financial econometric models i
DESCRIPTION
Financial Econometric Models, course I, Busines School MSc levelTRANSCRIPT
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Financial Econometric Models Vincent JEANNIN – ESGF 5IFM
Q1 2012
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Summary of the session (est 3h) • Introduction & Objectives • Bibliography • OLS & Exploration
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Introduction & Objectives
• What is a model?
• What the point writing models?
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Describe data behaviour
Modelise data behaviour
• Acquire theory knowledge on Econometrics & Statistics
• Step by step from OLS to ANOVA on residuals
• Usage of R and Excel
Forecast data behaviour
𝑂𝑏𝑠 = 𝑀𝑜𝑑𝑒𝑙 + 𝜀 with 𝜀 being a white noise
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Bibliography
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OLS & Exploration
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Linear regression model
Minimize the sum of the square vertical distances between the observations and the linear approximation
𝑦 = 𝑓 𝑥 = 𝑎𝑥 + 𝑏
Residual ε
OLS: Ordinary Least Square
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Two parameters to estimate: • Intercept α • Slope β
Minimising residuals
𝐸 = 𝜀𝑖2
𝑛
𝑖=1
= 𝑦𝑖 − 𝑎𝑥𝑖 + 𝑏 2
𝑛
𝑖=1
When E is minimal?
When partial derivatives i.r.w. a and b are 0
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𝐸 = 𝜀𝑖2
𝑛
𝑖=1
= 𝑦𝑖 − 𝑎𝑥𝑖 + 𝑏 2
𝑛
𝑖=1
= 𝑦𝑖 − 𝑎𝑥𝑖 − 𝑏 2
𝑛
𝑖=1
𝜕𝐸
𝜕𝑎= −2𝑥𝑖𝑦𝑖 + 2𝑎𝑥𝑖
2 + 2𝑏𝑥𝑖
𝑛
𝑖=1
= 0
𝑦𝑖 − 𝑎𝑥𝑖 − 𝑏 2 = 𝑦𝑖2 − 2𝑎𝑥𝑖𝑦𝑖 − 2𝑏𝑦𝑖 + 𝑎2𝑥𝑖
2 + 2𝑎𝑏𝑥𝑖 + 𝑏2
Quick high school reminder if necessary…
−𝑥𝑖𝑦𝑖 + 𝑎𝑥𝑖2 + 𝑏𝑥𝑖
𝑛
𝑖=1
= 0
𝑎 ∗ 𝑥𝑖2
𝑛
𝑖=1
+ 𝑏 ∗ 𝑥𝑖
𝑛
𝑖=1
= 𝑥𝑖𝑦𝑖
𝑛
𝑖=1
𝜕𝐸
𝜕𝑏= −2𝑦𝑖 + 2𝑏 + 2𝑎𝑥𝑖
𝑛
𝑖=1
= 0
−𝑦𝑖 + 𝑏 + 𝑎𝑥𝑖
𝑛
𝑖=1
= 0
𝑎 ∗ 𝑥𝑖
𝑛
𝑖=1
+ 𝑛𝑏 = 𝑦𝑖
𝑛
𝑖=1
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𝑎 ∗ 𝑥𝑖
𝑛
𝑖=1
+ 𝑛𝑏 = 𝑦𝑖
𝑛
𝑖=1
Leads easily to the intercept
𝑎𝑛𝑥 + 𝑛𝑏 = 𝑛𝑦
𝑎𝑥 + 𝑏 = 𝑦
The regression line is going through (𝑥 , 𝑦 )
The distance of this point to the line is 0 indeed
𝜕𝐸
𝜕𝑏
𝑏 = 𝑦 − 𝑎𝑥
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𝜕𝐸
𝜕𝑎= −2𝑥𝑖𝑦𝑖 + 2𝑎𝑥𝑖
2 + 2𝑏𝑥𝑖
𝑛
𝑖=1
= 0
y = 𝑎𝑥 + 𝑦 − 𝑎𝑥
y − 𝑦 = 𝑎(𝑥 − 𝑥 )
𝑏 = 𝑦 − 𝑎𝑥
𝑥𝑖 𝑦𝑖 − 𝑎𝑥𝑖 − 𝑏 = 0
𝑛
𝑖=1
𝜕𝐸
𝜕𝑏= −2𝑦𝑖 + 2𝑏 + 2𝑎𝑥𝑖 = 0
𝑛
𝑖=1
𝑦𝑖 − 𝑏 − 𝑎𝑥𝑖
𝑛
𝑖=1
= 0
𝑦𝑖 − 𝑦 + 𝑎𝑥 − 𝑎𝑥𝑖 = 0
𝑛
𝑖=1
(𝑦𝑖 − 𝑦 ) − 𝑎(𝑥𝑖 − 𝑥 )
𝑛
𝑖=1
= 0
