Applied Statistics I
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Applied Statistics I

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First course of Applied Statistics, MSc level in Buisiness School.

First course of Applied Statistics, MSc level in Buisiness School.

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Applied Statistics I Applied Statistics I Presentation Transcript

  • ESGF 4IFM Q1 2012 Applied StatisticsVincent JEANNIN – ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 1
  • ESGF 4IFM Q1 2012Summary of the session (est. 4.5h)• Introduction & Objectives• Bibliography• First approach: Descriptive Statistics vinzjeannin@hotmail.com• The Normal Distribution• Applications (GBM, B&S, Greeks, CRR) 2
  • Introduction & Objectives • What are statistics? ESGF 4IFM Q1 2012 • Why Should you use them? Describe data behaviour vinzjeannin@hotmail.com Modelise data behaviour Business decisions (pricing, investments,…)• Take the opportunity to remember financial mathematics basics• Acquire theory knowledge on statistics• Usage of R and Excel 3
  • Bibliography vinzjeannin@hotmail.com ESGF 4IFM Q1 20124
  • First Approach: Descriptivestatistics FCOJ Front Month: 31st Dec 2008 / 30th Sep 2011 ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 5
  • First step, calculate the linear returns = −1 −1 Then, the mean 1 = ESGF 4IFM Q1 2012 =1 Expected return, not average return!How to calculate average return on the period? (compound return) vinzjeannin@hotmail.comHow to obtain it by a sum? 2 1 = = ln 2 − ln 1 1 3 2 = = ln 3 − ln 2 2 1 −1 = −1 −1 = = ln −1 − ln −2 −2 = = ln − ln −1 −1 6 = = ln − ln 1 = −1 1 =2
  • Excel and R can give an idea of the distribution Excel, functions Min, Max, Average, Percentile ESGF 4IFM Q1 2012 • Free vinzjeannin@hotmail.com • Open Source • Developments shared by developers R, easier, faster,… Function summary 7
  • R can easily show the distribution of returns ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 8 Interesting shape but what next?
  • The four moments 1 Mean = =1 ESGF 4IFM Q1 2012 Expected Return Standard Deviation Dispersion from the mean vinzjeannin@hotmail.com Square root of the variance = − 2 SD is the square root of the mean of squared differences to the mean 9 1 2 = − =1
  • Quick check: what is the SD of {1,2,-3,0,-2,1,1}? ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com Excel function STDEVP R function sd 10 FCOJ has a SD of 2.16%
  • > xBar<-mean(FCOJ$V1)> SD <- sd(FCOJ$V1)> hist(FCOJ$V1, breaks=c(xBar-6*SD,xBar-5*SD,xBar-4*SD,xBar-3*SD,xBar-2*SD,xBar-SD,xBar,xBar+SD,xBar+2*SD,xBar+3*SD,xBar+4*SD,xBar+5*SD,xBar+6*SD),main="FCOJ Returns",xlab="Return",ylab="Occurence") ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com Histogram centred on the mean with SD multiples groups 11 Symmetric-ish
  • ESGF 4IFM Q1 2012693 data75.04% within ±1 vinzjeannin@hotmail.com94.23% within ±298.99% within ±3 12
  • Skewness, the third moment Asymmetry of the distribution ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com• Negative skew: long left tail, mass on the right, skew to the left• Positive skew: long right tail, mass on the left, skew to the right 3 − − 3 = = 13 − 2 3/2 Should I rather buy or sell a positive skewed asset?
  • ESGF 4IFM Q1 2012 vinzjeannin@hotmail.comExcel function SKEWR function skewness (package moments)> require(moments)> library(moments)> skewness(FCOJ$V1)[1] 0.2030842 14 FCOJ is positively skewed
  • Kurtosis, the fourth moment Peakedness of the distribution 4 − − 4 ESGF 4IFM Q1 2012 = = − 2 2 It’s a usage to deal with the excess kurtosis (relative to the normal distribution, subtracting 3 vinzjeannin@hotmail.com• Positive excess Kurtosis: high peak around the mean, fat tails• Negative excess Kurtosis: low peak around the mean, thin tails 15 Which distribution you’d rather buy or sell?
