applied statistics i
DESCRIPTION
First course of Applied Statistics, MSc level in Buisiness School.TRANSCRIPT
Applied Statistics Vincent JEANNIN – ESGF 4IFM
Q1 2012
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Summary of the session (est. 4.5h) • Introduction & Objectives • Bibliography • First approach: Descriptive Statistics • The Normal Distribution • Applications (GBM, B&S, Greeks, CRR)
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Introduction & Objectives
• What are statistics?
• Why Should you use them?
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Describe data behaviour
Modelise data behaviour
• Take the opportunity to remember financial mathematics basics
• Acquire theory knowledge on statistics
• Usage of R and Excel
Business decisions (pricing, investments,…)
Bibliography
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First Approach: Descriptive statistics
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FCOJ Front Month: 31st Dec 2008 / 30th Sep 2011
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First step, calculate the linear returns 𝑅𝑡 =𝑉𝑡𝑉𝑡−1− 1
Then, the mean 𝑅 =1
𝑛 𝑅𝑖
𝑛
𝑖=1
Expected return, not average return!
How to calculate average return on the period? (compound return) How to obtain it by a sum?
𝑅𝐶𝑜𝑚𝑝 =𝑉1𝑉𝑛
𝑛
− 1
𝑅𝐶𝑜𝑚𝑝 = 𝑙𝑛𝑉𝑖𝑉𝑖−1= ln 𝑉𝑛 − ln 𝑉1 = 𝑙𝑛
𝑉𝑛𝑉1
𝑛
𝑖=2
𝑅1 = 𝑙𝑛𝑉2𝑉1= ln 𝑉2 − ln 𝑉1
𝑅2 = 𝑙𝑛𝑉3𝑉2= ln 𝑉3 − ln 𝑉2
𝑅𝑛−1 = 𝑙𝑛𝑉𝑛−1𝑉𝑛−2= ln 𝑉𝑛−1 − ln 𝑉𝑛−2
𝑅𝑛 = 𝑙𝑛𝑉𝑛𝑉𝑛−1= ln 𝑉𝑛 − ln 𝑉𝑛−1
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Excel and R can give an idea of the distribution
R, easier, faster,… Function summary
Excel, functions Min, Max, Average, Percentile
• Free • Open Source • Developments shared by developers
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R can easily show the distribution of returns
Interesting shape but what next?
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The four moments
Mean 𝑅 =1
𝑛 𝑅𝑖
𝑛
𝑖=1
Standard Deviation
Expected Return
Dispersion from the mean
Square root of the variance
𝜎 = 𝐸 𝑋 − 𝑋 2
SD is the square root of the mean of squared differences to the mean
𝜎 =1
𝑛 𝑅𝑖 − 𝑅
2
𝑛
𝑖=1
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Quick check: what is the SD of {1,2,-3,0,-2,1,1}?
Excel function STDEVP
R function sd
FCOJ has a SD of 2.16%
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> xBar<-mean(FCOJ$V1)
> SD <- sd(FCOJ$V1)
> hist(FCOJ$V1, breaks=c(xBar-6*SD,xBar-5*SD,xBar-4*SD,xBar-3*SD,xBar-
2*SD,xBar-
SD,xBar,xBar+SD,xBar+2*SD,xBar+3*SD,xBar+4*SD,xBar+5*SD,xBar+6*SD),main="F
COJ Returns",xlab="Return",ylab="Occurence")
Histogram centred on the mean with SD multiples groups
Symmetric-ish
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693 data
75.04% within ±1𝜎
94.23% within ±2𝜎
98.99% within ±3𝜎
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Skewness, the third moment
𝑆𝐾𝐸𝑊 𝑋 = 𝐸𝑋 − 𝑋
𝜎
3
=𝐸 𝑋 − 𝑋 3
𝐸 𝑋 − 𝑋 2 3/2
Asymmetry of the distribution
• Negative skew: long left tail, mass on the right, skew to the left • Positive skew: long right tail, mass on the left, skew to the right
Should I rather buy or sell a positive skewed asset?
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Excel function SKEW
R function skewness (package moments)
FCOJ is positively skewed
> require(moments)
> library(moments)
> skewness(FCOJ$V1)
[1] 0.2030842
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Kurtosis, the fourth moment
Peakedness of the distribution
• Positive excess Kurtosis: high peak around the mean, fat tails • Negative excess Kurtosis: low peak around the mean, thin tails
𝐾𝑈𝑅𝑇 𝑋 = 𝐸𝑋 − 𝑋
𝜎
4
=𝐸 𝑋 − 𝑋 4
𝐸 𝑋 − 𝑋 2 2
It’s a usage to deal with the excess kurtosis (relative to the normal distribution, subtracting 3
Which distribution you’d rather buy or sell?
