session 9b. decision models -- prof. juran2 overview finance simulation models securities pricing...

Session 9b

Decision Models -- Prof. Juran

2

OverviewFinance Simulation Models• Securities Pricing

– Black-Scholes– Electricity Option

• Miscellaneous– Monte Carlo vs. Latin Hypercube– Review of Binomial


3

Finance Example

• A European call option on a stock earns the owner an amount equal to the price at expiration minus the exercise price, if the price of the stock on which the call is written exceeds the exercise price. Otherwise, the call pays nothing.

• A European put option earns the owner an amount equal to the exercise price minus the price at expiration, if the price at expiration is less than the exercise price. Otherwise the put pays nothing.


4

Finance Example

• The Black-Scholes formula calculates the price of a European options based on the following inputs: – today's stock price– the duration of the option (in years)– the option's exercise price– the risk-free rate of interest (per year)– the annual volatility (standard deviation) in

stock price


5

T h e B lack -Sch o les m o d el :

21 dNEedSNC rt

w h ere: S = cu r ren t stock p r ice E = exercise p r ice r = r isk -free rate o f retu rn σ2 = v ar ian ce o f th e stock ’s retu rn t = tim e to exp i ration

d1 =

t

trES

2

2

2ln

d2 = td 21

N (d) = p robabi l i ty th at z < d


6

1

234567

89

10111213

141516

A B C D E F G HInputs

Type of option (1 for call, 2 for put) 1Stock price 35Exercise price 40Duration (years) 0.5Riskfree interest rate 0.05Volatility 0.4

Quantities for Black-Scholes formulad1 -0.242 N(d1) 0.404d2 -0.525 N(d2) 0.300

Option price 2.456

=(LN(B3/ B4)+(B6+B7^2/ 2)*B5)/ (B7*SQRT(B5))

=B10-SQRT(B7^2*B5)

=IF(B2=1,NORMSDIST(B10),NORMSDIST(-B10))

=IF(B2=1,NORMSDIST(B11),NORMSDIST(-B11))

=IF(B2=1,B3*E10-B4*EXP(-B5*B6)*E11,-(B3*E10-B4*EXP(-B5*B6)*E11))

Notice the use of “if” statements in cells E10:E11 and B13, so that the same model can be used for both puts and calls.

A Black-Scholes calculator:


7

Example: Diageo (DEO)


8

Assume today is the first trading day of October and that DEO is selling for $57.98 per share.

What is a fair price for a six-month call option with a strike price of $60.00?

Assume the risk-free rate is 10%.

Two approaches:• Black-Scholes formula• Crystal Ball model


9

123456789101112

A B C D E F GDate Open High Low Close Volume Adj. Close*17-Dec-97 36.17 36.29 34.12 34.58 201080 25.53

2-Jan-98 34.12 35.37 32.81 33.26 254960 24.562-Feb-98 33.38 38.8 33.32 37.94 221678 28.022-Mar-98 38.74 45.42 38.4 44.33 181459 32.741-Apr-98 45.98 47.49 42.43 44.64 179619 33.6

1-May-98 44.87 46.44 41.91 42.03 206615 31.631-Jun-98 42.09 46.73 41.91 44.58 115309 33.551-Jul-98 44.41 48.07 43.19 44.93 136350 33.82

3-Aug-98 43.19 44.58 34.66 35.12 130133 26.431-Sep-98 37.62 39.47 30.65 33.67 160090 25.341-Oct-98 34.08 41.87 33.67 41.58 143327 31.92

Data file: deo-data.xlsx


10

12345678

A B C D EDate Adj. Close* Return17-Dec-97 25.53

2-Jan-98 24.56 -0.03802-Feb-98 28.02 0.14092-Mar-98 32.74 0.16851-Apr-98 33.6 0.0263

1-May-98 31.63 -0.05861-Jun-98 33.55 0.0607

=(B3-B2)/B2

Remove unnecessary columns and calculate monthly returns:

