Financial Econometric Models I

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Financial Econometric Models, course I, Busines School MSc level

Financial Econometric Models, course I, Busines School MSc level

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  • 1. ESGF 5IFM Q1 2012Financial Econometric Models Vincent JEANNIN – ESGF 5IFM Q1 2012 vinzjeannin@hotmail.com 1
  • 2. ESGF 5IFM Q1 2012Summary of the session (est 3h)• Introduction & Objectives• Bibliography• OLS & Exploration vinzjeannin@hotmail.com 2
  • 3. Introduction & Objectives • What is a model? = + with being a white noise ESGF 5IFM Q1 2012 • What the point writing models? Describe data behaviour vinzjeannin@hotmail.com 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
  • 4. Bibliography vinzjeannin@hotmail.com ESGF 5IFM Q1 20124
  • 5. 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 vinzjeannin@hotmail.com = = + Residual ε 5
  • 6. Two parameters to estimate: • Intercept α • Slope β ESGF 5IFM Q1 2012Minimising residuals = 2 = − + 2 vinzjeannin@hotmail.com =1 =1 When E is minimal? When partial derivatives i.r.w. a and b are 0 6
  • 7. = 2 = − + 2 = − − 2 =1 =1 =1 Quick high school reminder if necessary… ESGF 5IFM Q1 2012 − − 2 = 2 − 2 − 2 + 2 2 + 2 + 2 vinzjeannin@hotmail.com = −2 + 2 2 + 2 = 0 = −2 + 2 + 2 = 0 =1 =1 − + 2 + = 0 − + + = 0=1 =1 ∗ 2 + ∗ = ∗ + = =1 =1 =1 =1 =1 7
  • 8. Leads easily to the intercept ∗ + = =1 =1 ESGF 5IFM Q1 2012 + = + = vinzjeannin@hotmail.com = − The regression line is going through ( , ) The distance of this point to the line is 0 indeed 8
  • 9. = − y = + − y − = ( − ) ESGF 5IFM Q1 2012 = −2 + 2 2 + 2 = 0 = −2 + 2 + 2 = 0 =1 =1 vinzjeannin@hotmail.com − − = 0 − − = 0 =1 =1 − − + = 0 =1 − + − = 0 =1 ( − − − ) = 0 ( − ) − ( − ) = 0 =1 =1 9 ( − − − ) = 0 =1
  • 10. We have ( − − − ) = 0 and ( − − − ) = 0=1 =1 ESGF 5IFM Q1 2012 ( − − − ) = ( − − − ) =1 =1 vinzjeannin@hotmail.com ( − − − ) − − − − =0 =1 =1 ( − )( − − − ) = 0 =1 Finally… =1( − )( − ) 10 = 2 =1( − )
  • 11. Covariance =1( − )( − ) = 2 =1( − ) Variance ESGF 5IFM Q1 2012 = 2 vinzjeannin@hotmail.com = − You can use Excel function INTERCEPT and SLOPE 11
  • 12. Calculate the Variances and Covariance of X{1,2,3,3,1,2} and Y{2,3,1,1,3,2} ESGF 5IFM Q1 2012 vinzjeannin@hotmail.com 12 You can use Excel function VAR.P, COVARIANCE.P and STDEV.P
  • 13. Let’s asses the quality of the regressionLet’s calculate the correlation coefficient (aka Pearson Product-MomentCorrelation Coefficient – PPMCC): ESGF 5IFM Q1 2012 = Value between -1 and 1 = 1 vinzjeannin@hotmail.com Perfect dependence ~0 No dependence Give an idea of the dispersion of the scatterplot 13 You can use Excel function CORREL
  • 14. Poor quality R=0.62 R=0.96 High quality vinzjeannin@hotmail.com ESGF 5IFM Q1 201214
  • 15. What is good quality? ESGF 5IFM Q1 2012 Slightly discretionary… vinzjeannin@hotmail.comIf 3 ≥ = 0.8666 … 2 It’s largely admitted as the threshold for acceptable / poor 15
  • 16. The regression itself introduces a bias Let’s introduce the coefficient of determination R-Squared ESGF 5IFM Q1 2012Total Dispersion = Dispersion Regression + Dispersion Residual vinzjeannin@hotmail.com 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
  • 17. In a simple linear regression with intercept 2 = 2 ESGF 5IFM Q1 2012Is a good correlation coefficient and a good coefficient ofdetermination enough to accept the regression? vinzjeannin@hotmail.com Not necessarily! Residuals need to have no effect, in other word to be a white noise! 17
  • 18. vinzjeannin@hotmail.com ESGF 5IFM Q1 201218
  • 19. Don’t get fooled by numbers! ESGF 5IFM Q1 2012 For every dataset of the Quarter = 9 = 7.5 vinzjeannin@hotmail.com = 3 + 0.5 = 0.82 2 = 0.67 Can you say at this stage which regression is the best? 19Certainly not those on the right you need a LINEAR dependence
  • 20. ESGF 5IFM Q1 2012Is any linear regression useless? vinzjeannin@hotmail.com Think what you could do to the series Polynomial transformation, log transformation,… 20 Else, non linear regressions, but it’s another story