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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
Will be tested later with a Bayesian approach
𝑛 > 1 0 < 𝑝 < 1 with and 𝑝 ∈ ℝ and 𝑛 integer
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Excel function KURT
R function kurtosis (package moments)
FCOJ is leptokurtic
> require(moments)
> library(moments)
> kurtosis(FCOJ$V1)
[1] 6.34176
> kurtosis(FCOJ$V1)
[1] 3.34176
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Sum-up: • Positive expected return • Positive skew • Positive excess kurtosis
Buy or Sell?
Is that actually enough to take investment decision? What next? How different is the FCOJ distribution from the Normal Distribution?
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The Normal Distribution
Snapshot, 4 moments:
Mean
SD
Skewness
Kurtosis
0
1
0
3
Snapshot, Shape:
Let’s discuss about the standard normal first…
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𝑓 𝑥 =1
2𝜋𝜎2𝑒−(𝑥−𝜇)2
2𝜎2 Density
𝑁(𝜇, 𝜎) Notation
Distributions of zeros means with following SD: 0.5 / 0.75 / 1 / 1.5 / 2
Which one is which one?
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> x=seq(-4,4,length=500)
> y1=dnorm(x,mean=0,sd=0.5)
> y2=dnorm(x,mean=0,sd=0.75)
> y3=dnorm(x,mean=0,sd=1)
> y4=dnorm(x,mean=0,sd=1.5)
> y5=dnorm(x,mean=0,sd=2)
> plot(x,y1,type="l",lwd=3,col="red",
main="Normal Distributions", ylab="f(x)")
> lines(x,y2,type="l",lwd=3,col="blue")
> lines(x,y3,type="l",lwd=3,col="black")
> lines(x,y4,type="l",lwd=3,col="yellow")
> lines(x,y5,type="l",lwd=3,col="pink")
All other things equal, low SD is a high peak Values are more compacted around the mean
• FCOJ has a mean of 1.364% and a SD of 2.164% • Let’s compare the distribution with a normal distribution with the same
mean and SD
FCOJ<-
read.csv(file="C:/Users/Vinz/Desktop/FCOJStats.csv",head=FALSE,sep=",")
x=seq(-0.2,0.2,length=200)
y1=dnorm(x,mean=mean(FCOJ$V1),sd=sd(FCOJ$V1))
hist(FCOJ$V1, breaks=100,main="FCOJ Returns / Normal
Distribution",xlab="Return",ylab="Occurence")
lines(x,y1,type="l",lwd=3,col="red")
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The excess Kurtosis sign is obvious, isn’t it?
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Same SD, different mean, more straight forward
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Cumulative Distribution
This is the integral of the density function
Reminder: the CDF (Cumulative Distribution Function) is the probability of the random variable X given a distribution to be lower or equal to x
𝑃 𝑋 ≤ 𝑥 = 𝜙 𝑥 = 𝑓 𝑥 𝑑𝑥𝑥
−∞
Important Properties
𝑃 𝑋 = 𝑥 = 0
𝑃 𝑦 ≤ 𝑋 ≤ 𝑥 = 𝑃(𝑋 ≤ 𝑥)-𝑃(𝑋 ≤ 𝑦)
𝑃 𝑋 ≥ 𝑥 = 1 − 𝑃(𝑋 ≤ 𝑥)
lim𝑥→−∞𝑃 𝑋 ≤ 𝑥 = 0
lim𝑥→+∞𝑃 𝑋 ≤ 𝑥 = 1
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Can’t be expressed with elementary functions: - Help with tables - Help with calculator
Again, let’s discuss about the standard normal first…
𝑃 𝑋 ≤ 0 = 𝑃 𝑋 ≤ −1
𝑃 𝑋 ≤ −2
𝑃 𝑋 ≤ −3
𝑃 −1 ≤ 𝑋 ≤ 1
