Applied Statistics I

ESGF 4IFM Q1 2012
Applied Statistics
Vincent JEANNIN – ESGF 4IFM
Q1 2012

vinzjeannin@hotmail.com
1

ESGF 4IFM Q1 2012
Summary of the session (est. 4.5h)

• Introduction & Objectives
• Bibliography
• First approach: Descriptive Statistics

• 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

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 2012
4

First Approach: Descriptive
statistics
FCOJ Front Month: 31st Dec 2008 / 30th Sep 2011

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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)

How 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

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• Free

• Open Source
• Developments shared by developers

R, easier, faster,… Function summary

7

R can easily show the distribution of returns

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8

Interesting shape but what next?

The four moments
��
1
Mean �� = ��
��
��=1

ESGF 4IFM Q1 2012
Expected Return
Standard Deviation
Dispersion from the mean

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
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="F
COJ Returns",xlab="Return",ylab="Occurence")

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Histogram centred on the mean with SD multiples groups 11
Symmetric-ish

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693 data

75.04% within ±1��

94.23% within ±2��

98.99% within ±3��

12

Skewness, the third moment
Asymmetry of the distribution

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• 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?

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Excel function SKEW
R 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

• 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

> 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") 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 2012
On 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

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

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Excel function KURT
R 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?

Is that actually enough to take investment decision?
What next?
How different is the FCOJ distribution from the Normal Distribution?

19

The Normal Distribution
Let’s discuss about the standard normal first…

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Snapshot, 4 moments:

Mean 0
SD 1

Skewness 0
Kurtosis 3

Snapshot, Shape:

20

Notation ��(��, ��)
1 (��−��)2
−
Density �� = �� 2��2
2�� 2

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Distributions of zeros means with following SD: 0.5 / 0.75 / 1 / 1.5 / 2

Which one is which one?

21

> x=seq(-4,4,length=500)
> y1=dnorm(x,mean=0,sd=0.5)
> y3=dnorm(x,mean=0,sd=1)
> y5=dnorm(x,mean=0,sd=2)
> plot(x,y1,type="l",lwd=3,col="red",
main="Normal Distributions", ylab="f(x)")

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> 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)) 22
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? 23

Same SD, different mean, more straight forward

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24

Cumulative Distribution

Reminder: the CDF (Cumulative Distribution Function) is the probability
of the random variable X given a distribution to be lower or equal to x

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��
This is the integral of the density function �� ≤ �� = �� = ��
−∞

Important Properties

�� = �� = 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

> 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

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 -
CFD's")
>lines(x,pnorm(x,mean=0,sd=0.75
),col="red",type="l",lwd=3)
),col="pink",type="l",lwd=3) 27
),col="yellow",type="l",lwd=3)

Standardization

��~��(��, ��)

ESGF 4IFM Q1 2012
�� − ��
�� =
��

��~��(0,1)

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 =?

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$V1),do.points=FALSE, col="red", lwd=3, main="Normal
Distribution against FCOJ - CFD's", xlab="x", ylab="P(X<=x)")
>lines(x,pnorm(x,mean=mean(FCOJ$V1),sd=sd(FCOJ$V1)),col="blue",type="l",l

ESGF 4IFM Q1 2012
wd=3)

30

Where can you see the excess kurtosis?

>qqnorm(FCOJ$V1)
>qqline(FCOJ$V1)

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Fat Tail

• 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

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…

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

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

> 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[7]), type="l") 35

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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
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) method

library(sde)
require(sde)
nbsim<-252
Drift<-0.05
Volat<-0.05

ESGF 4IFM Q1 2012
Spot<-100
G1<-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) 37
lines(G8, col=8)
lines(G9, col=9)
lines(G10, col=10)

ESGF 4IFM Q1 2012
> 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

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")

39

ESGF 4IFM Q1 2012
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)
40
�� = ��−1 �� −1 = �� −��

Black & Scholes

Let’s look at the underling price diffusion process through another angle

ESGF 4IFM Q1 2012
��

�� + 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

• 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 Valuation

Based 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

(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 − �� − ��

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

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
• First derivative of the value of the delta with
• Pace of the delta movement
• Second order Greek

Gamma [0,1] if… Long option
Gamma [-1,0] if… Short option
Gamma=max if… ATMF

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
2
8.8003

Forgetting Gamma is dangerous, difference is 0.25% in our example!

