ESGF Financial Models R Session

ESGF 5IFM Q1 2012
Financial Econometric Models
Vincent JEANNIN – ESGF 5IFM
Q1 2012

vinzjeannin@hotmail.com
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ESGF 5IFM Q1 2012
Summary of the session

• R Step by Step
• Summary of last session
• OLS & Autocorrelation

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R Step by Step
Downloadable for free (open source)

ESGF 4IFM Q1 2012
http://www.r-project.org/

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Main screen

vinzjeannin@hotmail.com ESGF 4IFM Q1 2012
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Menu: File / New Script

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Step 1, upload your data

Excel CSV file easy to import

ESGF 4IFM Q1 2012
Path C:UsersvinDesktop

Note: 4 columns with headers

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DATA<-read.csv(file="C:/Users/vin/Desktop/DataFile.csv",header=T)

Run your instruction(s)

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You can call variables anytime you want

ESGF 4IFM Q1 2012
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summary(DATA) Shows a quick summary of the distribution of all variables

SPX SPXr AMEXr AMEX
Min. : 86.43 Min. :-0.0666344 Min. : 97.6 Min. :-0.0883287
1st Qu.: 95.70 1st Qu.:-0.0069082 1st Qu.:104.7 1st Qu.:-0.0094580
Median :100.79 Median : 0.0010016 Median :108.8 Median : 0.0013007

ESGF 4IFM Q1 2012
Mean : 99.67 Mean : 0.0001249 Mean :109.4 Mean : 0.0005891
3rd Qu.:103.75 3rd Qu.: 0.0075235 3rd Qu.:114.1 3rd Qu.: 0.0102923
Max. :107.21 Max. : 0.0474068 Max. :123.5 Max. : 0.0710967

summary(DATA$SPX) Shows a quick summary of the distribution of one variable

Min. 1st Qu. Median Mean 3rd Qu. Max.
86.43 95.70 100.80 99.67 103.80 107.20

min(DATA)
Careful using the following instructions max(DATA)
> min(DATA)
[1] -0.08832874
This will consider DATA as one variable > max(DATA)
[1] 123.4793

> sd(DATA)
SPX SPXr AMEXr AMEX
4.92763551 0.01468776 6.03035318 0.01915489 10
Mean & SD > mean(DATA)
SPX SPXr AMEXr AMEX
9.967046e+01 1.249283e-04 1.093951e+02 5.890780e-04

Easy to show histogram

ESGF 4IFM Q1 2012
hist(DATA$SPXr, breaks=25, main="Distribution of SPXr", ylab="Freq", 11
xlab="SPXr", col="blue")

Obvious Excess Kurtosis

ESGF 4IFM Q1 2012
Obvious Asymmetry

Functions doesn’t exists directly in R…

However some VNP (Very Nice Programmer) built and shared add-in

Package Moments 12

Menu: Packages / Install Package(s)

ESGF 4IFM Q1 2012
• Choose whatever mirror (server) you want
• Usually France (Toulouse) is very good as it’s a
University Server with all the packages available

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ESGF 4IFM Q1 2012
Once installed, you can load them with the
following instructions:
require(moments)
library(moments)

New functions can now be used!

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> require(moments)
> library(moments)
> skewness(DATA)
SPX SPXr AMEXr AMEX
-0.6358029 -0.4178701 0.1876994 -0.2453693

ESGF 4IFM Q1 2012
> kurtosis(DATA)
SPX SPXr AMEXr AMEX
2.411177 5.671254 2.078366 5.770583

Btw, you can store any result in a variable

> Kur<-kurtosis(DATA$SPXr)
> Kur
[1] 5.671254

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Lost?

Call the help! help(kurtosis)

ESGF 4IFM Q1 2012
Reminds you the package

Syntax

Arguments definition

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Let’s store a few values
SPMean<-mean(DATA$SPXr)
SPSD<-sd(DATA$SPXr) Package Stats

Build a sequence, the x axis

ESGF 4IFM Q1 2012
x<-seq(from=SPMean-4*SPSD,to=SPMean+4*SPSD,length=500)

Build a normal density on these x

Y1<-dnorm(x,mean=SPMean,sd=SPSD) Package Stats

Display the histogram
hist(DATA$SPXr, breaks=25,main="S&P Returns / Normal Package graphics
Distribution",xlab="Returns",ylab="Occurences", col="blue")

Display on top of it the normal density

lines(x,y1,type="l",lwd=3,col="red") Package graphics 17

ESGF 4IFM Q1 2012
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Positive Excess Kurtosis & Negative Skew

Let’s build a spread Spd<-DATA$SPXr-DATA$AMEX

What is the mean?