𝑥𝑖 𝑦𝑖 − 𝑎𝑥𝑖 − 𝑦 + 𝑎𝑥 = 0
𝑛
𝑖=1
𝑥𝑖(𝑦𝑖 − 𝑦 − 𝑎 𝑥𝑖 − 𝑥 )
𝑛
𝑖=1
= 0
𝑥 ( 𝑦𝑖 − 𝑦 − 𝑎 𝑥𝑖 − 𝑥 )
𝑛
𝑖=1
= 0
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𝑥𝑖(𝑦𝑖 − 𝑦 − 𝑎 𝑥𝑖 − 𝑥 )
𝑛
𝑖=1
= 0 𝑥 ( 𝑦𝑖 − 𝑦 − 𝑎 𝑥𝑖 − 𝑥 )
𝑛
𝑖=1
= 0
𝑥𝑖(𝑦𝑖 − 𝑦 − 𝑎 𝑥𝑖 − 𝑥 )
𝑛
𝑖=1
= 𝑥 ( 𝑦𝑖 − 𝑦 − 𝑎 𝑥𝑖 − 𝑥 )
𝑛
𝑖=1
𝑥𝑖(𝑦𝑖 − 𝑦 − 𝑎 𝑥𝑖 − 𝑥 )
𝑛
𝑖=1
− 𝑥 𝑦𝑖 − 𝑦 − 𝑎 𝑥𝑖 − 𝑥
𝑛
𝑖=1
= 0
(𝑥𝑖−𝑥 )(𝑦𝑖 − 𝑦 − 𝑎 𝑥𝑖 − 𝑥 )
𝑛
𝑖=1
= 0
𝑎 = (𝑥𝑖−𝑥 )(𝑦𝑖 − 𝑦 )𝑛
𝑖=1
(𝑥𝑖−𝑥 )2 𝑛𝑖=1
Finally…
We have
and
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𝑎 = (𝑥𝑖 − 𝑥 )(𝑦𝑖 − 𝑦 )𝑛
𝑖=1
(𝑥𝑖 − 𝑥 )2𝑛𝑖=1
Covariance
Variance
𝑎 =𝐶𝑜𝑣𝑥𝑦
𝜎2𝑥
𝑏 = 𝑦 − 𝑎 𝑥
You can use Excel function INTERCEPT and SLOPE
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Calculate the Variances and Covariance of X{1,2,3,3,1,2} and Y{2,3,1,1,3,2}
You can use Excel function VAR.P, COVARIANCE.P and STDEV.P
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Let’s asses the quality of the regression
Let’s calculate the correlation coefficient (aka Pearson Product-Moment Correlation Coefficient – PPMCC):
𝑟 =𝐶𝑜𝑣𝑥𝑦
𝜎𝑥𝜎𝑦 Value between -1 and 1
𝑟 = 1 Perfect dependence
𝑟 ~0 No dependence
Give an idea of the dispersion of the scatterplot
You can use Excel function CORREL
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R=0.96
High quality
R=0.62
Poor quality
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What is good quality?
Slightly discretionary…
𝑟 ≥3
2= 0.8666…
If
It’s largely admitted as the threshold for acceptable / poor
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The regression itself introduces a bias
Let’s introduce the coefficient of determination R-Squared
Total Dispersion = Dispersion Regression + Dispersion Residual
Dispersion Regression
Total Dispersion 𝑅2 =
In other words the part of the total dispersion explained by the regression
𝑦𝑖 − 𝑦 2 = 𝑦𝑖 − 𝑦𝑖 2 + 𝑦𝑖 − 𝑦 2
You can use Excel function RSQ
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In a simple linear regression with intercept 𝑅2 = 𝑟2
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!
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𝑦 = 7.5
𝑥 = 9
𝑦 = 3 + 0.5𝑥
𝑟 = 0.82
𝑅2 = 0.67
Don’t get fooled by numbers!
For every dataset of the Quarter
Can you say at this stage which regression is the best?
Certainly not those on the right you need a LINEAR dependence
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Is any linear regression useless?
