 18 / 39
Distributional properties of returns: What is the distribution of rit?
Moment of a random variable X with density f (x): r -th momentE (X r ) =
∫∞−∞ x r f (x)dx .
First moment: mean of X .
r -th central moment: E ((X − µX )r ) =∫∞−∞(x − µX )r f (x)dx
Second central moment: variance of X .
Skewness (symmetry): S(x) = E [ (X−µx )3
σ3x
].
Kurtosis (fat-tails) K (x) = E [ (X−µx )4
σ4x
], 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 orlong financial positions and in risk management. Moreover, highkurtosis implies heavy (or long) tails in distribution.
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 19 / 39
Estimation: Data {x1, ..., xT}.Sample mean: µx = 1
T
∑Tt=1 xt .
Sample variance: σ2x = 1
T−1
∑Tt=1(xt − µx)2.
Sample skewness: S(x) = 1(T−1)σ3
x
∑Tt=1(xt − µx)3.
Sample kurtosis: K (x) = 1(T−1)σ4
x
∑Tt=1(xt − µx)4.
Assume that xt is normally distributed, S(x) ∼ N(0, 6T ) and
K (x)− 3 ∼ N(0, 24T ).
Test for symmetry: S∗ = S(x)√6/T∼ N(0, 1) under normality. Decision
rule: Reject H0 of a symmetric distribution if |S∗| > Zα/2 or p-valueis less than α.
Test for tail thickness: K ∗ = K(x)−3√24/T
∼ N(0, 1) under normality.
Decision rule: Reject H0 of a symmetric distribution if |K ∗| > Zα/2 orp-value is less than α.
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 20 / 39
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 leftwith 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+ σ2
2 ), V (Y ) = exp(2µ+ σ2)[exp(σ2)− 1].
If Y is lognormal with mean µy and variance σ2y , then X = ln(Y ) is
normal with E (X ) = ln[µy√
1+σ2y
µ2y
], V (Y ) = ln[1 +σ2y
µ2y
].
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 21 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 22 / 39
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
−0.
2−
0.1
0.0
0.1
0.2
0.3
Index
ibm
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 23 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 24 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 25 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 26 / 39
hist(ibm,nclass=40) # Obtain histogram of IBM simple returns.
Histogram of ibm
ibm
Fre
quen
cy
−0.3 −0.2 −0.1 0.0 0.1 0.2 0.3
010
2030
4050
6070
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 27 / 39
d1=density(libm) # Compute density function of ibm log returns
plot(d1$x,d1$y,type='l') # Plot the sample density of log returns
mu=mean(libm); s1 = sd(libm) # compute the sample mean and standard deviation of IBM log returns.
x=seq(-0.4,0.4,0.01) # create a sequence of real numbers from -0.4 to 0.4 with increment 0.01.
y=dnorm(x,mean=mu,sd=s1) # obtain normal density with mean mu and standard deviation s1.
lines(x,y,lty=2) # impose a dashed line on the density plot for comparison with normal density.
lines(x,y,col="red") #will plot a red curve.
−0.3 −0.2 −0.1 0.0 0.1 0.2 0.3
01
23
45
6
d1$x
d1$y
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 28 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 29 / 39
Dependence: Consider two random variables X and Y , correlationcoefficient: ρ = Cov(X ,Y )
std(X )std(Y ) .
Kendall’s tau: Let (X , Y ) be an random copy of (X ,Y ).ρτ = 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 thecumulative distribution of X and Y . ρS = ρ(Fx(X ),Fy (Y )).
Spearman’s rho is the correlation of probability-transformed variables.It is the correlation coefficient of the ranks of the data.
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 30 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 31 / 39
Basic concepts of Time Series Models
Stationarity:- Strict: distributions are time-invariant.- Weak: first 2 moments (mean, variance and covariance) aretime-invariant.
Sample mean and sample variance are used to estimate the mean andvariance.
Lag-k autocovariance: let rt be log return,γk = Cov(rt , tt−k) = E [(rt − µ)(rt−k − µ)].
Autocorrelations: ρl =cov(rt ,rt−l )
var(rt). Note that ρ0 = 1 and ρk = ρ−k for
k 6= 0.
Existence of serial correlations implies that the return is predictableand indicating market inefficiency.
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 32 / 39
Basic concepts of Time Series Models
Sample autocorrelation function (ACF): ρl =∑T−l
t=1 (rt−r)(rt+l−r)∑T−lt=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= 0t = ρ1√
1/T=√T ρ1, which is asymptotically N(0, 1).
Decision rule: Reject H0 if |t| > Zα/2 or p-value is less than α.- Joint test (Ljung-Box statistics) :H0 : ρ1 = · · · = ρm = 0 v.s.H1 : ρi 6= 0
Q(m) = T (T + 2)∑m
l=1ρ2l
T−l , which is asymptotically χ2m.
Decision rule: Reject H0 if Q(m) > χ2m(α) or p-value is less than α.
Significant sample ACF does not necessarily imply market inefficiency.
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 33 / 39
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 34 / 39
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−
0.1
0.0
0.1
0.2
0.3
Monthly IBM returns
year
ibm
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 35 / 39
acf(ibm,lag=20) # specify the number of ACF to compute
0 5 10 15 20
0.0
0.2
0.4
0.6
0.8
1.0
Lag
AC
F
Series ibm
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 36 / 39
pacf(ibm,lag=20)
5 10 15 20
−0.
050.
000.
050.
10
Lag
Par
tial A
CF
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
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 37 / 39
Lag operator: Lrt = rt−1, L2rt = rt−2.
Available data at time t, Ft−1 ≡ {r1, r2, · · · , rt−1}.The return rt is decomposed into two parts given information Ft−1:rt = µt + at = E (rt |Ft−1) + σtεt .
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 .
εt is an i .i .d . sequence with mean zero and variance 1.
σt is conditional standard deviation. (volatility)
Model for µt : mean equation; Model for σ2t : volatility equation.
Two purposes for time series models:- A model for µt .- Understanding models for σ2
t . (properties, forcasting, etc.)
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 38 / 39
Linear time series
rt is linear if the predictable part is a linear function of Ft−1 and atsare i .i .d . distributed.
That is, rt can be written as rt = µ+∑∞
i=0 ψiat−i , where µ is aconstant, ψ0 = 1 and at is an i .i .d . sequence with mean zero andfinite variance.
at is shock or innovation at time t and ψi are the impulse responsesof rt .
Sheng-Kai Chang (NTU) Econ 7933: Financial Time Series Fall, 2015 39 / 39