Master QEM Summer 2008

Basics in Statistics Laurent Ferrara

Master QEM

L. Ferrara, 2008

Objective • Let a quantitative continuous random variable X with unknown distribution Pθ with density fθ . • Let (x1 , …, xn) a n-observation stemming from this variable • Our aim: fit a statistical model to those data → Inference : How to use this n-sample to estimate the empirical distribution and draw conclusions on the theoretical distribution of X : Pθ . ? Master QEM

L. Ferrara, 2008

Plan of the course 1. Graphical description of a distribution 2. Numerical description of a distribution 3. Recall on usual continous distribution 4. Tools to compare distributions Master QEM

L. Ferrara, 2008

1. Graphical description 1) Histogram Objective : estimation of the empirical distribution • We range the data : x(1) , …, x(n) . • We group the data into J equal classes of width h • A class is semi-open interval : B j =]b j −1 , b j ] • The center of each class j is : m j = (b j −1 + b j ) / 2 • The width of each class j is : h = b j − b j −1 thus B j =]( j − 1)h, jh] Master QEM

L. Ferrara, 2008

• Histogram is a function fH defined by, for all x in the support , 1 f H ( x) = × {nbre de xi dans la classe B j contenant x} nh

1 n 1 n 1{X i = xi ∈B j } = 1{xi ∈[ x ± h / 2 ]} f H ( x) = ∑ ∑ nh i =1 nh i =1

• Practical issue : – Choice of b0 ? – Choice of h ? = Choice of number of classes J ?

• Warning : Master QEM

b0 ≤ x(1)

et

x( n ) ≤ bJ L. Ferrara, 2008

1000

3000

5000

7000

Example : Daily returns of CAC 40 from 1987 to 2002 (n=3758)

0

1000

2000

3000

-8 -6 -4 -2

0

2

4

6

Time

Master QEM

0

1000

2000 Time

3000

L. Ferrara, 2008

1400

2000

1000

1500

600

1000

0 200

500 0

-5

0

5

-5

cac40.rdt

200

600

100

400

0

200 0

-5 Master QEM

5

300

cac40.rdt

0

0 cac40.rdt

5

-5

0 cac40.rdt

5 L. Ferrara, 2008

2) Non-parametric Kernel estimation →Same objective • 2 practical issues with the histogram – Choce of bandwidth h ? (= choice of number of classes J) – Choice of the origin b0?

• 2 main issues with histogram: – Loss of information by identifying all the points of the class to the center of the class – Density is supposed to be smoothed while histogram isn’t. (alternative: linear interpolation) Master QEM

L. Ferrara, 2008

We have :

1 n 1{ x − xi / h≤0.5} f H ( x) = ∑ nh i =1

We note :

K (u ) = 1{ u ≤0.5}

Then :

1 n f H ( x) = K ( x − xi / h) ∑ nh i =1

This function K is a Kernel function such that: – K positive : K >= 0 – K symmetric to zero : – +∞

K ( x) = K (− x)

∫ K ( x)dx = 1

−∞ Master QEM

L. Ferrara, 2008

0.0

0.2

0.4

yunif

0.6

0.8

1.0

• Idea : → histogram is associated to the uniform kernel

-1.0

-0.5

0.0

0.5

1.0

→ Find a smooth kernel → Find a kernel that gives more weight to values xi close to x x

Master QEM

L. Ferrara, 2008

0.6 0.4 0.0

0.2

ytriangle

0.8

1.0

• Other kernels: K (u ) = (1 − u )1{ u ≤1} Triangle :

-1.0

-0.5

0.0

0.5

1.0

ygauss

0.2 0.1

K (u ) =

0.0

Gaussien :

1 u2 exp( − ) 2 2π

0.3

0.4

x

-4

-2

0

2

0.4

yepanech

K (u ) = 34 (1 − u 2 )1{ u ≤1}

0.0

0.2

Epanechnikov:

0.6

x

-1.0

-0.5

0.0

0.5

1.0

0.5

1.0

0.6 yquartic

0.4

15 K (u ) = 16 (1 − u 2 ) 21{ u ≤1}

0.0

0.2

Quartic:

0.8

x

-1.0

-0.5

0.0 x

Master QEM

L. Ferrara, 2008

4

R commands

> x yunif yepanech plot(x,yepanech,type="l") #Quartic > yquartic plot(x,yquartic,type="l") # Gaussien > x ygauss plot(x,ygauss,type="l")

Master QEM

L. Ferrara, 2008

• We define a kernel estimate of the density by : n 1 fˆ ( x) = K ( x − xi / h) ∑ nh i =1

• We get a smooth estimate • Bandwidth h controls the smoothing. – h large: strongly smoothed estimate (small variance but strong bias) – h small : weakly smoothed estimate (strong variance but low bias) → Trade-off !!