𝑃 −2 ≤ 𝑋 ≤ 2
𝑃 −3 ≤ 𝑋 ≤ 3
𝑃 𝑋 ≤ −1.645
𝑃 𝑋 ≤ −2.326
0.5
> x=seq(-4,4,length=500)
>plot(x,pnorm(x,mean=0,sd=1),col=
"red",type="l",lwd=3,
xlab="x",ylab="P(X<=x)",
main="Normal Standard CFD")
= 0.05
= 0.01
= 0.158
= 0.023
= 0.001
= 0.682
= 0.954
= 0.996
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>x=seq(-4,4,length=500)
>plot(x,pnorm(x,mean=0,sd=1),co
l="black",type="l",lwd=3,
xlab="x",ylab="P(X<=x)",
main="Normal Distributions -
CFD's")
>lines(x,pnorm(x,mean=0,sd=0.75
),col="red",type="l",lwd=3)
>lines(x,pnorm(x,mean=0,sd=1.25
),col="pink",type="l",lwd=3)
>lines(x,pnorm(x,mean=1,sd=1.25
),col="yellow",type="l",lwd=3)
Identify: N(0,0.75) / N(0,1) / N(0,1.25) / N(1,1.25)
𝑃 𝑋 ≤ 𝜇 = 𝑃 𝑋 ≤ −𝜎 + 𝜇
𝑃 𝑋 ≤ −2 ∗ 𝜎 + 𝜇
𝑃 𝑋 ≤ −3 ∗ 𝜎 + 𝜇
𝑃 𝜇 − 𝜎 ≤ 𝑋 ≤ 𝜇 + 𝜎
𝑃 𝜇 − 2 ∗ 𝜎 ≤ 𝑋 ≤ 𝜇 + 2 ∗ 𝜎
𝑃 𝜇 − 3 ∗ 𝜎 ≤ 𝑋 ≤ 𝜇 + 3 ∗ 𝜎
𝑃 𝑋 ≤ −1.645 ∗ 𝜎 + 𝜇
𝑃 𝑋 ≤ −2.326 ∗ 𝜎 + 𝜇
0.5
General Case
= 0.05
= 0.01
= 0.159
= 0.023
= 0.001
= 0.682
= 0.954
= 0.996
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Standardization
𝑋~𝑁(𝜇, 𝜎)
𝑌 =𝑋 − 𝜇
𝜎
𝑌~𝑁(0,1)
Only one statistical table to use
𝑃 𝑋 ≤ 𝑥 = 𝑃 𝑌 ≤𝑥 − 𝜇
𝜎 𝑌~𝑁(0,1) with
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Let be X~N(2,4) Find:
𝑃 𝑋 ≤ −1.86
𝑃 𝑋 ≤ −1.86 =P 𝑌 ≤−1.86−2
4
With Y~N(0,1)
P 𝑌 ≤ −0.965 =?
Use the table!
Linear interpolation acceptable
P 𝑌 ≤ −0.96 =0.1685
P 𝑌 ≤ −0.97 =0.1660
P 𝑌 ≤ −0.965 =0.16725
P 𝑋 ≤ −1.86 =0.16725
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Back to FCOJ… Let’s compare FCOJ CFD with Normal Distribution (same mean/SD)
>x=seq(-4,4,length=500)
>plot(ecdf(FCOJ$V1),do.points=FALSE, col="red", lwd=3, main="Normal
Distribution against FCOJ - CFD's", xlab="x", ylab="P(X<=x)")
>lines(x,pnorm(x,mean=mean(FCOJ$V1),sd=sd(FCOJ$V1)),col="blue",type="l",l
wd=3)
Where can you see the excess kurtosis?
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>qqnorm(FCOJ$V1)
>qqline(FCOJ$V1)
• This is the QQ Plot to compare the quantiles to a normal distribution • If observations are not on the fitted line, it would suggest a normal distribution
Conclusion?
Following intuition is the first step of descriptive statistics, however, formally testing them is even better! Later step…
Fat Tail
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• Would you rather trade financial product with high or low SD? • Would you rather trade financial product which has return with a negative
mean?
Discussion
SD measures the risk, the volatility: depends on risk appetite
• Mean is irrelevant standalone and you could bet on mean reversion • Very often, the mean is fixed to 0 in finance whatever its real value is
Applications
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Discrete form 𝑑𝑠𝑡 = 𝜇𝑠𝑡𝑑𝑡 + 𝜎𝑠𝑡 𝑑𝑡𝜀
Geometric Brownian Motion
Based on Stochastic Differential Equation 𝑑𝑠𝑡 = 𝜇𝑠𝑡𝑑𝑡 + 𝜎𝑠𝑡𝑊𝑡
with 𝜀~N(0,1)
Used for random walk, martingale, Monte-Carlo, Black & Scholes…
It becomes easy to simulate the price process but what are problems?
