Financial Econometric Models I

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 (est 3h)

• Introduction & Objectives
• Bibliography
• OLS & Exploration

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Introduction & Objectives
• What is a model? = + with being a white noise

ESGF 5IFM Q1 2012
• What the point writing models?

Describe data behaviour

Modelise data behaviour
Forecast data behaviour

• Acquire theory knowledge on Econometrics & Statistics
• Step by step from OLS to ANOVA on residuals
• Usage of R and Excel 3

Bibliography

vinzjeannin@hotmail.com ESGF 5IFM Q1 2012
4

OLS & Exploration
OLS: Ordinary Least Square

ESGF 5IFM Q1 2012
Linear regression model
Minimize the sum of the square vertical distances
between the observations and the linear
approximation

= = +

Residual ε

5

Two parameters to estimate:
• Intercept α
• Slope β

ESGF 5IFM Q1 2012
Minimising residuals

= 2 = − + 2

=1 =1

When E is minimal?

When partial derivatives i.r.w. a and b are 0
6

= 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

∗ + =
=1 =1

ESGF 5IFM Q1 2012
+ =

+ =

= −

The regression line is going through ( , )

The distance of this point to the line is 0 indeed

8

= − 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
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( − − − ) = 0
=1

We have

( − − − ) = 0 and ( − − − ) = 0
=1 =1

ESGF 5IFM Q1 2012

( − − − ) = ( − − − )
=1 =1

( − − − ) − − − − =0
=1 =1

( − )( − − − ) = 0
=1

Finally…

=1( − )( − ) 10
= 2
=1( − )

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

ESGF 5IFM Q1 2012

=
2

= −

You can use Excel function INTERCEPT and SLOPE

11

Calculate the Variances and Covariance of X{1,2,3,3,1,2} and Y{2,3,1,1,3,2}

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You can use Excel function VAR.P, COVARIANCE.P and STDEV.P

Let’s asses the quality of the regression

Let’s calculate the correlation coefficient (aka Pearson Product-Moment
Correlation Coefficient – PPMCC):

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= Value between -1 and 1

= 1

Perfect dependence

~0 No dependence

Give an idea of the dispersion of the scatterplot
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You can use Excel function CORREL

Poor quality
R=0.62
R=0.96

High quality

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What is good quality?

ESGF 5IFM Q1 2012
Slightly discretionary…

If
3
≥ = 0.8666 …
2
It’s largely admitted as the threshold for acceptable / poor

15

The regression itself introduces a bias

Let’s introduce the coefficient of determination R-Squared

ESGF 5IFM Q1 2012
Total Dispersion = Dispersion Regression + Dispersion Residual

2 2 2
− = − + −

Dispersion Regression
2 =
Total Dispersion

In other words the part of the total dispersion explained by the regression 16

You can use Excel function RSQ

In a simple linear regression with intercept 2 = 2

ESGF 5IFM Q1 2012
Is a good correlation coefficient and a good coefficient of
determination enough to accept the regression?

Not necessarily!

Residuals need to have no effect, in other word to be a white noise!

17

18

Don’t get fooled by numbers!

ESGF 5IFM Q1 2012
For every dataset of the Quarter

= 9
= 7.5

= 3 + 0.5
= 0.82
2 = 0.67

Can you say at this stage which regression is the best?
19

Certainly not those on the right you need a LINEAR dependence

ESGF 5IFM Q1 2012
Is any linear regression useless?

Think what you could do to the series

Polynomial transformation, log transformation,…

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Else, non linear regressions, but it’s another story

First application on financial market

S&P / AmEx in 2011

ESGF 5IFM Q1 2012
21

,&
= = 0.8501
&

2 = 2 = 0.7227

ESGF 5IFM Q1 2012
Oups :-o
Is Excel wrong?

R-Squared has different calculation methods

Let’s accept the following regression then as the quality seems pretty good

= 0.06% + 1.1046 ∗ &

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How to use this?

ESGF 5IFM Q1 2012
• Forecasting? Not really…
Both are random variables

• Hedging? Yes but basis risk
Yes but careful to the residuals…

In theory, what is the daily result of the hedge?

Let’s have a try!

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Hedging $1.0M of AmEx Stocks with $1.1046M of S&P

ESGF 5IFM Q1 2012
It would have been too easy… Great differences… Why?