  • 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") 16> sum(toss)[1] 5001777
  • 50.01777% rate of success: fair or not fair? Trick coin ? Will 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 − ) 17 Easy to demonstrate if p=0.5 the Kurtosis will be the lowest Bit more complicated to demonstrate it for any distribution
  • ESGF 4IFM Q1 2012 vinzjeannin@hotmail.comExcel function KURTR function kurtosis (package moments)> require(moments)> library(moments)> kurtosis(FCOJ$V1)[1] 6.34176> kurtosis(FCOJ$V1)[1] 3.34176 18 FCOJ is leptokurtic
  • Sum-up: • Positive expected return ESGF 4IFM Q1 2012 • Positive skew • Positive excess kurtosis Buy or Sell? vinzjeannin@hotmail.comIs that actually enough to take investment decision?What next?How different is the FCOJ distribution from the Normal Distribution? 19
  • The Normal DistributionLet’s discuss about the standard normal first… ESGF 4IFM Q1 2012 Snapshot, 4 moments: Mean 0 SD 1 vinzjeannin@hotmail.com Skewness 0 Kurtosis 3 Snapshot, Shape: 20
  • Notation (, ) 1 (−)2 − Density = 22 2 2 ESGF 4IFM Q1 2012Distributions of zeros means with following SD: 0.5 / 0.75 / 1 / 1.5 / 2 Which one is which one? vinzjeannin@hotmail.com 21
  • > 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)") ESGF 4IFM Q1 2012 > 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") vinzjeannin@hotmail.com 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 SDFCOJ<-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)) 22hist(FCOJ$V1, breaks=100,main="FCOJ Returns / NormalDistribution",xlab="Return",ylab="Occurence")lines(x,y1,type="l",lwd=3,col="red")
  • ESGF 4IFM Q1 2012 vinzjeannin@hotmail.comThe excess Kurtosis sign is obvious, isn’t it? 23
  • Same SD, different mean, more straight forward ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 24
  • Cumulative DistributionReminder: the CDF (Cumulative Distribution Function) is the probabilityof the random variable X given a distribution to be lower or equal to x ESGF 4IFM Q1 2012 This is the integral of the density function ≤ = = −∞ Important Properties vinzjeannin@hotmail.com = = 0 ≥ = 1 − ( ≤ ) ≤ ≤ = ( ≤ )-( ≤ ) lim ≤ = 0 25 →−∞ lim ≤ = 1 →+∞
  • Can’t be expressed with elementary functions: - Help with tables - Help with calculator Again, let’s discuss about the standard normal first… ESGF 4IFM Q1 2012 ≤ 0 = 0.5 ≤ −1 = 0.158 −1 ≤ ≤ 1 = 0.682 ≤ −1.645 = 0.05 ≤ −2 = 0.023 −2 ≤ ≤ 2 = 0.954 ≤ −2.326 = 0.01 ≤ −3 = 0.001 −3 ≤ ≤ 3 = 0.996 vinzjeannin@hotmail.com > 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") 26
  • General Case ≤ = 0.5 − ≤ ≤ + = 0.682 ≤ − + = 0.159 ≤ −1.645 ∗ + = 0.05 − 2 ∗ ≤ ≤ + 2 ∗ = 0.954 ≤ −2 ∗ + = 0.023 ESGF 4IFM Q1 2012 ≤ −2.326 ∗ + = 0.01 − 3 ∗ ≤ ≤ + 3 ∗ = 0.996 ≤ −3 ∗ + = 0.001 vinzjeannin@hotmail.com Identify: N(0,0.75) / N(0,1) / N(0,1.25) / N(1,1.25) >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 - CFDs") >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) 27 >lines(x,pnorm(x,mean=1,sd=1.25 ),col="yellow",type="l",lwd=3)
  • Standardization ~(, ) ESGF 4IFM Q1 2012 − = ~(0,1) vinzjeannin@hotmail.com Only one statistical table to use − ≤ = ≤ with ~(0,1) 28
  • Let be X~N(2,4) Find: ≤ −1.86 ESGF 4IFM Q1 2012 −1.86−2 ≤ −1.86 =P ≤ 4 With Y~N(0,1) P ≤ −0.965 =? vinzjeannin@hotmail.com Use the table! Linear interpolation acceptable P ≤ −0.96 =0.1685 P ≤ −0.97 =0.1660 29 P ≤ −0.965 =0.16725 P ≤ −1.86 =0.16725
  • 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="NormalDistribution against FCOJ - CFDs", xlab="x", ylab="P(X<=x)")>lines(x,pnorm(x,mean=mean(FCOJ$V1),sd=sd(FCOJ$V1)),col="blue",type="l",l ESGF 4IFM Q1 2012wd=3) vinzjeannin@hotmail.com 30 Where can you see the excess kurtosis?