12345678

A B C D E FDate Adj. Close* Return 1+Return17-Dec-97 25.53

2-Jan-98 24.56 -0.0380 0.96202-Feb-98 28.02 0.1409 1.14092-Mar-98 32.74 0.1685 1.16851-Apr-98 33.6 0.0263 1.0263

1-May-98 31.63 -0.0586 0.94141-Jun-98 33.55 0.0607 1.0607

=C3+1


11

Black-Scholes assumes that the future price is the following random function of the current price:

tZtppt 20 5.0exp

t Some specific time in the future p0 Current price

pt Price at time t exp(x) ex e Root of the natural logarithms (about 2.718282) Mean percentage growth rate of the stock Standard deviation of the growth rate (a.k.a. Volatility) Z Normal random variable with mean 0 and standard deviation 1


12

Another way to look at it:

0log ln

pp

ROI t


13

12345678

A B C D E F GDate Adj. Close* Return 1+Return Ln(1+Return)17-Dec-97 25.53

2-Jan-98 24.56 -0.0380 0.9620 -0.03872-Feb-98 28.02 0.1409 1.1409 0.13182-Mar-98 32.74 0.1685 1.1685 0.15571-Apr-98 33.6 0.0263 1.0263 0.0259

1-May-98 31.63 -0.0586 0.9414 -0.06041-Jun-98 33.55 0.0607 1.0607 0.0589

=LN(D3)

93949596979899

100101102

A B C D E F G H1-Jul-05 54.44 -0.0612 0.9388 -0.0632

1-Aug-05 56.43 0.0366 1.0366 0.03591-Sep-05 58.01 0.0280 1.0280 0.02763-Oct-05 57.98 -0.0005 0.9995 -0.0005

Monthly Mean 0.0087StDev 0.0653

Annual Mean 0.1047StDev 0.2261

=AVERAGE(E3:E96)=STDEV(E3:E96)

=E98*12=E99*SQRT(12)


14

12345678910111213

A B C D EInputs

Type of option (1 for call, 2 for put) 1Stock price 57.98Exercise price 60Duration (years) 0.5Riskfree interest rate 0.1Volatility 0.226148277

Quantities for Black-Scholes formulad1 0.178 N(d1) 0.571d2 0.019 N(d2) 0.507

Option price 4.137


15

Now we create a new sheet that uses estimated parameters from the data to calculate the future price of DEO and the resulting cash flow from the option.

The present value of the expected payout is $0.17, but Black-Scholes says $4.14. Why?

123456789101112131415

A B C D E F GInputsCurrent Price 57.98$ Strike Price 60.00$ Mean Annual Return 0.1047StDev Annual Return 0.2261Risk-Free Rate 0.1Duriation (Years) 0.5

SimulationZ 0.0000Stock Price (at Risk-Free Rate) 60.18$ Option Cash Flow 0.18$

OutputDiscounted Value 0.17$

=data!E101=data!E102

=B2*EXP((B6-0.5*B5^2)*B7+B5*B10*SQRT(B7))=MAX(B11-B3,0)

=EXP(-B7*B6)*B12


16

Notes on the formulas:

B11:

(random future DEO price)

B12:

(payoff from the option)

B15:

(present value of the payoff)

tZtppt 20 5.0exp

ftreCPV

0,EpMAXC t


17

A green cell: B10 is now a standard normal random variable

A blue cell: B15 is now the present value of the random cash flow from the option

123456789101112131415

A B C D E F GInputsCurrent Price 57.98$ Strike Price 60.00$ Mean Annual Return 0.1047StDev Annual Return 0.2261Risk-Free Rate 0.1Duriation (Years) 0.5

SimulationZ 0.0000Stock Price (at Risk-Free Rate) 60.18$ Option Cash Flow 0.18$

OutputDiscounted Value 0.17$

=data!E101=data!E102

=B2*EXP((B6-0.5*B5^2)*B7+B5*B10*SQRT(B7))=MAX(B11-B3,0)

=EXP(-B7*B6)*B12


18

Simulation Results


19

OK, so simulation can do the same thing as Black-Scholes.

Maybe Black-Scholes is easier and/or quicker than running a simulation.