ESGF 4IFM Q1 2012
Mean is linear �� + �� = �� + ��(��)

�� − �� = �� − ��(��)

Let’s verify

> mean(DATA$SPXr)-mean(DATA$AMEX)-mean(Spd)
[1] 0

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What is the standard deviation?

Is standard deviation linear?
NO!

ESGF 4IFM Q1 2012
VAR �� + �� = ��2 �� + ��2 �� + 2��(��, ��)

> (var(DATA$SPXr)+var(DATA$AMEX)-2*cov(DATA$SPXr,DATA$AMEX))^0.5

[1] 0.01019212
> sd(Spd)
[1] 0.01019212

Let’s show the implication in a proper manner

Let’s create a portfolio containing half of each stocks 20

Portf<-0.5*DATA$SPXr+0.5*DATA$AMEX

plot(sd(DATA$SPXr),mean(DATA$SPXr),col="blue",ylim=c(0,0.0008),xlim=c(0.012
,0.022),ylab="Return",xlab="Vol")

points(sd(DATA$AMEX),mean(DATA$AMEX),col="red")

ESGF 4IFM Q1 2012
points(sd(Portf),mean(Portf),col="green")

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The efficient frontier

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points(sd(0.1*DATA$SPXr+0.9*DATA$AMEX),mean(0.1*DATA$SPXr+0.9*DATA$AMEX),c
ol="green")

ol="green")

ESGF 4IFM Q1 2012
ol="green")

ol="green")

ol="green")

ol="green")

ol="green")

ol="green")
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plot(DATA$AMEX,DATA$SPXr)
abline(lm(DATA$AMEX~DATA$SPXr), col="blue")

ESGF 4IFM Q1 2012
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LM stands for Linear Models

> lm(DATA$AMEX~DATA$SPXr)

ESGF 4IFM Q1 2012
Call:
lm(formula = DATA$AMEX ~ DATA$SPXr)

Coefficients:
(Intercept) DATA$SPXr
0.0004505 1.1096287

�� = 1.1096�� + 0.04%

Will be used later for linear regression and hedging

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Do you remember what is the most platykurtic distribution in the nature?

Toss Head = Success = 1 / Tail = Failure = 0

ESGF 4IFM Q1 2012
100 toss… Else memory issue…

> require(moments)
Loading required package: moments
> library(moments)

> toss<-rbinom(100,1,0.5)
> mean(toss)
[1] 0.52
> kurtosis(toss)
[1] 1.006410
> kurtosis(toss)-3
[1] -1.993590
> hist(toss, breaks=10,main="Tossing a
coin 100 times",xlab="Result of the
trial",ylab="Occurence")
> sum(toss)
[1] 52
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Let’s test the fairness

Density of a binomial distribution

�� + 1 ! ℎ
�� = ℎ, �� = �� = �� (1 − ��)��
ℎ! ��!

ESGF 4IFM Q1 2012
Let’s plot this density with

ℎ = 52
�� = 48

�� = 100
N<-100
h<-52
t<-48
r<-seq(0,1,length=500)
y<-
(factorial(N+1)/(factorial(h)*factori
al(t)))*r^h*(1-r)^t
plot(r,y,type="l",col="red",main="Pro
bability density to have 52 head out
100 flips")
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If the probability between 45% and 55% is significant we’ll accept the fairness

ESGF 4IFM Q1 2012
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What do you think?

What is the problem with this coin?

Obvious fake! Assuming the probability of head is 0.7

Toss it! Head = Success = 1 / Tail = Failure = 0

ESGF 4IFM Q1 2012
100 toss

> require(moments)
Loading required package: moments
> library(moments)

> toss<-rbinom(100,1,0.7)
> mean(toss)
[1] 0.72
> kurtosis(toss)
[1] 1.960317
> kurtosis(toss)-3
[1] -1.039683
> hist(toss, breaks=10,main="Tossing a
coin 100 times",xlab="Result of the
trial",ylab="Occurence")
> sum(toss)
[1] 72
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Let’s test the fairness (assuming you don’t know it’s a trick)

If the probability between 45% and 55% is significant we’ll accept the fairness
N<-100
h<-72
t<-28
r<-seq(0.2,0.8,length=500)
y<-(factorial(N+1)/(factorial(h)*factorial(t)))*r^h*(1-r)^t

ESGF 4IFM Q1 2012
plot(r,y,type="l",col="red",main="Probability density or r given 72
head out 100 flips")

Trick coin!