Sensitivity to the size of the sample
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Heteroscedasticity

Let’s have a similar approach using a proper statistics and econometrics software

ESGF 5IFM Q1 2012
• Free
• Open Source
• Developments shared by developers

Let’s begin with statistical exploration to get familiar with the series
and the software
> Val<-read.csv(file="C:/Users/Vinz/Desktop/Val.csv",head=TRUE,sep=",")
> summary(Val)

SPX AMEX
Min. :-0.0666344 Min. :-0.0883287
1st Qu.:-0.0069082 1st Qu.:-0.0094580
Median : 0.0010016 Median : 0.0013007 25
Mean : 0.0001249 Mean : 0.0005891
3rd Qu.: 0.0075235 3rd Qu.: 0.0102923
Max. : 0.0474068 Max. : 0.0710967

> hist(Val$AMEX, breaks=20, main="Distribution
AMEX Returns")
> sd(Val$AMEX)
[1] 0.01915489

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> hist(Val$SPX, breaks=20, main="Distribution
SPXX Returns")
> sd(Val$SPX)
[1] 0.01468776 26

These are obvious negatively skewed distributions

ESGF 5IFM Q1 2012
Reminders
3
− − 3
= =
− 2 3/2

• Negative skew: long left tail, mass on the right, skew to the left
• Positive skew: long right tail, mass on the left, skew to the right

> skewness(Val$AMEX)
[1] -0.2453693
> skewness(Val$SPX) 27
[1] -0.4178701

These are obvious leptokurtic distributions

ESGF 5IFM Q1 2012
Reminders

4
− − 4
= =
− 2 2

> library(moments)
> kurtosis(Val$AMEX) What is their K?
[1] 5.770583 (excess kurtosis)
> kurtosis(Val$SPX)
[1] 5.671254 28
Subtract 3 to make it relative to the
normal distribution…

Quick check: what are the Skewness and Kurtosis of {1,2,-3,0,-2,1,1}?

ESGF 4IFM Q1 2012
Excel function SKEW
R function skewness (package moments)
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ESGF 4IFM Q1 2012
Excel function KURT
R function kurtosis (package moments)
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By the way, what is the most platykurtic distribution in the nature?

Toss it!

ESGF 4IFM Q1 2012
Head = Success = 1 / Tail = Failure = 0

> require(moments)
> library(moments)
> toss<-rbinom(10000000,1,0.5)
> mean(toss)
[1] 0.5001777
> kurtosis(toss)
[1] 1.000001
> kurtosis(toss)-3
[1] -1.999999
> hist(toss, breaks=10,main="Tossing a
coin 10 millions times",xlab="Result
of the trial",ylab="Occurence") 31
> sum(toss)
[1] 5001777

50.01777% rate of success: fair or not fair? Trick coin ?

Can be tested later with a Bayesian approach

ESGF 4IFM Q1 2012
On a perfect 50/50, Kurtosis would be 1, Excess Kurtosis -2: the minimum!
This is a Bernoulli trial

(, ) with > 1 and 0 < < 1 ∈ ℝ and integer

Mean

SD (1 − )

Skewness 1 − 2
(1 − )

Kurtosis 1
−3
(1 − )
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Easy to demonstrate if p=0.5 the Kurtosis will be the lowest
Bit more complicated to demonstrate it for any distribution

Back to our series, a good tool is the BoxPlot

ESGF 5IFM Q1 2012
Too
Many
Outliers!

There should be 2 max
To be normal

Fatter tails than the
normal distribution

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boxplot(Val$AMEX,Val$SPX, main="AMEX & S&P BoxPlots",
names=c("AMEX","SPX"),col="blue")

Leptokurtic distributions

Negatively skewed distribution

ESGF 5IFM Q1 2012
Are they normal distributions?

Let’s compare them to normal distributions with same
standard deviation and mean and make the QQ Plots

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x=seq(-0.2,0.2,length=200)
y1=dnorm(x,mean=mean(Val$AMEX),sd=sd(
Val$AMEX))
hist(Val$AMEX, breaks=100,main="AmEx
Returns / Normal

ESGF 5IFM Q1 2012
Distribution",xlab="Return",ylab="Occ
urence")
lines(x,y1,type="l",lwd=3,col="red")

x=seq(-0.2,0.2,length=200)
y1=dnorm(x,mean=mean(Val$SPX),sd=sd(Val$S
PX))
hist(Val$SPX, breaks=20,main="S&P Returns
/ Normal
Distribution",xlab="Return",ylab="Occuren
ce")
lines(x,y1,type="l",lwd=3,col="red") 35

ESGF 5IFM Q1 2012
Excess kurtosis obvious

Fatter and longer tails

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Let’s have a look to their CDF through QQPlot

> qqnorm(Val$AMEX) > qqnorm(Val$SPX)
> qqline(Val$AMEX) > qqline(Val$SPX)