Volatility depends on the square root of the time, problem of extrapolation
1% Daily volatility is: • 4.58% Monthly • 7.94% Quarterly • 15.87% Yearly • 35.50% 5 Years • 50.20% 10Years
𝜎𝑇 = 𝜎𝑡 ∗𝑇
𝑡
Is this realistic?
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First Excel problem on the RAND function: • Random number generation is pseudo random • Uniform distribution [0,1] • No seed fixing = Heavy memory usage (new numbers generated when
spreadsheet is recalculated)
3 acceptable solutions: • Assume the generated number is a probability and the invert it with
NORM.INV(RAND(), mean, standard_dev) but fatter tails • Box-Muller method using SQRT(-2*LN(RAND()))*SIN(2*PI()*RAND()) but is
only exact with a perfect uniform random number generation • Central Limit Theorem, normal distribution is approached by 12 uniform
random variables [0,1] subtracting 6, so use RAND()+RAND()+RAND()+RAND()+RAND()+RAND()+RAND()+RAND()+RAND()+RAND()+RAND()+RAND()-6 but fatter tails
Actual normality of such methods will be tested later…
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So Excel is an hassle… Use R! • Proper random number generation on any chosen distribution • Seed fixable • Quicker
Let’s show why it’s better to use a discretisation • Let’s assume a stock with an annual drift (expected return) of 5%, a yearly
volatility of 5%, let’s simulate the price in one year by two methods • One year “one shot” • One year with daily (252 business days) steps
> Drift<-0.05
> Volat<-0.05
> Spot<-100
> Simul<-Spot+Drift*Spot+Volat*Spot*rnorm(10)
> plot(c(Spot,Simul[1]), type="l",
ylim=c(min(Simul)-1,max(Simul)+1),
main="Simulation one shot", xlab="T", ylab="S")
> lines(c(Spot,Simul[2]), type="l")
> lines(c(Spot,Simul[3]), type="l")
> lines(c(Spot,Simul[4]), type="l")
> lines(c(Spot,Simul[5]), type="l")
> lines(c(Spot,Simul[6]), type="l")
> lines(c(Spot,Simul[7]), type="l")
> lines(c(Spot,Simul[8]), type="l")
> lines(c(Spot,Simul[9]), type="l")
> lines(c(Spot,Simul[10]), type="l")
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> summary(Simul)
Min. 1st Qu. Median Mean 3rd Qu. Max.
96.51 105.10 107.00 106.60 108.80 116.50
> sd(Simul)
[1] 5.23066
Very sensitive to the random number picked
20.99 difference between the lowest and highest scenario
SD of 5.23 in the results
What would be the mean in a perfect situation?
Very sensitive to the number of trials
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library(sde)
require(sde)
nbsim<-252
Drift<-0.05
Volat<-0.05
Spot<-100
G1<-GBM(x=Spot,r=Drift, sigma=Volat,N=nbsim)
G2<-GBM(x=Spot,r=Drift, sigma=Volat,N=nbsim)
G3<-GBM(x=Spot,r=Drift, sigma=Volat,N=nbsim)
G4<-GBM(x=Spot,r=Drift, sigma=Volat,N=nbsim)
G5<-GBM(x=Spot,r=Drift, sigma=Volat,N=nbsim)
G6<-GBM(x=Spot,r=Drift, sigma=Volat,N=nbsim)
G7<-GBM(x=Spot,r=Drift, sigma=Volat,N=nbsim)
G8<-GBM(x=Spot,r=Drift, sigma=Volat,N=nbsim)
G9<-GBM(x=Spot,r=Drift, sigma=Volat,N=nbsim)
G10<-GBM(x=Spot,r=Drift, sigma=Volat,N=nbsim)
plot(G1,ylim=c(90,115), col=1, main="GBM day by day",
xlab="T", ylab="S")
lines(G2, col=2)
lines(G3, col=3)
lines(G4, col=4)
lines(G5, col=5)
lines(G6, col=6)
lines(G7, col=7)
lines(G8, col=8)
lines(G9, col=9)
lines(G10, col=10)
Use the package sde of R for the step by step (discrete) method
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> FinalS<-
c(G1[nbsim+1],G2[nbsim+1],G3[nbsim+1],G4[nbsim+1],G5[nbsim+1],G6[nbsim+1],G7[
nbsim+1],G8[nbsim+1],G9[nbsim+1],G10[nbsim+1])
> summary(FinalS)
Min. 1st Qu. Median Mean 3rd Qu. Max.