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Summary of the last session
�� = �� = �� + ��

ESGF 5IFM Q1 2012
Residual ε

��

�� = �� 2 = �� − �� + �� 2

��=1 ��=1

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��

�� = �� 2 = �� − �� + �� 2 = �� − �� − �� 2

��=1 ��=1 ��=1

Quick high school reminder if necessary…

ESGF 5IFM Q1 2012
�� − �� − �� 2 = �� 2 − 2�� − 2�� + �� 2 �� 2 + 2�� + ��2

��
��

= −2�� + 2�� 2 + 2�� = 0 = −2�� + 2�� + 2�� = 0
��
��=1 ��=1

��

−�� + �� 2 + �� = 0 −�� + �� + �� = 0
��=1 ��=1
��

�� ∗ �� 2 + �� ∗ �� = �� ∗ �� + �� = ��
��=1 ��=1 ��=1 ��=1 ��=1
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��
Leads easily to the intercept
��

��

�� ∗ �� + �� = ��

ESGF 5IFM Q1 2012
��=1 ��=1

�� + �� = ��

�� + �� = ��

�� = �� − ��

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�� = �� − �� y = �� + �� − ��

y − �� = ��(�� − �� )

ESGF 5IFM Q1 2012
��
��
= −2�� + 2�� 2 + 2�� = 0 = −2�� + 2�� + 2�� = 0
��
��=1 ��=1

��

�� − �� − �� = 0 �� − �� − �� = 0
��=1 ��=1
��
��
�� − �� − �� + �� = 0
��=1
�� − �� + �� − �� = 0
��=1
��

�� (�� − �� − �� − �� ) = 0 (�� − ��) − ��(�� − �� ) = 0
��=1 ��=1
�� 34
�� ( �� − �� − �� − �� ) = 0
��=1

We have
��

�� (�� − �� − �� − �� ) = 0 and �� ( �� − �� − �� − �� ) = 0
��=1 ��=1

ESGF 5IFM Q1 2012
��

�� (�� − �� − �� − �� ) = �� ( �� − �� − �� − �� )
��=1 ��=1

��

�� (�� − �� − �� − �� ) − �� − �� − �� − �� =0
��=1 ��=1

��

(�� −�� )(�� − �� − �� − �� ) = 0
��=1

Finally…

��
��=1(�� −�� )(�� − ��) 35
�� = �� 2
��=1(�� −�� )

�� Covariance
��=1(�� − �� )(�� − ��)
�� = �� 2
��=1(�� − �� ) Variance

ESGF 5IFM Q1 2012
��
�� =
��2��

�� = �� − ��

�� Dispersion Regression
�� = ��2 =
�� Total Dispersion
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Residuals need to be analysed

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3
�� − �� − �� 3
�� = �� =
�� − �� 2 3/2

ESGF 5IFM Q1 2012
4
�� − �� − �� 4
�� = �� =
�� − �� 2 2

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Too
Many
Outliers!

ESGF 5IFM Q1 2012
There should be 2 max
To be normal

Fatter tails than the
normal distribution

Excess kurtosis obvious

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Fatter and longer tails

ESGF 5IFM Q1 2012
Resid<-resid(Reg)
ks.test(Resid, "pnorm")
Fatter tails

One-sample Kolmogorov-Smirnov test

data: Resid 40
D = 0.4889, p-value < 2.2e-16 Reject H0 (Normality)
alternative hypothesis: two-sided

OLS & Autocorrelation
New idea… No intercept

ESGF 5IFM Q1 2012
Only one parameters to estimate:
• Slope β

Minimising residuals

��

�� = �� 2 = �� − �� 2

��=1 ��=1

When E is minimal?

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When partial derivatives i.r.w. a is 0

��

�� = �� 2 = �� − �� 2

��=1 ��=1

Quick high school reminder if necessary…

ESGF 5IFM Q1 2012
�� − �� 2 = �� 2 − 2�� + ��2 �� 2

��
��

= −2�� + 2�� 2 = 0
��
��=1

��
��=1 ��
�� − �� 2 = 0 �� = �� 2
��=1 ��
��=1
��
��
�� ∗ �� =2 �� =
�� 2
��=1 ��=1
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Any better?

Simply… lm(Val$AMEX~Val$SPX-1)

Call:
lm(formula = Val$AMEX ~ Val$SPX - 1)

ESGF 5IFM Q1 2012
Coefficients:
Val$SPX
1.11

Let’s follow the same process

layout(matrix(1:4,2,2))
plot(lm(Val$AMEX~Val$SPX-1))

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Not much better…

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ks.test(resid(lm(Val$AMEX~Val$SPX-1)), "pnorm")


ESGF 5IFM Q1 2012
data: resid(lm(Val$AMEX ~ Val$SPX - 1))
D = 0.4887, p-value < 2.2e-16

H0 rejected

Not much better

It’s the way statistics are… You look for, but sometimes you don’t find!
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However you can now regress without intercept and that’s great!