  • >qqnorm(FCOJ$V1) >qqline(FCOJ$V1) ESGF 4IFM Q1 2012Fat Tail vinzjeannin@hotmail.com• 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? 31 Following intuition is the first step of descriptive statistics, however, formally testing them is even better! Later step…
  • Discussion ESGF 4IFM Q1 2012• Would you rather trade financial product with high or low SD?• Would you rather trade financial product which has return with a negative vinzjeannin@hotmail.com mean? 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 32
  • Applications Geometric Brownian Motion ESGF 4IFM Q1 2012 Based on Stochastic Differential Equation = + Discrete form = + with ~N(0,1) Used for random walk, martingale, Monte-Carlo, Black & Scholes… vinzjeannin@hotmail.com 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 33 • 50.20% 10Years Is this realistic?
  • 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) ESGF 4IFM Q1 2012 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 vinzjeannin@hotmail.com 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… 34
  • So Excel is an hassle… Use R! • Proper random number generation on any chosen distribution • Seed fixable • Quicker ESGF 4IFM Q1 2012 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 vinzjeannin@hotmail.com> 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") 35> lines(c(Spot,Simul[8]), type="l")> lines(c(Spot,Simul[9]), type="l")> lines(c(Spot,Simul[10]), type="l")
  • ESGF 4IFM Q1 2012> 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 vinzjeannin@hotmail.com Very sensitive to the random number picked Very sensitive to the number of trials 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? 36
  • Use the package sde of R for the step by step (discrete) methodlibrary(sde)require(sde)nbsim<-252Drift<-0.05Volat<-0.05 ESGF 4IFM Q1 2012Spot<-100G1<-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) vinzjeannin@hotmail.comG6<-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) 37lines(G8, col=8)lines(G9, col=9)lines(G10, col=10)
  • ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com> 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 38 Still sensitive to the number of trials
  • Introduction to LogNormaility Do you remember the slide number 6? ESGF 4IFM Q1 2012 = −1 ∗ (1 + ) = − 1 = −1 ∗ = +1 vinzjeannin@hotmail.comFCOJ<-read.csv(file="S:/Vincent/FCOJStats.csv",head=FALSE,sep=",")FCOJ$V1<-log(FCOJ$V1+1)hist(FCOJ$V1,breaks=100, main="FCOJ LogReturns / NormalDistribution",xlab="LogReturn",ylab="Occurence")x=seq(-0.2,0.2,length=200)y1=dnorm(x,mean=mean(FCOJ$V1),sd=sd(FCOJ$V1))lines(x,y1,type="l",lwd=3,col="red") 39
  • ESGF 4IFM Q1 2012 vinzjeannin@hotmail.comThe LogReturns seem normal (ish) distributedIf LogReturns are normally distributed, the stock price is lognormally distributed (useful property as it’s bounded by 0and it allows to use continuous compounded returns) 40 = −1 −1 = −
  • Black & Scholes Let’s look at the underling price diffusion process through another angle ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com + μ Time Job done, isn’t it? 41
  • Pricing Principle Price distribution of the underlying at maturity Payoff distribution of the option at maturity can be deducted Expected Payoff can be calculated ESGF 4IFM Q1 2012 Present value of the expected payoff is the option price! Assumptions vinzjeannin@hotmail.com• 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. This is obviously not true… Very strong assumptions! 42
  • Geometric Brownian Motion & Black & Scholes Option ValuationBased on Stochastic Differential Equation = + ESGF 4IFM Q1 2012 is a Brownian Motion, in other word a random walk following a normal distribution (zero mean) Demonstration based on integration with = + Ito Lemma and risk neutral probability vinzjeannin@hotmail.com (11.6 / 12.7 in John Hull) A small variation of price has an expected return of (known, drift) and a standard deviation of (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 43 basis is risk free and portfolio earns the risk free rate. Drift then has no impact
  • Pricing Formulas = 1 ∗ − 2 ∗ ∗ −∗(−) = − −1 ∗ + −2 ∗ ∗ −∗(−) 2 ESGF 4IFM Q1 2012 + + ∗ ( − ) 2 1 = − 2 = 1 − − vinzjeannin@hotmail.com Buy the Call, Sell the Put… Arbitrage? − = 1 ∗ − 2 ∗ ∗ −∗ − + −1 ∗ − −2 ∗ ∗ −∗(−) − = 1 ∗ − 2 ∗ ∗ −∗ − + 1 − 1 ∗ − 1 − 2 ∗ ∗ −∗(−) − = − ∗ −∗(−) The price difference is the present value of the difference to the strike 44 No arbitrage opportunity! When does C=P?