97.81 101.80 103.00 103.70 105.80 109.00
> sd(FinalS)
[1] 3.535826
Lower sensitive to the random numbers chosen
11.29 difference between the lowest and highest scenario
SD of 3.54
Still sensitive to the number of trials
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Introduction to LogNormaility
Do you remember the slide number 6?
𝑅𝑙𝑖𝑛 = 𝑒𝑅𝑙𝑛 − 1 𝑉𝑡 = 𝑉𝑡−1 ∗ (1 + 𝑅𝑙𝑖𝑛)
𝑉𝑡 = 𝑉𝑡−1 ∗ 𝑒𝑅𝑙𝑛 𝑅𝑙𝑛 = 𝑙𝑛
𝑅𝑙𝑖𝑛+1
FCOJ<-read.csv(file="S:/Vincent/FCOJStats.csv",head=FALSE,sep=",")
FCOJ$V1<-log(FCOJ$V1+1)
hist(FCOJ$V1,breaks=100, main="FCOJ LogReturns / Normal
Distribution",xlab="LogReturn",ylab="Occurence")
x=seq(-0.2,0.2,length=200)
y1=dnorm(x,mean=mean(FCOJ$V1),sd=sd(FCOJ$V1))
lines(x,y1,type="l",lwd=3,col="red")
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The LogReturns seem normal (ish) distributed
If LogReturns are normally distributed, the stock price is log normally distributed (useful property as it’s bounded by 0 and it allows to use continuous compounded returns)
𝑆𝑡 = 𝑆𝑡−1𝑒𝑟𝑑𝑡 𝑆𝑡−1 = 𝑆𝑡𝑒
−𝑟𝑑𝑡
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Black & Scholes
Let’s look at the underling price diffusion process through another angle
𝑅𝑓 + μ
Time
𝑆
Job done, isn’t it?
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Price distribution of the underlying at maturity
Payoff distribution of the option at maturity can be deducted
Expected Payoff can be calculated
Present value of the expected payoff is the option price!
Pricing Principle
• No arbitrage opportunity (no free lunch). • Existence of a risk-free rate (borrower and lender). • No liquidity problem on long and short positions. • No fees or costs. • Market efficiency. • Stock price follows a geometric Brownian motion with constant drift and volatility. • No dividend.
Assumptions
This is obviously not true… Very strong assumptions!
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Geometric Brownian Motion & Black & Scholes Option Valuation
Based on Stochastic Differential Equation 𝑑𝑠𝑡 = 𝜇𝑠𝑡𝑑𝑡 + 𝜎𝑠𝑡𝑊𝑡
is a Brownian Motion, in other word a random walk following a normal distribution (zero mean)
𝑊𝑡
𝑑𝑆
𝑆= 𝜇𝑑𝑡 + 𝜎𝑑𝑊
A small variation of price has an expected return of 𝜇 (known, drift) and a standard deviation of 𝜎𝑑𝑊 (uncertain, diffusion)
Over longer horizons, the price is lognormally distributed (then it can’t go below 0, we’ll come back to this)
Risk neutral probability: an option perfectly hedge on continuous basis is risk free and portfolio earns the risk free rate. Drift then has no impact
Demonstration based on integration with Ito Lemma and risk neutral probability (11.6 / 12.7 in John Hull)
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Pricing Formulas
𝐶 = 𝑁 𝑑1 ∗ 𝑆 − 𝑁 𝑑2 ∗ 𝐾 ∗ 𝑒−𝑟∗(𝑇−𝑡)
𝑑1 =𝑙𝑛𝑆𝐾+ 𝑟 +
𝜎2
2∗ (𝑇 − 𝑡)
𝜎 𝑇 − 𝑡
𝑑2 = 𝑑1 − 𝜎 𝑇 − 𝑡
𝑃 = −𝑁 −𝑑1 ∗ 𝑆 + 𝑁 −𝑑2 ∗ 𝐾 ∗ 𝑒−𝑟∗(𝑇−𝑡)
Buy the Call, Sell the Put… Arbitrage?