  • 21. First application on financial market S&P / AmEx in 2011 ESGF 5IFM Q1 2012 vinzjeannin@hotmail.com 21
  • 22. ,& = = 0.8501 & 2 = 2 = 0.7227 ESGF 5IFM Q1 2012 Oups :-o Is Excel wrong? vinzjeannin@hotmail.com R-Squared has different calculation methodsLet’s accept the following regression then as the quality seems pretty good = 0.06% + 1.1046 ∗ & 22
  • 23. How to use this? ESGF 5IFM Q1 2012 • Forecasting? Not really… Both are random variables vinzjeannin@hotmail.com • 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! 23
  • 24. Hedging $1.0M of AmEx Stocks with $1.1046M of S&P ESGF 5IFM Q1 2012 vinzjeannin@hotmail.com It would have been too easy… Great differences… Why? Sensitivity to the size of the sample 24 Heteroscedasticity
  • 25. Let’s have a similar approach using a proper statistics and econometrics software ESGF 5IFM Q1 2012 • Free • Open Source • Developments shared by developers vinzjeannin@hotmail.com Let’s begin with statistical exploration to get familiar with the series and the software > Val<-read.csv(file="C:/Users/Vinz/Desktop/Val.csv",head=TRUE,sep=",") > summary(Val) SPX AMEX Min. :-0.0666344 Min. :-0.0883287 1st Qu.:-0.0069082 1st Qu.:-0.0094580 Median : 0.0010016 Median : 0.0013007 25 Mean : 0.0001249 Mean : 0.0005891 3rd Qu.: 0.0075235 3rd Qu.: 0.0102923 Max. : 0.0474068 Max. : 0.0710967
  • 26. > hist(Val$AMEX, breaks=20, main="Distribution AMEX Returns") > sd(Val$AMEX) [1] 0.01915489 ESGF 5IFM Q1 2012 vinzjeannin@hotmail.com> hist(Val$SPX, breaks=20, main="DistributionSPXX Returns")> sd(Val$SPX)[1] 0.01468776 26
  • 27. These are obvious negatively skewed distributions ESGF 5IFM Q1 2012 Reminders 3 − − 3 = = − 2 3/2 vinzjeannin@hotmail.com• Negative skew: long left tail, mass on the right, skew to the left• Positive skew: long right tail, mass on the left, skew to the right > skewness(Val$AMEX) [1] -0.2453693 > skewness(Val$SPX) 27 [1] -0.4178701
  • 28. These are obvious leptokurtic distributions ESGF 5IFM Q1 2012 Reminders 4 − − 4 = = − 2 2 vinzjeannin@hotmail.com> library(moments)> kurtosis(Val$AMEX) What is their K?[1] 5.770583 (excess kurtosis)> kurtosis(Val$SPX)[1] 5.671254 28 Subtract 3 to make it relative to the normal distribution…
  • 29. Quick check: what are the Skewness and Kurtosis of {1,2,-3,0,-2,1,1}? ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com Excel function SKEW R function skewness (package moments) 29
  • 30. ESGF 4IFM Q1 2012 vinzjeannin@hotmail.comExcel function KURTR function kurtosis (package moments) 30
  • 31. By the way, what is the most platykurtic distribution in the nature? Toss it! ESGF 4IFM Q1 2012 Head = Success = 1 / Tail = Failure = 0 vinzjeannin@hotmail.com> require(moments)> library(moments)> toss<-rbinom(10000000,1,0.5)> mean(toss)[1] 0.5001777> kurtosis(toss)[1] 1.000001> kurtosis(toss)-3[1] -1.999999> hist(toss, breaks=10,main="Tossing acoin 10 millions times",xlab="Resultof the trial",ylab="Occurence") 31> sum(toss)[1] 5001777
  • 32. 50.01777% rate of success: fair or not fair? Trick coin ? Can be tested later with a Bayesian approach ESGF 4IFM Q1 2012On a perfect 50/50, Kurtosis would be 1, Excess Kurtosis -2: the minimum!This is a Bernoulli trial (, ) with > 1 and 0 < < 1 ∈ ℝ and integer vinzjeannin@hotmail.com Mean SD (1 − ) Skewness 1 − 2 (1 − ) Kurtosis 1 −3 (1 − ) 32 Easy to demonstrate if p=0.5 the Kurtosis will be the lowest Bit more complicated to demonstrate it for any distribution
  • 33. Back to our series, a good tool is the BoxPlot ESGF 5IFM Q1 2012TooManyOutliers! vinzjeannin@hotmail.comThere should be 2 maxTo be normalFatter tails than thenormal distribution 33 boxplot(Val$AMEX,Val$SPX, main="AMEX & S&P BoxPlots", names=c("AMEX","SPX"),col="blue")
  • 34. Leptokurtic distributionsNegatively skewed distribution ESGF 5IFM Q1 2012 Are they normal distributions? vinzjeannin@hotmail.com Let’s compare them to normal distributions with same standard deviation and mean and make the QQ Plots 34