Think what you could do to the series
Polynomial transformation, log transformation,…
Else, non linear regressions, but it’s another story
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First application on financial market
S&P / AmEx in 2011
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𝑅𝐴𝑚𝑒𝑥 = 0.06% + 1.1046 ∗ 𝑅𝑆&𝑃
𝑟 =𝐶𝑜𝑣𝐴𝑚𝐸𝑥,𝑆&𝑃
𝜎𝐴𝑚𝐸𝑥𝜎𝑆&𝑃= 0.8501
𝑅2 = 𝑟2 = 0.7227
Oups :-o
Is Excel wrong?
R-Squared has different calculation methods
Let’s accept the following regression then as the quality seems pretty good
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How to use this?
• Forecasting? Not really… Both are random variables
• Hedging? Yes but basis risk Yes but careful to the residuals…
Let’s have a try!
In theory, what is the daily result of the hedge? 𝑎
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Hedging $1.0M of AmEx Stocks with $1.1046M of S&P
It would have been too easy… Great differences… Why?
Sensitivity to the size of the sample
Heteroscedasticity
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Let’s have a similar approach using a proper statistics and econometrics software
• Free • Open Source • Developments shared by developers
> Val<-read.csv(file="C:/Users/Vinz/Desktop/Val.csv",head=TRUE,sep=",")
> summary(Val)
SPX AMEX
Min. :-0.0666344 Min. :-0.0883287
1st Qu.:-0.0069082 1st Qu.:-0.0094580
Median : 0.0010016 Median : 0.0013007
Mean : 0.0001249 Mean : 0.0005891
3rd Qu.: 0.0075235 3rd Qu.: 0.0102923
Max. : 0.0474068 Max. : 0.0710967
Let’s begin with statistical exploration to get familiar with the series and the software
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> hist(Val$AMEX, breaks=20, main="Distribution
AMEX Returns")
> sd(Val$AMEX)
[1] 0.01915489
> hist(Val$SPX, breaks=20, main="Distribution
SPXX Returns")
> sd(Val$SPX)
[1] 0.01468776
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These are obvious negatively skewed distributions
> skewness(Val$AMEX)
[1] -0.2453693
> skewness(Val$SPX)
[1] -0.4178701
Reminders
• Negative skew: long left tail, mass on the right, skew to the left • Positive skew: long right tail, mass on the left, skew to the right
𝑆𝐾𝐸𝑊 𝑋 = 𝐸𝑋 − 𝑋
𝜎
3
=𝐸 𝑋 − 𝑋 3
𝐸 𝑋 − 𝑋 2 3/2
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These are obvious leptokurtic distributions
> library(moments)
> kurtosis(Val$AMEX)
[1] 5.770583
> kurtosis(Val$SPX)
[1] 5.671254
Reminders
What is their K? (excess kurtosis)
Subtract 3 to make it relative to the normal distribution…
𝐾𝑈𝑅𝑇 𝑋 = 𝐸𝑋 − 𝑋
𝜎
4
=𝐸 𝑋 − 𝑋 4
𝐸 𝑋 − 𝑋 2 2
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Excel function SKEW
R function skewness (package moments)
Quick check: what are the Skewness and Kurtosis of {1,2,-3,0,-2,1,1}?
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Excel function KURT
R function kurtosis (package moments)
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By the way, what is the most platykurtic distribution in the nature?
Toss it!
Head = Success = 1 / Tail = Failure = 0
> require(moments)
> library(moments)
> toss<-rbinom(10000000,1,0.5)
> mean(toss)
[1] 0.5001777
> kurtosis(toss)
[1] 1.000001
> kurtosis(toss)-3
[1] -1.999999
> hist(toss, breaks=10,main="Tossing a
coin 10 millions times",xlab="Result
of the trial",ylab="Occurence")
> sum(toss)
[1] 5001777
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50.01777% rate of success: fair or not fair? Trick coin ?
On a perfect 50/50, Kurtosis would be 1, Excess Kurtosis -2: the minimum!
This is a Bernoulli trial
𝐵(𝑛, 𝑝)
𝑝 Mean
SD 𝑝(1 − 𝑝)
Skewness 1 − 2𝑝
𝑝(1 − 𝑝)
Kurtosis 1
𝑝(1 − 𝑝)− 3
Easy to demonstrate if p=0.5 the Kurtosis will be the lowest Bit more complicated to demonstrate it for any distribution
Can be tested later with a Bayesian approach
𝑛 > 1 0 < 𝑝 < 1 with and 𝑝 ∈ ℝ and 𝑛 integer
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Back to our series, a good tool is the BoxPlot
boxplot(Val$AMEX,Val$SPX, main="AMEX & S&P BoxPlots",
names=c("AMEX","SPX"),col="blue")
Too Many Outliers!