𝐶 − 𝑃 = 𝑁 𝑑1 ∗ 𝑆 − 𝑁 𝑑2 ∗ 𝐾 ∗ 𝑒−𝑟∗ 𝑇−𝑡 + 𝑁 −𝑑1 ∗ 𝑆 − 𝑁 −𝑑2 ∗ 𝐾 ∗ 𝑒
−𝑟∗(𝑇−𝑡)
𝐶 − 𝑃 = 𝑁 𝑑1 ∗ 𝑆 − 𝑁 𝑑2 ∗ 𝐾 ∗ 𝑒−𝑟∗ 𝑇−𝑡 +
1 − 𝑁 𝑑1 ∗ 𝑆 − 1 − 𝑁 𝑑2 ∗ 𝐾 ∗ 𝑒−𝑟∗(𝑇−𝑡)
𝐶 − 𝑃 = 𝑆 − 𝐾 ∗ 𝑒−𝑟∗(𝑇−𝑡)
The price difference is the present value of the difference to the strike No arbitrage opportunity!
When does C=P?
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Greeks - Delta
∆𝐶=𝜕𝐶
𝜕𝑆= 𝑁(𝑑1)
• First derivative of the value of the option with respect to the underlying price S
• Underlying equivalent position • Probability of the option to be at the money at
expiry
Delta ~0.5 if…
Delta [0,1] if…
Delta [-1,0] if…
S is the present value of the strike for a call
For a Call
For a Put
∆𝑃= 𝑁(𝑑1) − 1
What is the exact delta of a Long Call ATMF?
What is the delta of a combined Long Call and Long Put ATMF?
What is the delta of a combined Long Call and Short Put ATMF?
What is the new price of the Call ($7.9683) if S moves up $1.5 with delta=0.5398?
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Greeks - Gamma
• Second derivative of the value of the option with respect to the underlying price S
• First derivative of the value of the delta with respect to the underlying price S
• Pace of the delta movement • Second order Greek
Gamma [0,1] if…
Gamma [-1,0] if…
Gamma=max if…
Long option
Short option
ATMF
𝛾 =𝜕∆
𝜕𝑆=𝑁′(𝑑1)
𝑠𝜎 𝑇 − 𝑡
What is the new price of the Call ($7.9683) if S moves up $1.5 with delta=0.5398 and a gamma of 0.0198?
Need to use second order central finite difference (Taylor Series)
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Greeks – Delta/Gamma
Forgetting Gamma is dangerous, difference is 0.25% in our example!
𝑑𝐶 = 𝐶 + ∆ ∗ 𝑑𝑆 +1
2∗ 𝛾 ∗ 𝑑𝑆2
What is the new delta?
8.8003
0.5695
How to delta hedge and gamma hedge?
Third order known as Speed, hardly used…
Write the Taylor Development until the Speed level… 1
6∗ 𝑆𝑝𝑒𝑒𝑑 ∗ 𝑑𝑆3
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Greeks - Vega
• First derivative of the value of the option with respect to the implied volatility
• Volatility sensitivity • First order Greek
Vega [0,1] if…
Vega [-1,0] if…
Long option
Short option
𝜏 =𝜕𝐶
𝜕𝜎= 𝑆𝑁′(𝑑1) 𝑇 − 𝑡
What is the new price of the Call ($7.9683) if the volatility moves up 1.5 point with a 0.7942 Vega?
Note, it’s not an actual Greek letter! Tau is used…
Second order exists as Vanna, third order as Vomma… Hardly used as it can’t be hedged easily. Volatility of the volatility is THE BIG problem in finance!
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Greeks - Theta
𝜃𝐶 = −𝑆𝑁′ 𝑑1 𝜎
2 𝑇 − 𝑡− 𝑟𝐾𝑒−𝑟 𝑇−𝑡 𝑁(𝑑2)
𝜃𝑃 = −𝑆𝑁′ 𝑑1 𝜎
2 𝑇 − 𝑡+ 𝑟𝐾𝑒−𝑟 𝑇−𝑡 𝑁(−𝑑2)
Simply the time decay 𝜃 =𝜕𝑉
𝜕𝑡
Theta >0 if…
Theta <0 if…
Short option
Long option
Time has as well noticeable effects on Delta (Charm), Gamma (Color) and Vega (DvegaDtime)
What is the new price of the Call ($7.9683) in 2 days with -0.9920 Theta?
Theta is am annual value
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Greeks - Rho
𝜌𝐶 =𝜕𝐶
𝜕𝑟= 𝐾 𝑇 − 𝑡 𝑒−𝑟(𝑇−𝑇)𝑁(𝑑2)
• First derivative of the value of the option with respect to the interest rate
𝜌𝑃 = −𝐾 𝑇 − 𝑡 𝑒−𝑟 𝑇−𝑡 𝑁(−𝑑2)
What is the new price of the Call ($7.9683) if r moves up 1 basis point with Rho=184.1895?