  • Greeks - Delta ∆ = = (1 ) • First derivative of the value of the option with∆ = (1 ) − 1 ESGF 4IFM Q1 2012 respect to the underlying price S • Underlying equivalent position • Probability of the option to be at the money at expiry vinzjeannin@hotmail.com Delta ~0.5 if… S is the present value of the strike for a call Delta [0,1] if… For a Call Delta [-1,0] if… For a Put 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? 45 What is the new price of the Call ($7.9683) if S moves up $1.5 with delta=0.5398?
  • Greeks - Gamma ∆ ′ (1 ) = = • Second derivative of the value of the option with − ESGF 4IFM Q1 2012 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 vinzjeannin@hotmail.com Gamma [0,1] if… Long option Gamma [-1,0] if… Short option Gamma=max if… ATMF 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? 46 Need to use second order central finite difference (Taylor Series)
  • Greeks – Delta/Gamma 1 = + ∆ ∗ + ∗ ∗ 2 ESGF 4IFM Q1 2012 28.8003Forgetting Gamma is dangerous, difference is 0.25% in our example! vinzjeannin@hotmail.comWhat is the new delta?0.5695Third order known as Speed, hardly used… 1 47 Write the Taylor Development until the Speed level… ∗ ∗ 3 6How to delta hedge and gamma hedge?
  • Greeks - Vega Note, it’s not an actual Greek letter! Tau is used… = = ′ (1 ) − • First derivative of the value of the option with ESGF 4IFM Q1 2012 respect to the implied volatility • Volatility sensitivity • First order Greek vinzjeannin@hotmail.com Vega [0,1] if… Long option Vega [-1,0] if… Short option What is the new price of the Call ($7.9683) if the volatility moves up 1.5 point with a 0.7942 Vega? 48 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!
  • Greeks - Theta Simply the time decay = ESGF 4IFM Q1 2012 ′ 1 = − − − − (2 ) 2 − ′ 1 = − + − − (−2 ) vinzjeannin@hotmail.com 2 − Theta >0 if… Short optionTheta <0 if… Long option Theta is am annual value Time has as well noticeable effects on Delta (Charm), Gamma (Color) and Vega (DvegaDtime) 49 What is the new price of the Call ($7.9683) in 2 days with -0.9920 Theta?