ESGF 5IFM Q1 2012
Fatter tails 37
Let’s properly test the normality

Can use many tests…

• Kolmogorov-Smirnov
• Jarque Bera
• Chi Square
•

ESGF 5IFM Q1 2012
Shapiro Wilk

Let’s try Kolmogorov-Smirnov

It compares the distance between the empirical

CDF and the CFD of the reference distribution

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ESGF 5IFM Q1 2012
x=seq(-4,4,length=1000)
plot(ecdf(Val$AMEX),do.points=FALSE, col="red", lwd=3,
main="Normal Distribution against AMEX - CFD's", xlab="x",
ylab="P(X<=x)")
lines(x,pnorm(x,mean=mean(Val$AMEX),sd=sd(Val$AMEX)),col="blue",t
ype="l",lwd=3)

x=seq(-4,4,length=1000)
plot(ecdf(Val$SPX),do.points=FALSE, col="red", lwd=3,
main="Normal Distribution against S&P - CFD's", xlab="x",
ylab="P(X<=x)")
lines(x,pnorm(x,mean=mean(Val$SPX),sd=sd(Val$SPX)),col="blue",typ
e="l",lwd=3)

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> ks.test(Val$SPX, "pnorm") > ks.test(Val$AMEX, "pnorm")

One-sample Kolmogorov- One-sample Kolmogorov-Smirnov
Smirnov test test

data: Val$SPX data: Val$AMEX
D = 0.4811, p-value < 2.2e-16 D = 0.4742, p-value < 2.2e-16
alternative hypothesis: two-sided alternative hypothesis: two-sided

ESGF 5IFM Q1 2012
The 0 hypothesis is the distribution is normal

Do we accept or reject the hypothesis 0 with a 95%
confidence interval?

The hypothesis regarding the distributional
form is rejected if the test statistic, D, is greater
than the critical value obtained from a table

40

1.36
Sample size: 251 = 0.086
251

Rejected or not? 41

P-Value was giving
Rejected! Series aren’t fitting a normal distribution
the answer

Ok, we now know a bit more the 2 series we want to regress
> lm(Val$AMEX~Val$SPX)

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

ESGF 5IFM Q1 2012
Coefficients:
(Intercept) Val$SPX
0.0004505 1.1096287

plot(Val$SPX,Val$AMEX, main="S&P / AmEx", xlab="S&P", ylab="AmEx",
col="red")

abline(lm(Val$AMEX~Val$SPX), col="blue")

= 110.96% ∗ + 0.045%

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The next important step is no analyse the residuals

> Reg<-lm(Val$AMEX~Val$SPX)
> summary(Reg)

ESGF 5IFM Q1 2012
Call:
lm(formula = Val$AMEX ~ Val$SPX)

Residuals:
Min 1Q Median 3Q Max
-0.030387 -0.006072 -0.000114 0.006624 0.027824

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.0004505 0.0006365 0.708 0.48
Val$SPX 1.1096287 0.0434231 25.554 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’
1

Residual standard error: 0.01008 on 249 degrees of freedom
Multiple R-squared: 0.7239, Adjusted R-squared: 0.7228
F-statistic: 653 on 1 and 249 DF, p-value: < 2.2e-16

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They need to be a white noise, you can have a first assessment with quartiles

plot(Reg)
layout(matrix(1:4,2,2))

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QQ Plot compares the CDF

ESGF 5IFM Q1 2012
A perfect fit is a line

Left tail noticeably different

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ESGF 5IFM Q1 2012
Residuals should be randomly distributed around the 0 horizontal line

You don’t want to see a trend, a dependence

To accept or reject the regression you need residuals to be a white noise

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Their mean should be 0

ESGF 5IFM Q1 2012
Nothing suggesting a white noise

• Square root of the standardized residuals as a function of the
fitted values
• There should be no obvious trend in this plot

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Showing now leverage

Marginal importance of a point in the regression

ESGF 5IFM Q1 2012
Far points suggest outlier or poor model

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So do we accept the regression?

Probably not… But let’s check…
Kolmogorov-Smirnov on residuals

ESGF 5IFM Q1 2012
1.36 Higher bound value for the
= = 0.086
251 H0 to be accepted

Resid<-resid(Reg)
ks.test(Resid, "pnorm")

One-sample Kolmogorov-Smirnov test

data: Resid
D = 0.4889, p-value < 2.2e-16
alternative hypothesis: two-sided

Rejected! Regression between 2 different asset are very often poor
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Heteroscedasticity

Basis risk if you hedge anyway

Conclusion

ESGF 5IFM Q1 2012
OLS

Residuals

Normality

Heteroscedasticity

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