There should be 2 max To be normal
Fatter tails than the normal distribution
![Page 34: Financial Econometric Models I](https://reader031.vdocument.in/reader031/viewer/2022020101/556876b5d8b42a3b7b8b4e0e/html5/thumbnails/34.jpg)
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Leptokurtic distributions
Negatively skewed distribution
Are they normal distributions?
Let’s compare them to normal distributions with same standard deviation and mean and make the QQ Plots
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x=seq(-0.2,0.2,length=200)
y1=dnorm(x,mean=mean(Val$AMEX),sd=sd(
Val$AMEX))
hist(Val$AMEX, breaks=100,main="AmEx
Returns / Normal
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$SPX, breaks=20,main="S&P Returns
/ Normal
Distribution",xlab="Return",ylab="Occuren
ce")
lines(x,y1,type="l",lwd=3,col="red")
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Excess kurtosis obvious
Fatter and longer tails
Let’s have a look to their CDF through QQPlot
![Page 37: Financial Econometric Models I](https://reader031.vdocument.in/reader031/viewer/2022020101/556876b5d8b42a3b7b8b4e0e/html5/thumbnails/37.jpg)
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Let’s properly test the normality
> qqnorm(Val$AMEX)
> qqline(Val$AMEX)
> qqnorm(Val$SPX)
> qqline(Val$SPX)
Fatter tails
![Page 38: Financial Econometric Models I](https://reader031.vdocument.in/reader031/viewer/2022020101/556876b5d8b42a3b7b8b4e0e/html5/thumbnails/38.jpg)
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Can use many tests…
• Kolmogorov-Smirnov • Jarque Bera • Chi Square • Shapiro Wilk
Let’s try Kolmogorov-Smirnov
It compares the distance between the empirical CDF and the CFD of the reference distribution
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x=seq(-4,4,length=1000)
plot(ecdf(Val$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$SPX)),col="blue",typ
e="l",lwd=3)
![Page 40: Financial Econometric Models I](https://reader031.vdocument.in/reader031/viewer/2022020101/556876b5d8b42a3b7b8b4e0e/html5/thumbnails/40.jpg)
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> ks.test(Val$SPX, "pnorm")
One-sample Kolmogorov-
Smirnov test
data: Val$SPX
D = 0.4811, p-value < 2.2e-16
alternative hypothesis: two-sided
> ks.test(Val$AMEX, "pnorm")
One-sample Kolmogorov-Smirnov
test
data: Val$AMEX
D = 0.4742, p-value < 2.2e-16
alternative hypothesis: two-sided
The 0 hypothesis is the distribution is normal
Do we accept or reject the hypothesis 0 with a 95% confidence interval?
The hypothesis regarding the distributional form is rejected if the test statistic, D, is greater than the critical value obtained from a table
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Sample size: 251 1.36
251= 0.086
Rejected or not?
Rejected! Series aren’t fitting a normal distribution P-Value was giving the answer
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Ok, we now know a bit more the 2 series we want to regress
> lm(Val$AMEX~Val$SPX)
Call:
lm(formula = Val$AMEX ~ Val$SPX)
Coefficients:
(Intercept) Val$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")
𝑦 = 110.96% ∗ 𝑥 + 0.045%
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> Reg<-lm(Val$AMEX~Val$SPX)
> summary(Reg)
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
The next important step is no analyse the residuals
They need to be a white noise, you can have a first assessment with quartiles
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layout(matrix(1:4,2,2))
plot(Reg)
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QQ Plot compares the CDF
A perfect fit is a line
Left tail noticeably different
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Residuals should be randomly distributed around the 0 horizontal line
To accept or reject the regression you need residuals to be a white noise
Their mean should be 0
You don’t want to see a trend, a dependence
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• Square root of the standardized residuals as a function of the fitted values
• There should be no obvious trend in this plot
Nothing suggesting a white noise
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Showing now leverage
Marginal importance of a point in the regression
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
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
𝐷 =1.36
251= 0.086
Higher bound value for the H0 to be accepted
Rejected! Regression between 2 different asset are very often poor
Heteroscedasticity
Basis risk if you hedge anyway
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Conclusion
OLS
Residuals
Normality
Heteroscedasticity