Master QEM

L. Ferrara, 2008

0.30 0.25 0.20 0.15 0.10 0.0

0.05

density(cac40.rdt)$y

Densité empirique NP du CAC avec un noyau gaussien

-10

-5

0

5

0.15 0.10 0.0

0.05

Comparaison densité empirique NP et densité théorique Gaussienne

0.20

0.25

0.30

density(cac40.rdt)$x

-10

Master QEM

-5

0

5

L. Ferrara, 2008

Problèmes de mise en œuvre : → Choix du support ? du noyau ? de la fenêtre h ?

• Choix du support – Création d’un support pour la distribution empirique à partir d ’un maillage équidistant de xmin à xmax tq :

xmin ≤ x(1)

et

x( n ) ≤ xmax

– Par défaut, on choisit le support légèrement étendu d ’une fraction de la fenêtre h

• Choix du Noyau – Choix empirique. – Noyau Gaussien préféré car il existe alors un h optimal sous certaines conditions. Master QEM

L. Ferrara, 2008

• Choice of the bandwidth h – Theory of estimation shows that the optimal bandwidth, in the sense of a deviation to the true distribution criteria, is a function of n-1/5 , ie: hopt = c × n −1/ 5 – In practice, Silverman (1986) proves that if the data are Gaussian the optimal bandwidth is given by : • with K Gaussian : • with K Quartic : • with K Epanechnikov :

G hopt = 1.06 × s × n −1/ 5 Q hopt = 2.78 × s × n −1/ 5 E hopt = 2.34 × s × n −1/ 5

where s is the empirical standard error. Master QEM

L. Ferrara, 2008

This empirical rule (« rule of thumb ») stays valid even if the distribution is not too far away from the Gaussian

Other methods to choose h Cross validation Bootstrap

Conclusion To estimate the density distribution of a continuous variable, we argue in favour of a kernel estimate rather than the histogram.

Master QEM

L. Ferrara, 2008

3) Cumulative distribution function (cdf) • Let X a random variable with density fθ . CDF: F ( x) = ∫

x

−∞

fθ (u )du

• Growing function between 0 and 1 • True cdf is estimated by the empirical cdf 1 n Fn ( x) = ∑1{x ≤ x} n i =1 i

Master QEM

L. Ferrara, 2008

3) cdf • More complicated to interpret than density • Useful function to compute some statistics such that the quantiles

Master QEM

L. Ferrara, 2008

Example Cac 40 : empirical cdf and Gaussian cdf

0.0

0.2

0.4

0.6

0.8

1.0

Empirical and Hypothesized normal CDFs

-5

0

5

solid line is the empirical d.f. Master QEM

L. Ferrara, 2008

2. Numerical description • Let (x1 , …, xn ) an observed n-sample from a random variable X with unknown distribution. The main characteristics of the distribution of X are 2.1 position 2.2 dispersion 2.3 symmetry 2.4 tail behaviour 2.5 multi-modes

Master QEM

L. Ferrara, 2008

2.1 Position measures • Mean:

n

X = ∑ ω i xi où i =1

n

∑ω i =1

i

=1

Arithmetical mean → uniform weights : ω i = 1 / n, ∀i Mean = estimate of E(X) (first order moments) • Median : med(X) med(X) is the value that splits the sample into 2 sub-samples, values in the first sub-sample being lower than med(X) ; values in the other being greater. Is is a rank statistics. Master QEM

L. Ferrara, 2008

2.1 Position measures Let x(1) , …, x(n) , the ranked sample n odd : med(X) = x((n+1)/2) , n even : med(X) = (x(n/2) + x(n/2+1) ) /2.

Fn ( x) ≤ 0.5 ⇔ x < med( X )

Median satisfies :

0.0

0.2

0.4

0.6

0.8

1.0

Empirical and Hypothesized normal CDFs

Master QEM

-5

0 solid line is the empirical d.f.

5

L. Ferrara, 2008

2.1 Position measures • Quantiles Generalization of the median that allows to split the sample into a finite number of sub-samples: 4 parts → quartiles 10 parts → déciles For 0 < α < 1, quantile of order α is defined by : −1

qα = Fn (α )

Master QEM

L. Ferrara, 2008

2.1 Position measures • Mode Value that appears the most frequently Mainly used for qualitative data For a continuous variable, we choose the value for which the distribution is maximum Maybe several local maxima, implying several modes: multimodal distribution

Master QEM

L. Ferrara, 2008

Why diverse position measures ?