So, why do we need the simulation at all?


20

• Black-Scholes is an analytical result; if specific assumptions hold true, then we can calculate the expected value of the payout on an option.

• Analytical solutions do not exist in general for all types of financial instruments.

• In the absence of analytical results, Monte Carlo simulation offers an alternative approach.

• Analytical solutions may exist for expected value, but not for other important parameters.


21

Example: Asian Option

George Brickfield’s business is highly exposed to volatility in the cost of electricity.

He has asked his investment banker, Lisa Siegel, to propose an option whereby he can hedge himself against changes in the cost of a kilowatt hour of electricity over the next twelve months.


22

Lisa thinks that an Asian option would work nicely for George’s situation.

An Asian option is based on the average price of a kilowatt hour (or other underlying commodity) over a specified time period.


23

In this case, Lisa wants to offer George a one year Asian option with a target price of $0.059.

• If the average price per kilowatt hour over the next twelve months is greater than this target price, then Lisa will pay George the difference.

• If the average price per kilowatt hour over the next twelve months is less than this target price, then George loses the price he paid for the option (but he is happy, because he ends up buying relatively cheap electricity).


24

What is a fair price for Lisa to charge for 1 million kwh worth of these options?

Use the historical data provided and Monte Carlo simulation to arrive at a fair price.


25

12345678910111213

A B C D E F G H IMonth $/kwh Return mean stdev

Jan 0.051$ 0.1768% 7.3462%Feb 0.056$ 0.09804Mar 0.054$ -0.03571Apr 0.052$ -0.03704

May 0.052$ 0.00000Jun 0.057$ 0.09615Jul 0.067$ 0.17544

Aug 0.064$ -0.04478Sep 0.064$ 0.00000Oct 0.058$ -0.09375Nov 0.054$ -0.06897Dec 0.057$ 0.05556

=(B3-B2)/B2 =AVERAGE(C2:C121)

=STDEV(C2:C121)

Analysis of historical data: Our model will be based not on the actual prices, but on monthly percent changes in price (a.k.a. returns):


26

Histogram of Electricity Returns

0

5

10

15

20

25

30

-0.200 -0.175 -0.150 -0.125 -0.100 -0.075 -0.050 -0.025 0.000 0.025 0.050 0.075 0.100 0.125 0.150 0.175 0.200 0.225 0.250 0.275 0.300

Monthly Price Change

Fre

qu

en

cy


27

• Returns are approximately normal.

• We’ll use the sample mean and sample standard deviation from the data (0.001768 and 0.073462, respectively).


28

12345678910111213141516171819

A B C D E F G HInitial Electricity Price $0.05684Target Price $0.05900Mean monthly return 0.18%Std dev monthly return 7.35%# kwh per option 1,000,000

Month Return Price Average PriceJan 1.13% 0.05748 0.06486Feb -0.21% 0.05736Mar 7.82% 0.06185 PayoutApr -5.77% 0.05828 5,861.89$

May 6.54% 0.06209Jun 8.31% 0.06725Jul -12.05% 0.05915

Aug 9.71% 0.06489Sep 1.56% 0.06590Oct 6.84% 0.07041Nov 9.39% 0.07703Dec -0.50% 0.07664

=AVERAGE(C8:C26)

=C5*MAX(E8-C2,0)

=C15*(1+B16)


29

In B8:B19 we have 12 Crystal Ball assumption cells, normally distributed with the mean and standard deviation from our sample data (C3 and C4).

In C8:C19 we use the random percent returns to calculate monthly prices, which are averaged in E8 for the whole year.

E11 calculates the payout on the option (a Crystal Ball forecast cell). The average value of E11 over many trials will be a reasonable estimate of the fair price for this option.