  • 35. x=seq(-0.2,0.2,length=200) y1=dnorm(x,mean=mean(Val$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") vinzjeannin@hotmail.comx=seq(-0.2,0.2,length=200)y1=dnorm(x,mean=mean(Val$SPX),sd=sd(Val$SPX))hist(Val$SPX, breaks=20,main="S&P Returns/ NormalDistribution",xlab="Return",ylab="Occurence")lines(x,y1,type="l",lwd=3,col="red") 35
  • 36. ESGF 5IFM Q1 2012 Excess kurtosis obvious vinzjeannin@hotmail.comFatter and longer tails 36Let’s have a look to their CDF through QQPlot
  • 37. > qqnorm(Val$AMEX) > qqnorm(Val$SPX)> qqline(Val$AMEX) > qqline(Val$SPX) ESGF 5IFM Q1 2012 vinzjeannin@hotmail.com Fatter tails 37 Let’s properly test the normality
  • 38. Can use many tests…• Kolmogorov-Smirnov• Jarque Bera• Chi Square• ESGF 5IFM Q1 2012 Shapiro WilkLet’s try Kolmogorov-Smirnov It compares the distance between the empirical vinzjeannin@hotmail.com CDF and the CFD of the reference distribution 38
  • 39. ESGF 5IFM Q1 2012x=seq(-4,4,length=1000)plot(ecdf(Val$AMEX),do.points=FALSE, col="red", lwd=3,main="Normal Distribution against AMEX - CFDs", xlab="x",ylab="P(X<=x)")lines(x,pnorm(x,mean=mean(Val$AMEX),sd=sd(Val$AMEX)),col="blue",type="l",lwd=3) vinzjeannin@hotmail.comx=seq(-4,4,length=1000)plot(ecdf(Val$SPX),do.points=FALSE, col="red", lwd=3,main="Normal Distribution against S&P - CFDs", xlab="x",ylab="P(X<=x)")lines(x,pnorm(x,mean=mean(Val$SPX),sd=sd(Val$SPX)),col="blue",type="l",lwd=3) 39
  • 40. > ks.test(Val$SPX, "pnorm") > ks.test(Val$AMEX, "pnorm") One-sample Kolmogorov- One-sample Kolmogorov-SmirnovSmirnov test testdata: Val$SPX data: Val$AMEXD = 0.4811, p-value < 2.2e-16 D = 0.4742, p-value < 2.2e-16alternative hypothesis: two-sided alternative hypothesis: two-sided ESGF 5IFM Q1 2012 The 0 hypothesis is the distribution is normal vinzjeannin@hotmail.com Do we accept or reject the hypothesis 0 with a 95% confidence interval? The hypothesis regarding the distributional form is rejected if the test statistic, D, is greater than the critical value obtained from a table 40
  • 41. vinzjeannin@hotmail.com 1.36 Sample size: 251 = 0.086 251 Rejected or not? 41 P-Value was givingRejected! Series aren’t fitting a normal distribution the answer
  • 42. Ok, we now know a bit more the 2 series we want to regress > lm(Val$AMEX~Val$SPX) Call: lm(formula = Val$AMEX ~ Val$SPX) ESGF 5IFM Q1 2012 Coefficients: (Intercept) Val$SPX 0.0004505 1.1096287plot(Val$SPX,Val$AMEX, main="S&P / AmEx", xlab="S&P", ylab="AmEx",col="red") vinzjeannin@hotmail.comabline(lm(Val$AMEX~Val$SPX), col="blue") = 110.96% ∗ + 0.045% 42
  • 43. 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 vinzjeannin@hotmail.com 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 43They need to be a white noise, you can have a first assessment with quartiles
  • 44. plot(Reg) layout(matrix(1:4,2,2)) vinzjeannin@hotmail.com ESGF 5IFM Q1 201244
  • 45. QQ Plot compares the CDF ESGF 5IFM Q1 2012A perfect fit is a line vinzjeannin@hotmail.com Left tail noticeably different 45
  • 46. ESGF 5IFM Q1 2012 vinzjeannin@hotmail.comResiduals should be randomly distributed around the 0 horizontal lineYou don’t want to see a trend, a dependenceTo accept or reject the regression you need residuals to be a white noise 46 Their mean should be 0
  • 47. ESGF 5IFM Q1 2012Nothing suggesting a white noise vinzjeannin@hotmail.com • Square root of the standardized residuals as a function of the fitted values • There should be no obvious trend in this plot 47
  • 48. Showing now leverage Marginal importance of a point in the regression ESGF 5IFM Q1 2012 vinzjeannin@hotmail.comFar points suggest outlier or poor model 48
  • 49. 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 vinzjeannin@hotmail.com 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-sidedRejected! Regression between 2 different asset are very often poor 49 Heteroscedasticity Basis risk if you hedge anyway
  • 50. Conclusion ESGF 5IFM Q1 2012 OLS Residuals vinzjeannin@hotmail.com Normality Heteroscedasticity 50