→ Find a reasonable estimate in case of non-Gaussianity Master QEM

L. Ferrara, 2008

Why diverse Position measures ? Median or Mean? – – – –

For symmetric distribution: median = mean Assymmetry to the right : mean > median Assymetry to the left : mean < median Median is robust to outliers or rare avents

• Alternative measures: – Mid-mean : mean for the data between quantiles 0.25 et 0.75 – Trimmed-mean

Master QEM

L. Ferrara, 2008

2.2 Dispersion measures Let x1 , …, xn . 2 questions : 1) How are dispersed the values close to the center of the distribution ? 2) How are dispersed the values in the tails of the distribution ? The following measures give more or less weight to each of those 2 components

Master QEM

L. Ferrara, 2008

2.2 Dispersion measures • Empirical variance : (Centered moment of order 2)

• Empirical standard-error :

• Range :

1 n 2 s( X ) = ( x − X ) ∑ i n − 1 i =1

x( n) − x(1)

• Average Absolute Deviation : Master QEM

n 1 2 s2 ( X ) = ( x − X ) ∑ i n − 1 i =1

1 n AAD = ∑ xi − X n i =1 L. Ferrara, 2008

2.2 Dispersion measures • Median Absolute Deviation :

• Ecart Inter-Quartile :

MAD = med( xi − med( X ) )

IQ = q0.75 − q0.25

Variance, Standard-error, AAD and MAD measure the 2 components of the variability AAD and MAD do not over-weight the tail beahaviour Range only measures the tail variability IQ only measures the central variability Master QEM

L. Ferrara, 2008

Why diverse dispersion measures ?

→ Find a reasonable estimate in case of non-Gaussianity Master QEM

L. Ferrara, 2008

2.3 Symmetry measure Skewness measures the asymetry, based on the centered order 3 moement m3(X), such that : 1 n m3 ( X ) = ∑ ( xi − X )3 n i =1 et : m3 ( X ) Sk ( X ) = 3 s (X ) Sk(X) = 0 if symmetry Positive asymmetrie = Sk > 0 = right tail fat Negative asymmetrie = Sk < 0 = left tail fat Master QEM

L. Ferrara, 2008

2.4 Tail thickness measure Kurtosis measures the thickness of the tails, based on the centered moment of order 4 m4(X) :

and :

1 n m4 ( X ) = ∑ ( xi − X ) 4 n i =1

m4 ( X ) K(X ) = 4 s (X )

K(X) = 3 if the distribution is Gaussian. A useful measure is the Excess Kurtosis defined by: K(X) - 3 EK(X) > 0 indicates that the tails are thicker thant the Normal and conversely. Master QEM L. Ferrara, 2008

Master QEM

L. Ferrara, 2008

2.5 Multi-modes • Distribution with several local maxima : approximation by a mixture of Normal Reveals the presence of non-linearities in the data → Use of non-linear models or piecewise linear models

Master QEM

L. Ferrara, 2008

Example of bi-modal distribution 8

6

4

2

0

-2

-4

-6

-8 100

110

120

130

140

150

160

170

180

190

200

0.18

0.16

0.14

0.12

0.10

0.08

0.06

0.04

0.02

0.00 -7.5

Master QEM

-5.0

-2.5

0.0

2.5

5.0

7.5

10.0

L. Ferrara, 2008

Example de tri-modal distribution : IPI Euro-zone 130 120 110 100 90 80 70 60 1970 1972 1974 1976 1978 1980 1982 1984 1986 1988 1990 1992 1994 1996 1998 2000 2002 0.015 0.010 0.005 0.000 -0.005 -0.010 -0.015 -0.020 1970 1972 1974 1976 1978 1980 1982 1984 1986 1988 1990 1992 1994 1996 1998 2000 2002

Master QEM

L. Ferrara, 2008

Example 140 120 100 80 60 40 20 0 -0.010

Master QEM

-0.005

0.000

0.005

0.010

L. Ferrara, 2008

Exemple : Daily returns of CAC 40 from 1987 to 2004 (n=4337) > summary(cac40.rdt) Regular Time Series: Observations: 4337 Min. 1st Qu. -0.07678 -0.00681