30

W e’ll add a graph, to show the change in electricity prices over the course of each simulated year:

123456789

1011121314151617181920212223

A B C D E F G H I J K LInitia l E lectric ity Price $0.05684Target Price $0.05900M ean m onthly return 0.18%Std dev m onthly return 7.35%# kwh per option 1,000,000

M onth Return Price Average PriceJan 1.13% 0.05748 0.06486Feb -0.21% 0.05736M ar 7.82% 0.06185 PayoutApr -5.77% 0.05828 5,861.89$

M ay 6.54% 0.06209Jun 8.31% 0.06725Jul -12.05% 0.05915

Aug 9.71% 0.06489Sep 1.56% 0.06590O ct 6.84% 0.07041Nov 9.39% 0.07703Dec -0.50% 0.07664

Price

0.04

0.05

0.06

0.07

0.08

0.09

0.10

0.11

0.12

M onth

Pri

ce ($

/kw

h)


31


32

123456789101112131415

A B C D E F G HStatistics Payout

Trials 1000000 alpha 0.05Base Case $5,861.89 z-value 1.959964Mean $2,927.16Median $0.00 UCL $2,937.45Mode $0.00 LCL $2,916.87Standard Deviation $5,249.37Variance $27,555,886.95Skewness 2.31Kurtosis 9.16Coeff. of Variation 1.79Minimum $0.00Maximum $63,307.77Range Width $63,307.77Mean Std. Error $5.25

=-NORMSINV(E2/2)

=$B$4+$E$3*$B$15=$B$4-$E$3*$B$15

X ns

96.1

$2,927.16 000,000,1

67.249,5$96.1

25.5$96.1

29.10$

95% confident that the true fair price is between $2,916.87 and $2,937.45.

Could narrow the interval by running a longer simulation.


33

Monte Carlo vs. Latin Hypercube


34

-4 -3 -2 -1 0 1 2 3 4

Example: Standard normal distribution

(mean = 0, standard deviation = 1)

Divided into 8 equal-probability ranges


35

-4 -3 -2 -1 0 1 2 3 4

*

Probability that the first random independent observation falls into any one range is 0.125.


36

-4 -3 -2 -1 0 1 2 3 4

*

Probability that the first two observations fall into any one range is 0.1252 = 0.01563.

*


37

-4 -3 -2 -1 0 1 2 3 4

*

Probability that the first three observations fall into any one range is 0.1253 = 0.001953.

A small (but not zero!) chance of an unrepresentative sample.

**


38

-4 -3 -2 -1 0 1 2 3 4*

Latin Hypercube ensures that each range gets one observation before any range gets a second observation (but with more than 8 ranges).

** *** * *


39

Monte Carlo vs. Latin Hypercube

12345

A B C

0

0=B2


40


41


42


43


44


45

Binomial Random Variable


46

The number of successes in a Bernoulli process is a binomial random variable. The probability that a binomial variable X will take on any particular value x is given by the binomial formula:

p,nxXP xnx ppx

n

1

xnx pp

!xn!x!n

1

If X is a Binomial(n, p) random variable, then

Expected Value: E(X) = np

Variance: Var(X) = np(1 - p)

Standard Deviation: (X) = )p(np 1


47

A Bernoulli process is a sequence of n identical trials of a random experiment such that each trial:

(1) produces one of two possible complimentary outcomes that are conventionally called success and failure and

(2) is independent of any other trial so that the probability of success or failure is constant from trial to trial.

Note that the success and failure probabilities are assumed to be constant from trial to trial, but they are not necessarily equal to each other.


48

12345678910111213141516171819202122

A B C D E F G H I Jn 10p 0.44

X0 0.00301 0.02382 0.08433 0.17654 0.24275 0.22896 0.14997 0.06738 0.01989 0.0035

10 0.0003

Binomial Distribution

0.0

0.1

0.1

0.2

0.2

0.3

0.3

0 1 2 3 4 5 6 7 8 9 10

Number of "Successes"

Pro

bab

ility

=BINOMDIST(A5,$C$1,$C$2,0)


49

SummaryFinance Simulation Models• Securities Pricing

– Black-Scholes– Electricity Option

• Miscellaneous– Monte Carlo vs. Latin Hypercube– Review of Binomial

session 9b. decision models -- prof. juran2 overview finance simulation models securities pricing...

Documents

decision models

option slide

stock price slide

payoff slide

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