  • Greeks - Rho = = − −(−) (2 ) = − − − − (−2 ) ESGF 4IFM Q1 2012 • First derivative of the value of the option with vinzjeannin@hotmail.com respect to the interest rateWhat is the new price of the Call ($7.9683) if r moves up 1 basis point withRho=184.1895?Careful, high convexity. Need a second order for extreme movement. 50
  • Sum Up - ExampleWhat is the new price of the Call ($7.9683) if S moves up $1.5 with ESGF 4IFM Q1 2012delta=0.5398 and a gamma of 0.0198, volatility moves up 1.5 pointwith a 0.7942 Vega, r moves up 1 basis point with Rho=184.1895 andplacing you 2 days after with a final Theta of -0.9920? vinzjeannin@hotmail.com 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! 51
  • Sum Up Greeks/TimeCall 100, S=105, r=5%, Maturity from 4y, Vol=10% ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 52
  • Sum Up Greeks/Spot PriceCall 100, r=5%, Maturity 4y, Vol=10% ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 53
  • Sum Up Greeks/StrikeS=105, r=5%, Maturity 4y, Vol=10% ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 54
  • Sum Up Greeks/VolCall 100, S=105, r=5%, Maturity 4y ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 55
  • Conclusion on B&SGreat, easy, quick ESGF 4IFM Q1 2012Strong assumptions, continuousOnly European option vinzjeannin@hotmail.comWe need a path dependant method!It will allow to include early exercise, dividend, pricingEuropean digital,… 56
  • Binomial Model (Cox, Ross, Rubinstein, 1979) ESGF 4IFM Q1 2012Why? Path dependent (valuation of European options, American options, Digital,…) May include dividends vinzjeannin@hotmail.comHow? Discretisation of the continuous random walk 57
  • Binomial Model: principles ESGF 4IFM Q1 20123 Steps “Slice” maturity in a predefined number of steps Construct a tree lattice representing the stock price vinzjeannin@hotmail.com following a GBM Price the option by backwards induction 58
  • Let’s assume the maturity is divided by 2 ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 59
  • Cox Ross Rubinstein up and down factors based on GBMAt each node, S goes up or down by one SD ESGF 4IFM Q1 2012 = vinzjeannin@hotmail.com 1 = = − Do you see other methods? Which? Why? Which one are better? = 1 60
  • Let’s build a tree with 3 steps, with S=100, σ=10%, 1.5 year to maturity = = 0.1 0.5 = 1.073271 = − = −0.1 0.5 = 0.931731 ESGF 4IFM Q1 2012 123.63 115.19 107.33 107.33 vinzjeannin@hotmail.com100 100 93.17 93.17 86.81 80.89 Be clever building it! 61 What happened to the drift implied by the risk free rate?
  • What is the price of the stock at any given node? = 0 ∗ − ESGF 4IFM Q1 2012How many nodes do you have at the end of the tree? vinzjeannin@hotmail.com + 1If number of steps are even, what’s the value of the middle node on the last step? 62
  • Having S at maturity, it’s easy to have the price of a EU Call 105 at maturity ESGF 4IFM Q1 2012 123.63 115.19 18.63 107.33 107.33 vinzjeannin@hotmail.com 100 100 2.33 93.17 93.17 86.81 0 80.89 0Backward inductions, we have the probabilities, let’s assume a 63risk free rate of 5%
  • u 123.63 115.19 18.63 ESGF 4IFM Q1 2012 107.33 d 2.33 Need to calculate the new probabilities integrating the Risk Free Rate to comply with the risk neutrality assumption vinzjeannin@hotmail.com S = + 1 − = + 1 − − = − Therfore: BV= OpUp ∗ p + OpDown ∗ 1 − p ∗ − 64 12.78
  • 123.63 ESGF 4IFM Q1 2012 115.19 18.63 107.33 12.78 107.33 100 8.74 100 2.33 vinzjeannin@hotmail.com5.96 1.5 93.17 93.17 0.97 86.81 0 0 80.89 0 65
  • 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 ESGF 4IFM Q1 2012How much with B&S? 6.22 vinzjeannin@hotmail.com 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! 66
  • B&S / CRR Convergence: usually 40 steps are reasonable 6.46.35 ESGF 4IFM Q1 2012 6.36.25 6.2 vinzjeannin@hotmail.com6.15 CRB BS 6.16.05 65.95 67 5.9 1 6 11 16 21 26 31 36 41 46 51 56 61 66 71
  • ESGF 4IFM Q1 2012 I meant American Option! Let’s start all over again… vinzjeannin@hotmail.com 68CRR main advantage is the ability to price American Options