Median 0.00041

Mean 0.00032

3rd Qu. 0.00784

Max. 0.07002

Time Parameters : start deltat frequency 2 1 1 > skewness(cac40.rdt) [1] -0.1245464 > kurtosis(cac40.rdt) [1] 2.833693

Master QEM

L. Ferrara, 2008

3. Usual continuous distributions From an observed n-sample, we try to identify the distribution of X to a known distribution, from its empirical characteristics (stylized facts) • Gaussian • Student • Uniform • Chi-2 • Fischer • Log-Normal • Exponential Master QEM

L. Ferrara, 2008

• Gauss (or Normal) distribution La va X suivant une loi Normale N (m,σ2) a la densité suivante: 1 (u − m) 2 exp(− ) f (u ) = 2 2σ σ 2π Loi Normale standard pour m = 0 et σ =1 cdf:

Master QEM

1 Φ( x) = P( X ≤ x) = σ 2π

(u − m) 2 ∫−∞ exp(− 2σ 2 )du x

L. Ferrara, 2008

• La loi de Gauss (ou Normale) Propriétés (Rappel) : Φ (− x) = 1 − Φ ( x) P( − x ≤ X ≤ x) = 2Φ ( x) − 1 X ≈ N (m, σ ) ⇔ 2

X −m

σ

≈ N (0,1)

P(m − σ ≤ X ≤ m + σ ) ≈ 68% P(m − 2σ ≤ X ≤ m + 2σ ) ≈ 95% Master QEM

L. Ferrara, 2008

Distributions de Gauss standard

cdf de Gauss standard

Master QEM

L. Ferrara, 2008

• Caractéristiques de la loi de Gauss Moyenne = Mediane = Mode = m Ecart-type = σ Skewness = 0 Kurtosis = 3

Master QEM

L. Ferrara, 2008

• La loi Uniforme La v.a. X suivant une loi Uniforme dans l ’intervalle [a,b] a la densité suivante: f (u ) =

1 1[ a ,b ] b−a

a est le paramètre de position b-a est le paramètre de dispersion Distribution standard uniforme : a=0, b=1 cdf: F ( x) = P( X ≤ x) = x Master QEM

L. Ferrara, 2008

Distribution Uniforme

Master QEM

L. Ferrara, 2008

• Loi Uniforme Moyenne = Mediane = (a+b)/2 Variance =

(b − a) 2 12

Skewness = 0 Kurtosis = 9/5

Master QEM

L. Ferrara, 2008

• La loi de Student Soit X0 , X1 , …, Xn , n+1 v.a. iid selon la loi Normale standard. Alors la v.a. T tq: X0 T= 1 ∑X2 i n suit une loi de Student à n degrés de liberté Densité: f (u ) = Master QEM

Γ((n + 1) / 2) (1 + u 2 / n) −( n +1) / 2 Γ ( n / 2) nπ L. Ferrara, 2008

• La loi de Student avec la fonction Gamma, pour a > 0, tq: ∞

Γ(a) = ∫ x a −1 exp(− x)dx 0

On rappelle que : Γ(a ) = (a − 1)Γ(a − 1) Γ(1 / 2) = π Γ(n) = (n − 1)!, n ∈ N

Master QEM

L. Ferrara, 2008

Distributions de Student

Master QEM

L. Ferrara, 2008

• La loi de Student Moyenne = Mediane = 0 n n−2

Variance =

n>2

Skewness = 0 Kurtosis =

Master QEM

3(n − 2) n−4

n>4

L. Ferrara, 2008

• Loi du Chi-2 Soit X1 , …, Xn , n v.a. iid selon la loi Normale standard. Alors la v.a. Z tq: n Z = ∑ X i2 i =1

suit une loi du Chi-2 à n degrés de liberté Densité pour u ≥ 0:

Master QEM

1 f (u ) = n / 2 exp(−u / 2)u ( n / 2−1) 2 Γ ( n / 2) L. Ferrara, 2008

Distributions du Chi-2

Master QEM

L. Ferrara, 2008

• La loi du Chi-2 Moyenne = n Mediane = n - 2/3 , lorsque n grand Mode = n - 2 , pour n > 2 Variance =

2n

Skewness =

21/ 5 n

Kurtosis =

12 3+ n

Master QEM

L. Ferrara, 2008

• Loi de Fischer Soit X1 , …, Xn , Xn+1 , …, Xn+m , n+m v.a. iid selon la loi N(0,1) Alors la v.a. F tq: n 2 1 X i n ∑ i =1 F = m 2 1 X i m ∑ i = n +1

suit une loi de Fischer à (n,m) degrés de liberté Densité pour u ≥ 0: Γ((n + m) / 2) n / 2 m / 2 n / 2−1 f (u ) = n m u (m + nu ) −( n + m ) / 2 Γ(n / 2)Γ(m / 2) Master QEM