What is the new delta?

0.5695

Third order known as Speed, hardly used…

1 47
Write the Taylor Development until the Speed level… ∗ �� ∗ �� 3
6
How 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

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 )

2 �� − ��
Theta >0 if… Short option
Theta <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

respect to the interest rate

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.
50

Sum Up - Example


ESGF 4IFM Q1 2012
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!

51

Sum Up Greeks/Time
Call 100, S=105, r=5%, Maturity from 4y, Vol=10%

ESGF 4IFM Q1 2012
52

Sum Up Greeks/Spot Price
Call 100, r=5%, Maturity 4y, Vol=10%

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53

Sum Up Greeks/Strike
S=105, r=5%, Maturity 4y, Vol=10%

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54

Sum Up Greeks/Vol
Call 100, S=105, r=5%, Maturity 4y

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55

Conclusion on B&S

Great, easy, quick

ESGF 4IFM Q1 2012
Strong assumptions, continuous

Only European option

We need a path dependant method!

It will allow to include early exercise, dividend, pricing
European digital,…

56

Binomial Model (Cox, Ross, Rubinstein, 1979)

ESGF 4IFM Q1 2012
Why?
Path dependent (valuation of European
options, American options, Digital,…)

May include dividends

How?

Discretisation of the continuous random walk

57

Binomial Model: principles

ESGF 4IFM Q1 2012
3 Steps
“Slice” maturity in a predefined number of steps

Construct a tree lattice representing the stock price

following a GBM

Price the option by backwards induction

58

Let’s assume the maturity is divided by 2

ESGF 4IFM Q1 2012
59

Cox Ross Rubinstein up and down factors based on GBM

At each node, S goes up or down by one SD

ESGF 4IFM Q1 2012
�� = ��

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

100 100
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 2012
How many nodes do you have at the end of the tree?

�� + 1

If 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

100 100 2.33

93.17 93.17
86.81 0

80.89
0

Backward inductions, we have the probabilities, let’s assume a 63
risk 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

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

5.96 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 2012
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!
66

B&S / CRR Convergence: usually 40 steps are reasonable

6.4

6.35

ESGF 4IFM Q1 2012
6.3

6.25

6.2

6.15 CRB
BS

6.1

6.05

6

5.95
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…

68

CRR 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

5.96 1.5 0
93.17 93.17
0.97 0
86.81 0
0 0
80.89
0

But sometimes holding is better than exercising
Binomial Value and in this case no early exercise worth and price 69
Intrinsic 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
70

Binomial Value at the next to last and last node (i.e. Valuating as if
it was a European Put)

ESGF 4IFM Q1 2012
0

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

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

0
0 0
0.64 0
2.16 1.30 0
4.49 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

Difference 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

75

Last node pay off is then straight forward

ESGF 4IFM Q1 2012
0

0

0

15

15
76
15

Then method doesn’t change… Backward induction.

ESGF 4IFM Q1 2012
0

0
0 0
1.75 0
4.38 3.58 0
7.00 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

78

Any early exercise worth?

ESGF 4IFM Q1 2012
0

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

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
21.93
A “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

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 2012
Tree of stock price impacted of dividends

82

ESGF 4IFM Q1 2012
Pricing by the usual backward induction (don’t forget potential early exercise)

0

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-Up

The American Put worth 4.49

ESGF 4IFM Q1 2012
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

The 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

KO

0
5.45

85

You can’t tell how much worth the option on this final node: 0 or 5.45?

CRR Extension

How to converge faster to the correct option price?

ESGF 4IFM Q1 2012
Put a third factor

• Up
• Down

• Stable

Careful, the tree has to recombine!

YES NO 86

Conclusion

ESGF 4IFM Q1 2012
Normal Distribution

GBM

B&S
CRR

87

Applied Statistics I

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Applied Statistics I