Careful, high convexity. Need a second order for extreme movement.
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Sum Up - Example
What is the new price of the Call ($7.9683) if S moves up $1.5 with delta=0.5398 and a gamma of 0.0198, volatility moves up 1.5 point with a 0.7942 Vega, r moves up 1 basis point with Rho=184.1895 and placing you 2 days after with a final Theta of -0.9920?
10.0147
Real pricing: 10.0094
Difference of only 0.05% mainly due to the other effects on Greeks by time decay but it’s pretty close!
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Sum Up Greeks/Time
Call 100, S=105, r=5%, Maturity from 4y, Vol=10%
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Sum Up Greeks/Spot Price
Call 100, r=5%, Maturity 4y, Vol=10%
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Sum Up Greeks/Strike
S=105, r=5%, Maturity 4y, Vol=10%
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Sum Up Greeks/Vol
Call 100, S=105, r=5%, Maturity 4y
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Conclusion on B&S
Great, easy, quick
Strong assumptions, continuous
Only European option
We need a path dependant method!
It will allow to include early exercise, dividend, pricing European digital,…
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Binomial Model (Cox, Ross, Rubinstein, 1979)
Path dependent (valuation of European options, American options, Digital,…)
May include dividends
Discretisation of the continuous random walk
Why?
How?
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Binomial Model: principles
Construct a tree lattice representing the stock price following a GBM
Price the option by backwards induction
“Slice” maturity in a predefined number of steps
3 Steps
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Let’s assume the maturity is divided by 2
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Cox Ross Rubinstein up and down factors based on GBM
Do you see other methods? Which? Why? Which one are better?
At each node, S goes up or down by one SD
𝑢 = 𝑒𝜎 𝑡
𝑑 =1
𝑢= 𝑒−𝜎 𝑡
𝑓𝑎𝑐𝑡𝑜𝑟𝑠 = 1
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Let’s build a tree with 3 steps, with S=100, σ=10%, 1.5 year to maturity
100
𝑢 = 𝑒𝜎 𝑡 = 𝑒0.1 0.5 = 1.073271
𝑑 = 𝑒−𝜎 𝑡 = 𝑒−0.1 0.5 = 0.931731
107.33
93.17
115.19
100
86.81
107.33
93.17
123.63
80.89
Be clever building it!
What happened to the drift implied by the risk free rate?
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𝑆𝑛 = 𝑆0 ∗ 𝑢𝑁𝑢−𝑁𝑑
What is the price of the stock at any given node?
𝑛 + 1
How many nodes do you have at the end of the tree?
𝑆
If number of steps are even, what’s the value of the middle node on the last step?
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100
107.33
93.17
115.19
100
86.81
107.33
93.17
123.63
80.89
Having S at maturity, it’s easy to have the price of a EU Call 105 at maturity
0
0
2.33
18.63
Backward inductions, we have the probabilities, let’s assume a risk free rate of 5%
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115.19
107.33
123.63
2.33
18.63
u
d
Need to calculate the new probabilities integrating the Risk Free Rate to comply with the risk neutrality assumption
S𝑒𝑟𝑡 = 𝑝𝑆𝑢 + 1 − 𝑝 𝑆𝑑 𝑒𝑟𝑡 = 𝑝𝑢 + 1 − 𝑝 𝑑
𝑝 =𝑒𝑟𝑡 − 𝑑
𝑢 − 𝑑
Therfore:
BV= OpUp ∗ p + OpDown ∗ 1 − p ∗ 𝑒−𝑟𝑡
12.78
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100
107.33
93.17
115.19
100
86.81
107.33
93.17
123.63
80.89
0
0
2.33
18.63
12.78
1.5
0
8.74
0.97
5.96
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A European 105 Call option with 1.5 years to Maturity, a Volatility of 10% and a risk free rate of 5% with three steps worth 5.96
How much with B&S? 6.22
Significant difference, why?
Sensitivity to the number of steps
The more step, the less discrete, the more continuous
Extrapolated to the infinite, you’d find your GBM and so B&S!