L. Ferrara, 2008

Distributions de Fischer

Master QEM

L. Ferrara, 2008

• Loi de Fischer Moyenne = Mode =

m m−2

m ( n − 2) n ( m + 2)

,m>2 ,n>2

Ecart-type =

Master QEM

L. Ferrara, 2008

• Loi exponentielle La va X suivant une loi exponentielle a la densité suivante: f (u ) =

1

β

exp(−

(u − μ )

β

)

pour u ≥ μ et β > 0. μ : paramètre de position et β : paramètre de dispersion Loi exponentielle standard pour μ = 0 et β = 1 cdf: x ≥ 0 et β > 0 F ( x) = P ( X ≤ x) = 1 − exp( − x / β ) Master QEM

L. Ferrara, 2008

Distribution exponentielle standard

Densité

cdf

Master QEM

L. Ferrara, 2008

• La loi exponentielle Moyenne = β Mediane = β ln(2) Mode = 0 Variance = β2 Skewness = 2 Kurtosis = 9 Master QEM

L. Ferrara, 2008

• Loi log-Normale La va X suit une loi log-normale si log(X) suit une loi Normale. Sa densité est la suivante : 1 1 (log(u ) − m) 2 exp(− ) f (u ) = 2 2σ σ 2π u

u ≥ 0 et σ > 0 (m = position, σ = dispersion) cdf: x ≥ 0 et σ > 0

Master QEM

L. Ferrara, 2008

• Loi Log-normale standard Moyenne =

exp(

σ2 2

)

Variance =

exp(σ 2 )(exp(σ 2 ) − 1)

Skewness =

exp(σ 2 + 2) (exp(σ 2 ) − 1)

Kurtosis =

Master QEM

exp(σ 2 ) 4 + 2 exp(σ 2 ) 3 + 3 exp(σ 2 ) 2 − 3

L. Ferrara, 2008

Distributions log-Normale

Master QEM

L. Ferrara, 2008

• Loi Skewed-Normal Sa densité est la suivante : avec :

→ Si α = 0, on retrouve la Normale → Quand α augmente, le skewness augmente aussi → Quand α change de signe, la densité prend la forme opposée

Master QEM

L. Ferrara, 2008

α

=2

α

Master QEM

=5

L. Ferrara, 2008

0.30 0.25 0.20

=0

0.0

0.05

0.10

0.15

α

-10

α

-5

0

5

= -5

Master QEM

L. Ferrara, 2008

4. Comparaison de distributions Outils graphiques de comparaisons de distributions : 1) Box-Plot 2) QQ- Plot Outils plus formels : Tests d ’hypothèses

Master QEM

L. Ferrara, 2008

4.1 le Box-Plot • Permet une représentation graphique de la distribution basée sur les résumés numériques de position et dispersion • Utile pour comparer les distribution de 2 populations • Mise en évidence de valeurs aberrantes Principe : Les quartiles encadrent la médiane. Est outlier : toute valeur supérieure à q(0.75)+1.5*IQ toute valeur inférieure à q(0.25)-1.5*IQ Master QEM

L. Ferrara, 2008

2 1 -2

-1

t(20)

0

1

2

3

-1

0

N(0,1)

Master QEM

L. Ferrara, 2008

Exemple : 4 types d’appareils à produire de l’énergie

Master QEM

L. Ferrara, 2008

-5

0

5

Exemple : Rendements du CAC 40

Master QEM

L. Ferrara, 2008

4.2 le QQ-Plot • Permet de comparer la distribution d’une variable avec celle d’une autre variable ou d’une loi théorique à partir des quantiles • Diagramme (« scatter-plot ») des quantiles de X1 contre X2 – Quantiles empiriques de X1 contre quantiles empiriques X2 – Quantiles empiriques de X1 contre quantiles théoriques d’une loi donnée

• Si les 2 lois sont proches, le diagramme est proche d’une ligne droite de référence Avantage: les 2 jeux de données n’ont pas besoin d’être de même taille Master QEM

L. Ferrara, 2008

Exemples

Master QEM

L. Ferrara, 2008

5 -5 -10 0

2

-2

N(0,1) vs N(0,2)

N(0,1) vs N(0,5)