The purpose was to see if the market as effect an effect on a particular stock

The dependence is obvious but residuals too volatile for any stable application

ESGF 5IFM Q1 2012
But attention!
We are looking for causation, not correlation!

Causation implies correlation

Reciprocity is not true!

DON’T BE FOOLED BY PRETTY NUMBERS
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Let prove this…

ESGF 5IFM Q1 2012
Perfect linear dependence

Excellent R-Squared
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Residuals are a white noise

What’s the problem then?

ESGF 5IFM Q1 2012
Do you really think fresh lemon reduces car fatalities?
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How to use OLS to make predictions?

ESGF 5IFM Q1 2012
Not possible with 2 random variables

Need to find a variable with any future value known
That would leave the randomness to only 1 variable

The time is easily predictable, isn't it?

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Spread<-
read.csv(file="C:/Users/Vin/Desktop/SparkSpd.csv",head=TRUE,sep=",")
plot(Spread$Nb,Spread$Spark, main="Spark Spread Aug07-Aug08",
xlab="Time", ylab="Dirty Spark Spread", col="red", type="l")

ESGF 5IFM Q1 2012
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Clear trend with seasonality

plot(Spread$Nb,Spread$Spark, main="Spark Spread Aug07-Aug08", xlab="Time",
ylab="Dirty Spark Spread", col="red")
TReg<-lm(Spread$Spark~Spread$Nb)
summary(TReg)
abline(TReg, col="blue")

Call:

ESGF 5IFM Q1 2012
lm(formula = Spread$Spark ~ Spread$Nb)

Residuals:
Min 1Q Median 3Q Max
-1.19775 -0.51046 0.01431 0.50554 1.44953

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.137e+01 8.873e-02 128.19 <2e-16 ***
Spread$Nb 2.173e-02 7.618e-04 28.53 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.6266 on 199 degrees of freedom
Multiple R-squared: 0.8035, Adjusted R-squared: 0.8025
F-statistic: 813.9 on 1 and 199 DF, p-value: < 2.2e-16

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eps<-resid(TReg)
ks.test(eps, "pnorm")
plot(TReg)

ESGF 5IFM Q1 2012
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data: eps
D = 0.1333, p-value = 0.001578

ESGF 5IFM Q1 2012
Normality rejected

Regression rejected

What would be the next step?

Mistake in methodology?
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What else could we regress?

lag.plot(Spread$Spark, 9, do.lines=FALSE)

ESGF 5IFM Q1 2012
56

What is this?

Lag Plots!

Maybe the series is auto-correlated to itself…

Seems the case with 1 to 6 lags

ESGF 5IFM Q1 2012
par(mfrow=c(2,1))
acf(Spread$Spark,20)
pacf(Spread$Spark,20)

This will show correlogram (ACF)

�� = ��(�� − ��−�� ) Correlation between pairs of values of {Yt},
separated by an interval of length k.

This will show the partial auto-correlation (PACF)
Correlation between the current value and the value k
periods ago, after controlling for observations at
intermediate lags (identifying intermediate lag effects) 57

ESGF 5IFM Q1 2012
ACF is decreasing slowly

Propagation of autocorrelation due to step 1
58
Main character of non stationary time series

Heteroscedasticity

Need some differentiation

The series has three components

• Trend

ESGF 5IFM Q1 2012
• Seasonality
• Residual

First order differentiation may be useful

plot(diff(Spread$Spark), type="l")

Stationary?

Seasonality? 59

No more seasonality

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Differentiation bring new horizons…

ESGF 5IFM Q1 2012
The whole point is to show you the methodology

Step by steps…

Sometimes unsuccessfully (bad results) 61

Let’s go back to the stage one of the OLS

Differentiation can happen before the OLS

NonLin<-

ESGF 5IFM Q1 2012
read.csv(file="C:/Users/Vinz/Desktop/ExExp.csv",head=TRUE,sep=",")
plot(NonLin$X,NonLin$VarEp)

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What do you suggest?

Let’s create a new variable

�� = ln(��)

ESGF 5IFM Q1 2012
Lin<-log(NonLin$VarEp)
plot(NonLin$X,Lin)

Magic!

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> lm(Lin~NonLin$X)

Call:
lm(formula = Lin ~ NonLin$X)
Coefficients: plot(lm(Lin~NonLin$X))

ESGF 5IFM Q1 2012
(Intercept) NonLin$X
-4.605 1.000

Do not hesitate to
transform

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Conclusion
R

ESGF 5IFM Q1 2012
OLS

Autocorrelation

Residuals

Normality

Heteroscedasticity

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