  • On each node you need to check any early exercise possibility 123.63 ESGF 4IFM Q1 2012 115.19 18.63 107.33 13.84 10.19 107.33 8.74 2.33 100 100 2.33 vinzjeannin@hotmail.com5.96 1.5 0 93.17 93.17 0.97 0 86.81 0 0 0 80.89 0 But sometimes holding is better than exercisingBinomial Value and in this case no early exercise worth and price 69Intrinsic value of the European Call and American Call will be the same
  • 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 ESGF 4IFM Q1 2012 vinzjeannin@hotmail.com 70
  • Binomial Value at the next to last and last node (i.e. Valuating as ifit was a European Put) ESGF 4IFM Q1 2012 0 vinzjeannin@hotmail.com 0 0 0 0 2.66 5.45 9.90 14.64 18.08 71 21.93
  • Any early exercise worth? ESGF 4IFM Q1 2012 0 vinzjeannin@hotmail.com 0 0 0 0 0 0 2.66 0 5.45 9.90 10.31 14.64 18.08 18.5 72 21.93
  • Finally… ESGF 4IFM Q1 2012 0 vinzjeannin@hotmail.com 0 0 0 0.64 0 2.16 1.30 04.49 3.77 2.66 6.96 6.38 5.45 10.36 10.31 14.64 14.64 18.5 73 21.93
  • ESGF 4IFM Q1 2012 The American Put worth 4.49 The European Put worth 4.32 vinzjeannin@hotmail.comDifference can be non negligible 74
  • 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. The pay-off at maturity is binary: 0 if out of the money, Q if in the money ESGF 4IFM Q1 2012 Tree of stock price vinzjeannin@hotmail.com 75
  • Last node pay off is then straight forward ESGF 4IFM Q1 2012 0 vinzjeannin@hotmail.com 0 0 15 15 76 15
  • Then method doesn’t change… Backward induction. ESGF 4IFM Q1 2012 0 vinzjeannin@hotmail.com 0 0 0 1.75 0 4.38 3.58 07.00 7.15 7.33 9.81 10.96 15 12.72 14.88 14.75 15 14.88 77 15
  • 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. Let’s suppose this Bermuda can only be exercised between the 4th and 5th step ESGF 4IFM Q1 2012 Tree of stock price vinzjeannin@hotmail.com 78
  • Any early exercise worth? ESGF 4IFM Q1 2012 0 vinzjeannin@hotmail.com 0 0 0 0 0 0 2.66 0 5.45 9.90 10.31 14.64 18.08 18.5 79 21.93 No exercises on lower steps
  • Finally… ESGF 4IFM Q1 2012 0 vinzjeannin@hotmail.com 0 0 0 0.64 0 2.16 1.30 0 4.44 3.77 2.66 6.86 6.38 5.45 10.16 10.31 14.22 14.64 18.5 80 21.93A “full” American option would have been exercised, not this one
  • 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. 3 Steps ESGF 4IFM Q1 2012 Construct the usual tree Subtract the present value of the dividend on each node before it occurs Pricing can continue as usual vinzjeannin@hotmail.com The dividend occurs between the 3rd and 4th step 3.5 −10%∗ Value at step 0 = 2.06 ∗ 12 =2 3.5 0.4167 −10%∗ − Value at step 1 = 2.06 ∗ 12 5 = 2.02 3.5 0.4167∗2 Value at step 2 −10%∗ 12 − = 2.06 ∗ 5 = 2.03 81 3.5 0.4167∗3 −10%∗ − Value at step 3 = 2.06 ∗ 12 5 = 2.05
  • ESGF 4IFM Q1 2012Tree of stock price impacted of dividends vinzjeannin@hotmail.com 82
  • ESGF 4IFM Q1 2012Pricing by the usual backward induction (don’t forget potential early exercise) 0 vinzjeannin@hotmail.com 0 0 0 0.64 0 2.16 1.30 0 4.44 3.77 2.66 6.86 6.38 5.45 10.16 10.31 14.22 14.64 18.50 21.93 83
  • CRR Sum-UpThe American Put worth 4.49 ESGF 4IFM Q1 2012The European Put worth 4.32The Digital Put paying 15 worth 7.00 vinzjeannin@hotmail.comThe Bermuda Put with exercise on the lath fifth of the maturity worth 4.44The American Put paying a 2.06 dividend worth 4.44 You can virtually price anything you want! 84 What can’t you price?
  • 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 The option is cancelled if S goes to 60 Way to reduce the price of the option ESGF 4IFM Q1 2012 Tree of stock price vinzjeannin@hotmail.com KO 0 5.45 85 You can’t tell how much worth the option on this final node: 0 or 5.45?
  • CRR ExtensionHow to converge faster to the correct option price? ESGF 4IFM Q1 2012 Put a third factor • Up • Down vinzjeannin@hotmail.com • Stable Careful, the tree has to recombine! YES NO 86
  • Conclusion ESGF 4IFM Q1 2012 Normal Distribution GBM vinzjeannin@hotmail.com B&S CRR 87