2

5 0 -10

-10

-5

0

xnorm3

5

10

xnorm

-5

xnorm2

0

xnorm

10

-2

-2 Master QEM

0

xnorm1

0 -10

-5

xnorm

5

10

N(0,1) vs N(5,1)

10

N(0,1) vs N(0,1)

0 xnorm

2

-2

0 xnorm

2 L. Ferrara, 2008

5 -5 -10 0

2

-2

N(0,1) vs t(20)

N(0,1) vs t(50)

2

5 0 -10

-10

-5

0

xt50

5

10

xnorm

-5

xt20

0

xnorm

10

-2

-2 Master QEM

0

xt5

0 -10

-5

xt2

5

10

N(0,1) vs t(5)

10

N(0,1) vs t(2)

0 xnorm

2

-2

0 xnorm

2 L. Ferrara, 2008

-20

-10

0

xt2

10

20

Exemple : t(5) vs t(2)

-4

-2

0

2

4

6

xt5 Master QEM

L. Ferrara, 2008

0 -5

cac40.rdt

5

Exemple : Rendements du CAC 40

-2

0

2

Quantiles of Standard Normal

Master QEM

L. Ferrara, 2008

2

4

2

4

2

5 -5 5 6

-6

0

2

4

-4

2

2

4

6

-2

0

2

4

5 sort(cac40.rdt)

-5 -4

-2

0

2

4

-4

qt(ppoints(cac40.rdt), df = i)

4

0

qt(ppoints(cac40.rdt), df = i)

-2

0

2

4

qt(ppoints(cac40.rdt), df = i)

5 0

-2

5 -2

qt(ppoints(cac40.rdt), df = i)

sort(cac40.rdt) -2

qt(ppoints(cac40.rdt), df = i)

-4

qt(ppoints(cac40.rdt), df = i)

-5 -4

10

0

sort(cac40.rdt) 4

sort(cac40.rdt) -4

4

5

-5 2

5 0

0

-5

4

sort(cac40.rdt) -2

0

sort(cac40.rdt)

5 0

5 2

0

5 -2

qt(ppoints(cac40.rdt), df = i)

sort(cac40.rdt) 0

-2

qt(ppoints(cac40.rdt), df = i)

-5 -2

qt(ppoints(cac40.rdt), df = i)

-4

-5 -4

5 -4

0 5 -6

5 sort(cac40.rdt) 0

0

sort(cac40.rdt) 6

-5 -2

0

sort(cac40.rdt) 4

4

qt(ppoints(cac40.rdt), df = i)

-5 2

2

sort(cac40.rdt) -4

qt(ppoints(cac40.rdt), df = i)

5

0

0

-5

4

5

-4

0

-2

qt(ppoints(cac40.rdt), df = i)

-2

qt(ppoints(cac40.rdt), df = i)

0

sort(cac40.rdt) 4

-5 -4

-4

-5

qt(ppoints(cac40.rdt), df = i)

-5 -6

-5 2

qt(ppoints(cac40.rdt), df = i)

-10

5

2

10

sort(cac40.rdt)

0

5

0

6

0

0

4

5 -2

-5

qt(ppoints(cac40.rdt), df = i)

-5

2

qt(ppoints(cac40.rdt), df = i)

5

0

-10

0

0

sort(cac40.rdt) -4

0

-2

20

-5

4

-5

-4

10

5 sort(cac40.rdt) -2

5 2

0

-5 -4

-5 0

-10

qt(ppoints(cac40.rdt), df = i)

0

sort(cac40.rdt)

5

-2

-5 -20

0

60

qt(ppoints(cac40.rdt), df = i)

0

sort(cac40.rdt)

40

5

-6

qt(ppoints(cac40.rdt), df = i)

sort(cac40.rdt)

20

-5 5

-5

-4

sort(cac40.rdt)

0

0

sort(cac40.rdt)

5 sort(cac40.rdt)

0

0

qt(ppoints(cac40.rdt), df = i)

Master QEM

-20

0

-40

qt(ppoints(cac40.rdt), df = i)

-5

-5

sort(cac40.rdt)

5 -5 -60

0

2000

0

1000

0

0

0

-1000

qt(ppoints(cac40.rdt), df = i)

0

-2000

0

sort(cac40.rdt)

5 0

sort(cac40.rdt)

-5

0 -5

sort(cac40.rdt)

5

Exemple : Rendements du CAC 40

-4

-2

0

2

qt(ppoints(cac40.rdt), df = i)

4

-4

-2

0

2

4

qt(ppoints(cac40.rdt), df = i)

L. Ferrara, 2008