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5.9
5.95
6
6.05
6.1
6.15
6.2
6.25
6.3
6.35
6.4
1 6 11 16 21 26 31 36 41 46 51 56 61 66 71
CRB
BS
B&S / CRR Convergence: usually 40 steps are reasonable
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I meant American Option! Let’s start all over again…
CRR main advantage is the ability to price American Options
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100
107.33
93.17
115.19
100
86.81
107.33
93.17
123.63
80.89
0
0
2.33
18.63
13.84
1.5
0
Binomial Value
On each node you need to check any early exercise possibility
Intrinsic value
10.19
0
0
But sometimes holding is better than exercising and in this case no early exercise worth and price of the European Call and American Call will be the same
8.74
0.97
2.33
0
5.96
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Pricing of an American Put option, S=50, K=50 with a 10% risk free rate, a 40% volatility, 5 steps and time to maturity 0.4167 year.
Tree of stock price
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Binomial Value at the next to last and last node (i.e. Valuating as if it was a European Put)
0
0
0
5.45
14.64
21.93
0
0
2.66
9.90
18.08
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Any early exercise worth?
0
0
0
5.45
14.64
21.93
0
0
2.66
9.90
18.08
0
0
0
10.31
18.5
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Finally…
0
0
0
5.45
14.64
21.93
0
0
2.66
10.31
18.5
0
1.30
6.38
14.64
0.64
3.77
10.36
6.96
2.16 4.49
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The American Put worth 4.49
The European Put worth 4.32
Difference can be non negligible
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Pricing of an European Digital Put option, Q=15, S=50, K=50 with a 10% risk free rate, a 40% volatility, 5 steps and time to maturity 0.4167 year.
Tree of stock price
The pay-off at maturity is binary: 0 if out of the money, Q if in the money
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Last node pay off is then straight forward
0
0
0
15
15
15
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Then method doesn’t change… Backward induction.
0
0
0
15
15
15
0
0
7.33
14.88
14.88
14.75
10.96
3.58
0
1.75
7.15
12.72
9.81
4.38
7.00
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Pricing of an Bermuda Put option, S=50, K=50 with a 10% risk free rate, a 40% volatility, 5 steps and time to maturity 0.4167 year.
Tree of stock price
Let’s suppose this Bermuda can only be exercised between the 4th and 5th step
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Any early exercise worth?
0
0
0
5.45
14.64
21.93
0
0
2.66
9.90
18.08
0
0
0
10.31
18.5
No exercises on lower steps
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Finally…
0
0
0
5.45
14.64
21.93
0
0
2.66
10.31
18.5
0
1.30
6.38
14.22
0.64
3.77
10.16
6.86
2.16 4.44
A “full” American option would have been exercised, not this one
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Pricing of an Put option, S=50, K=50 with a 10% risk free rate, a 40% volatility, 5 steps and time to maturity 0.4167 year, paying a $2.06 dividend on the in 3.5 months.
Construct the usual tree
Subtract the present value of the dividend on each node before it occurs
Pricing can continue as usual
3 Steps
The dividend occurs between the 3rd and 4th step
Value at step 0
Value at step 1
Value at step 2
Value at step 3
𝑃𝑉 = 2.06 ∗ 𝑒−10%∗3.512 = 2
𝑃𝑉 = 2.06 ∗ 𝑒−10%∗
3.512−0.41675 = 2.02
𝑃𝑉 = 2.06 ∗ 𝑒−10%∗
3.512−0.4167∗25 = 2.03
𝑃𝑉 = 2.06 ∗ 𝑒−10%∗
3.512−0.4167∗35 = 2.05
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Tree of stock price impacted of dividends
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Pricing by the usual backward induction (don’t forget potential early exercise)
0
0
0
0
0 0
2.66
10.31
18.50
1.30
6.38
14.22
10.16
3.77
6.86
2.16
4.44 5.45
14.64
21.93
0.64
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The American Put worth 4.49
The European Put worth 4.32
The Digital Put paying 15 worth 7.00
The Bermuda Put with exercise on the lath fifth of the maturity worth 4.44
You can virtually price anything you want!
CRR Sum-Up
What can’t you price?
The American Put paying a 2.06 dividend worth 4.44
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Pricing of an Barrier Put option, S=50, K=50 with a 10% risk free rate, a 40% volatility, 5 steps and time to maturity 0.4167 year with a knock out barrier at 60
Tree of stock price
The option is cancelled if S goes to 60
Way to reduce the price of the option
KO
You can’t tell how much worth the option on this final node: 0 or 5.45?
0
5.45
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CRR Extension
How to converge faster to the correct option price?
Put a third factor
• Up • Down • Stable
Careful, the tree has to recombine!
YES NO
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Conclusion
Normal Distribution
GBM
B&S
CRR